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This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.
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{{task}}
There are four types of common coins in [https://en.wikipedia.org/wiki/United_States US] currency: :::# quarters (25 cents) :::# dimes (10 cents) :::# nickels (5 cents), and :::# pennies (1 cent)
There are six ways to make change for 15 cents: :::# A dime and a nickel :::# A dime and 5 pennies :::# 3 nickels :::# 2 nickels and 5 pennies :::# A nickel and 10 pennies :::# 15 pennies
;Task: How many ways are there to make change for a dollar using these common coins? (1 dollar = 100 cents).
;Optional: Less common are dollar coins (100 cents); and very rare are half dollars (50 cents). With the addition of these two coins, how many ways are there to make change for $1000?
(Note: the answer is larger than 232).
;Reference:
- [http://mitpress.mit.edu/sicp/full-text/book/book-Z-H-11.html#%_sec_Temp_52 an algorithm from MIT Press].
11l
{{trans|Python}}
F changes(amount, coins)
V ways = [Int64(0)] * (amount + 1)
ways[0] = 1
L(coin) coins
L(j) coin .. amount
ways[j] += ways[j - coin]
R ways[amount]
print(changes(100, [1, 5, 10, 25]))
print(changes(100000, [1, 5, 10, 25, 50, 100]))
Output:
242
13398445413854501
360 Assembly
{{trans|AWK}}
* count the coins 04/09/2015
COINS CSECT
USING COINS,R12
LR R12,R15
L R8,AMOUNT npenny=amount
L R4,AMOUNT
SRDA R4,32
D R4,=F'5'
LR R9,R5 nnickle=amount/5
L R4,AMOUNT
SRDA R4,32
D R4,=F'10'
LR R10,R5 ndime=amount/10
L R4,AMOUNT
SRDA R4,32
D R4,=F'25'
LR R11,R5 nquarter=amount/25
SR R1,R1 count=0
SR R4,R4 p=0
LOOPP CR R4,R8 do p=0 to npenny
BH ELOOPP
SR R5,R5 n=0
LOOPN CR R5,R9 do n=0 to nnickle
BH ELOOPN
SR R6,R6
LOOPD CR R6,R10 do d=0 to ndime
BH ELOOPD
SR R7,R7 q=0
LOOPQ CR R7,R11 do q=0 to nquarter
BH ELOOPQ
LR R3,R5 n
MH R3,=H'5'
LR R2,R4 p
AR R2,R3
LR R3,R6 d
MH R3,=H'10'
AR R2,R3
LR R3,R7 q
MH R3,=H'25'
AR R2,R3 s=p+n*5+d*10+q*25
C R2,=F'100' if s=100
BNE NOTOK
LA R1,1(R1) count=count+1
NOTOK LA R7,1(R7) q=q+1
B LOOPQ
ELOOPQ LA R6,1(R6) d=d+1
B LOOPD
ELOOPD LA R5,1(R5) n=n+1
B LOOPN
ELOOPN LA R4,1(R4) p=p+1
B LOOPP
ELOOPP XDECO R1,PG+0 edit count
XPRNT PG,12 print count
XR R15,R15
BR R14
AMOUNT DC F'100' start value in cents
PG DS CL12
YREGS
END COINS
{{out}}
242
Ada
{{Works with|gnat/gcc}}
with Ada.Text_IO;
procedure Count_The_Coins is
type Counter_Type is range 0 .. 2**63-1; -- works with gnat
type Coin_List is array(Positive range <>) of Positive;
function Count(Goal: Natural; Coins: Coin_List) return Counter_Type is
Cnt: array(0 .. Goal) of Counter_Type := (0 => 1, others => 0);
-- 0 => we already know one way to choose (no) coins that sum up to zero
-- 1 .. Goal => we do not (yet) other ways to choose coins
begin
for C in Coins'Range loop
for Amount in 1 .. Cnt'Last loop
if Coins(C) <= Amount then
Cnt(Amount) := Cnt(Amount) + Cnt(Amount-Coins(C));
-- Amount-Coins(C) plus Coins(C) sums up to Amount;
end if;
end loop;
end loop;
return Cnt(Goal);
end Count;
procedure Print(C: Counter_Type) is
begin
Ada.Text_IO.Put_Line(Counter_Type'Image(C));
end Print;
begin
Print(Count( 1_00, (25, 10, 5, 1)));
Print(Count(1000_00, (100, 50, 25, 10, 5, 1)));
end Count_The_Coins;
Output:
242
13398445413854501
ALGOL 68
{{works with|ALGOL 68G|Any - tested with release 2.4.1}} {{trans|Haskell}} This corresponds to a "naive" Haskell version; to do the larger problem will require a better approach.
#
Rosetta Code "Count the coins"
This is a direct translation of a Haskell version, using an array rather than
a list. LWB, UPB, and array slicing makes the mapping very simple:
LWB > UPB <=> []
LWB = UPB <=> [x]
a[LWB a] <=> head xs
a[LWB a + 1:] <=> tail xs
#
BEGIN
PROC ways to make change = ([] INT denoms, INT amount) INT :
BEGIN
IF amount = 0 THEN
1
ELIF LWB denoms > UPB denoms THEN
0
ELIF LWB denoms = UPB denoms THEN
(amount MOD denoms[LWB denoms] = 0 | 1 | 0)
ELSE
INT sum := 0;
FOR i FROM 0 BY denoms[LWB denoms] TO amount DO
sum +:= ways to make change(denoms[LWB denoms + 1:], amount - i)
OD;
sum
FI
END;
[] INT denoms = (25, 10, 5, 1);
print((ways to make change(denoms, 100), newline))
END
Output:
+242
Arturo
{{trans|Python}}
changes [amount,coins]{
ways #(0)*(amount+1)
ways.0 1
loop coins [coin]{
loop $(range coin amount) [j]{
ways.[j] ways.[j]+ways.[j-coin]
}
}
return ways.[amount]
}
print $(changes 100 #(1 5 10 25))
print $(changes 100000 #(1 5 10 25 50 100))
{{out}}
242
13398445413854501
AutoHotkey
{{trans|Go}} {{Works with|AutoHotkey_L}}
countChange(amount){
return cc(amount, 4)
}
cc(amount, kindsOfCoins){
if ( amount == 0 )
return 1
if ( amount < 0 ) || ( kindsOfCoins == 0 )
return 0
return cc(amount, kindsOfCoins-1)
+ cc(amount - firstDenomination(kindsOfCoins), kindsOfCoins)
}
firstDenomination(kindsOfCoins){
return [1, 5, 10, 25][kindsOfCoins]
}
MsgBox % countChange(100)
AWK
Iterative implementation, derived from Run BASIC:
#!/usr/bin/awk -f
BEGIN {
print cc(100)
exit
}
function cc(amount, coins, numPennies, numNickles, numQuarters, p, n, d, q, s, count) {
numPennies = amount
numNickles = int(amount / 5)
numDimes = int(amount / 10)
numQuarters = int(amount / 25)
count = 0
for (p = 0; p <= numPennies; p++) {
for (n = 0; n <= numNickles; n++) {
for (d = 0; d <= numDimes; d++) {
for (q = 0; q <= numQuarters; q++) {
s = p + n * 5 + d * 10 + q * 25;
if (s == 100) count++;
}
}
}
}
return count;
}
Run time: time ./change-itr.awk 242
real 0m0.065s user 0m0.063s sys 0m0.002s
Recursive implementation (derived from Scheme example):
#!/usr/bin/awk -f
BEGIN {
COINSEP = ", "
coins = 1 COINSEP 5 COINSEP 10 COINSEP 25
print cc(100, coins)
exit
}
function cc(amt, coins) {
if (length(coins) == 0) return 0
if (amt < 0) return 0
if (amt == 0) return 1
return cc(amt, tail(coins)) + cc(amt - head(coins), coins)
}
function tail(coins, koins, s, c) {
split(coins, koins, COINSEP)
s = ""
for (c = 2; c <= length(koins); c++) s = s (s == "" ? "" : COINSEP) koins[c]
return s;
}
function head(coins, koins) {
split(coins, koins, COINSEP)
return koins[1]
}
Run time: time ./change-rec.awk 242
real 0m0.081s user 0m0.079s sys 0m0.002s
While the recursive version is slower for small amounts, about 2 bucks it gets faster than the iterative version, at least until is segfaults from exhausting the stack.
BBC BASIC
Non-recursive solution:
DIM uscoins%(3)
uscoins%() = 1, 5, 10, 25
PRINT FNchange(100, uscoins%()) " ways of making $1"
PRINT FNchange(1000, uscoins%()) " ways of making $10"
DIM ukcoins%(7)
ukcoins%() = 1, 2, 5, 10, 20, 50, 100, 200
PRINT FNchange(100, ukcoins%()) " ways of making £1"
PRINT FNchange(1000, ukcoins%()) " ways of making £10"
END
DEF FNchange(sum%, coins%())
LOCAL C%, D%, I%, N%, P%, Q%, S%, table()
C% = 0
N% = DIM(coins%(),1) + 1
FOR I% = 0 TO N% - 1
D% = coins%(I%)
IF D% <= sum% IF D% >= C% C% = D% + 1
NEXT
C% *= N%
DIM table(C%-1)
FOR I% = 0 TO N%-1 : table(I%) = 1 : NEXT
P% = N%
FOR S% = 1 TO sum%
FOR I% = 0 TO N% - 1
IF I% = 0 IF P% >= C% P% = 0
IF coins%(I%) <= S% THEN
Q% = P% - coins%(I%) * N%
IF Q% >= 0 table(P%) = table(Q%) ELSE table(P%) = table(Q% + C%)
ENDIF
IF I% table(P%) += table(P% - 1)
P% += 1
NEXT
NEXT
= table(P%-1)
Output (BBC BASIC does not have large enough integers for the optional task):
242 ways of making $1
142511 ways of making $10
4563 ways of making £1
321335886 ways of making £10
C
Using some crude 128-bit integer type.
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
// ad hoc 128 bit integer type; faster than using GMP because of low
// overhead
typedef struct { uint64_t x[2]; } i128;
// display in decimal
void show(i128 v) {
uint32_t x[4] = {v.x[0], v.x[0] >> 32, v.x[1], v.x[1] >> 32};
int i, j = 0, len = 4;
char buf[100];
do {
uint64_t c = 0;
for (i = len; i--; ) {
c = (c << 32) + x[i];
x[i] = c / 10, c %= 10;
}
buf[j++] = c + '0';
for (len = 4; !x[len - 1]; len--);
} while (len);
while (j--) putchar(buf[j]);
putchar('\n');
}
i128 count(int sum, int *coins)
{
int n, i, k;
for (n = 0; coins[n]; n++);
i128 **v = malloc(sizeof(int*) * n);
int *idx = malloc(sizeof(int) * n);
for (i = 0; i < n; i++) {
idx[i] = coins[i];
// each v[i] is a cyclic buffer
v[i] = calloc(sizeof(i128), coins[i]);
}
v[0][coins[0] - 1] = (i128) {{1, 0}};
for (k = 0; k <= sum; k++) {
for (i = 0; i < n; i++)
if (!idx[i]--) idx[i] = coins[i] - 1;
i128 c = v[0][ idx[0] ];
for (i = 1; i < n; i++) {
i128 *p = v[i] + idx[i];
// 128 bit addition
p->x[0] += c.x[0];
p->x[1] += c.x[1];
if (p->x[0] < c.x[0]) // carry
p->x[1] ++;
c = *p;
}
}
i128 r = v[n - 1][idx[n-1]];
for (i = 0; i < n; i++) free(v[i]);
free(v);
free(idx);
return r;
}
// simple recursive method; slow
int count2(int sum, int *coins)
{
if (!*coins || sum < 0) return 0;
if (!sum) return 1;
return count2(sum - *coins, coins) + count2(sum, coins + 1);
}
int main(void)
{
int us_coins[] = { 100, 50, 25, 10, 5, 1, 0 };
int eu_coins[] = { 200, 100, 50, 20, 10, 5, 2, 1, 0 };
show(count( 100, us_coins + 2));
show(count( 1000, us_coins));
show(count( 1000 * 100, us_coins));
show(count( 10000 * 100, us_coins));
show(count(100000 * 100, us_coins));
putchar('\n');
show(count( 1 * 100, eu_coins));
show(count( 1000 * 100, eu_coins));
show(count( 10000 * 100, eu_coins));
show(count(100000 * 100, eu_coins));
return 0;
}
output (only the first two lines are required by task):
4563 10056050940818192726001 99341140660285639188927260001 992198221207406412424859964272600001
## C++
```cpp
#include <iostream>
#include <stack>
#include <vector>
struct DataFrame {
int sum;
std::vector<int> coins;
std::vector<int> avail_coins;
};
int main() {
std::stack<DataFrame> s;
s.push({ 100, {}, { 25, 10, 5, 1 } });
int ways = 0;
while (!s.empty()) {
DataFrame top = s.top();
s.pop();
if (top.sum < 0) continue;
if (top.sum == 0) {
++ways;
continue;
}
if (top.avail_coins.empty()) continue;
DataFrame d = top;
d.sum -= top.avail_coins[0];
d.coins.push_back(top.avail_coins[0]);
s.push(d);
d = top;
d.avail_coins.erase(std::begin(d.avail_coins));
s.push(d);
}
std::cout << ways << std::endl;
return 0;
}
{{out}}
242
C#
// Adapted from http://www.geeksforgeeks.org/dynamic-programming-set-7-coin-change/
class Program
{
static long Count(int[] C, int m, int n)
{
var table = new long[n + 1];
table[0] = 1;
for (int i = 0; i < m; i++)
for (int j = C[i]; j <= n; j++)
table[j] += table[j - C[i]];
return table[n];
}
static void Main(string[] args)
{
var C = new int[] { 1, 5, 10, 25 };
int m = C.Length;
int n = 100;
Console.WriteLine(Count(C, m, n)); //242
Console.ReadLine();
}
}
Clojure
(def denomination-kind [1 5 10 25])
(defn- cc [amount denominations]
(cond (= amount 0) 1
(or (< amount 0) (empty? denominations)) 0
:else (+ (cc amount (rest denominations))
(cc (- amount (first denominations)) denominations))))
(defn count-change
"Calculates the number of times you can give change with the given denominations."
[amount denominations]
(cc amount denominations))
(count-change 15 denomination-kind) ; = 6
COBOL
{{trans|C#}}
identification division.
program-id. CountCoins.
data division.
working-storage section.
77 i pic 9(3).
77 j pic 9(3).
77 m pic 9(3) value 4.
77 n pic 9(3) value 100.
77 edited-value pic z(18).
01 coins-table value "01051025".
05 coin pic 9(2) occurs 4.
01 ways-table.
05 way pic 9(18) occurs 100.
procedure division.
main.
perform calc-count
move way(n) to edited-value
display function trim(edited-value)
stop run
.
calc-count.
initialize ways-table
move 1 to way(1)
perform varying i from 1 by 1 until i > m
perform varying j from coin(i) by 1 until j > n
add way(j - coin(i)) to way(j)
end-perform
end-perform
.
{{out}}
242
Coco
{{trans|Python}}
changes = (amount, coins) ->
ways = [1].concat [0] * amount
for coin of coins
for j from coin to amount
ways[j] += ways[j - coin]
ways[amount]
console.log changes 100, [1 5 10 25]
Commodore BASIC
'''Example 1:''' Base example in Commodore BASIC (works on PET, C64, VIC20, etc.)
This example is based on the Spectrum ZX BASIC example found below. Direct copy of that algorithm and executed on an emulated Commodore 64 in VICE resulted in a timed performance of 46 minutes and 37 seconds (46:37) as measured by the C64 BASIC system clock (TIME$ or TI$, times are approximate within a few seconds). Some improvements were made as follows:
Reversed the order of the loops to start counting with the largest denomination > smallest denomination. Result: 44:45
It makes no sense to check with anything other than a multiple of 5 pennies, since the other denominations value a multiple of 5. Adding "step 5" to the penny for loop skips over a good portion of useless iteration. Result: about 9:44.
Not printing any of the individual results speeds up total time to 9:30.
Removing the specific variables used in the NEXT statements helps the interpreter speed up. Result: 9:10.
Now that the denominations were reordered, it makes sense that each sub-loop with the next lower denomination should loop only through the remaining money not accounted for by the larger denomination. Result: 2:12.
5 m=100:rem money = $1.00 or 100 pennies.
10 print chr$(147);chr$(14);"This program will calculate the number"
11 print "of combinations of 'change' that can be"
12 print "given for a $1 bill."
13 print:print "The coin values are:"
14 print "0.01 = Penny":print "0.05 = Nickle"
15 print "0.10 = Dime":print "0.25 = Quarter"
16 print
20 print "Would you like to see each combination?"
25 get k$:yn=(k$="y"):if k$="" then 25
100 p=m:ti$="000000"
130 q=int(m/25)
140 count=0:ps=1
147 if yn then print "Count P N D Q"
150 for qc=0 to q:d=int((m-qc*25)/10)
160 for dc=0 to d:n=int((m-dc*10)/5)
170 for nc=0 to n:p=m-nc*5
180 for pc=0 to p step 5
190 s=pc+nc*5+dc*10+qc*25
200 if s=m then count=count+1:if yn then gosub 1000
210 next:next:next:next
245 en$=ti$
250 print:print count;"different combinations found in"
260 print tab(len(str$(count))+1);
265 print left$(en$,2);":";mid$(en$,3,2);":";right$(en$,2);"."
270 end
1000 print count;tab(6);pc;tab(11);nc;tab(16);dc;tab(21);qc:return
'''Example 2:''' Commodore 64 with Screen Blanking
Make the following changes on a Commodore 64 to enable screen blanking. This will give the CPU a few extra cycles normally held by the VIC-II. Add line 145 and change line 245 as shown.
Enabling screen blanking (and therefore not printing each result) results in a total time of 1:44.
145 if not yn then poke 53265,peek(53265) and 239
245 en$=ti$:if not yn then poke 53265,peek(53265) or 16
'''Example 3:''' Commodore 128 with VIC-II blanking, 2MHz fast mode.
Similar to above, however the Commodore 128 is capable of using a faster clock speed at the expense of any VIC-II graphics display. Timed result is 1:18. Add/change the following lines on the Commodore 128:
145 if not yn then fast
245 en$=ti$:if not yn then slow
Common Lisp
Recursive Version With Cache
(defun count-change (amount coins
&optional
(length (1- (length coins)))
(cache (make-array (list (1+ amount) (length coins))
:initial-element nil)))
(cond ((< length 0) 0)
((< amount 0) 0)
((= amount 0) 1)
(t (or (aref cache amount length)
(setf (aref cache amount length)
(+ (count-change (- amount (first coins)) coins length cache)
(count-change amount (rest coins) (1- length) cache)))))))
; (compile 'count-change) ; for CLISP
(print (count-change 100 '(25 10 5 1))) ; = 242
(print (count-change 100000 '(100 50 25 10 5 1))) ; = 13398445413854501
(terpri)
Iterative Version
(defun count-change (amount coins &aux (ways (make-array (1+ amount) :initial-element 0)))
(setf (aref ways 0) 1)
(loop for coin in coins do
(loop for j from coin upto amount
do (incf (aref ways j) (aref ways (- j coin)))))
(aref ways amount))
D
Basic Version
{{trans|Go}}
import std.stdio, std.bigint;
auto changes(int amount, int[] coins) {
auto ways = new BigInt[amount + 1];
ways[0] = 1;
foreach (coin; coins)
foreach (j; coin .. amount + 1)
ways[j] += ways[j - coin];
return ways[$ - 1];
}
void main() {
changes( 1_00, [25, 10, 5, 1]).writeln;
changes(1000_00, [100, 50, 25, 10, 5, 1]).writeln;
}
{{out}}
242
13398445413854501
Safe Ulong Version
This version is very similar to the precedent, but it uses a faster ulong type, and performs a checked sum to detect overflows at run-time.
import std.stdio, core.checkedint;
auto changes(int amount, int[] coins, ref bool overflow) {
auto ways = new ulong[amount + 1];
ways[0] = 1;
foreach (coin; coins)
foreach (j; coin .. amount + 1)
ways[j] = ways[j].addu(ways[j - coin], overflow);
return ways[amount];
}
void main() {
bool overflow = false;
changes( 1_00, [25, 10, 5, 1], overflow).writeln;
if (overflow)
"Overflow".puts;
overflow = false;
changes( 1000_00, [100, 50, 25, 10, 5, 1], overflow).writeln;
if (overflow)
"Overflow".puts;
}
The output is the same.
Faster Version
{{trans|C}}
import std.stdio, std.bigint;
BigInt countChanges(in int amount, in int[] coins) pure /*nothrow*/ {
immutable n = coins.length;
int cycle;
foreach (immutable c; coins)
if (c <= amount && c >= cycle)
cycle = c + 1;
cycle *= n;
auto table = new BigInt[cycle];
table[0 .. n] = 1.BigInt;
int pos = n;
foreach (immutable s; 1 .. amount + 1) {
foreach (immutable i; 0 .. n) {
if (i == 0 && pos >= cycle)
pos = 0;
if (coins[i] <= s) {
immutable int q = pos - (coins[i] * n);
table[pos] = (q >= 0) ? table[q] : table[q + cycle];
}
if (i)
table[pos] += table[pos - 1];
pos++;
}
}
return table[pos - 1];
}
void main() {
immutable usCoins = [100, 50, 25, 10, 5, 1];
immutable euCoins = [200, 100, 50, 20, 10, 5, 2, 1];
foreach (immutable coins; [usCoins, euCoins]) {
countChanges( 1_00, coins[2 .. $]).writeln;
countChanges( 1000_00, coins).writeln;
countChanges( 10000_00, coins).writeln;
countChanges(100000_00, coins).writeln;
writeln;
}
}
{{out}}
242
13398445413854501
1333983445341383545001
133339833445334138335450001
4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001
===128-bit Version=== A much faster version that mixes high-level and low-level style programming. This version uses basic 128-bit unsigned integers, like the C version. The output is the same as the second D version. {{trans|C}}
import std.stdio, std.bigint, std.algorithm, std.conv, std.functional;
struct Ucent { /// Simplified 128-bit integer (like ucent).
ulong hi, lo;
static immutable one = Ucent(0, 1);
void opOpAssign(string op="+")(in ref Ucent y) pure nothrow @nogc @safe {
this.hi += y.hi;
if (this.lo >= ~y.lo)
this.hi++;
this.lo += y.lo;
}
string toString() const /*pure nothrow @safe*/ {
return text((this.hi.BigInt << 64) + this.lo);
}
}
Ucent countChanges(in int amount, in int[] coins) pure nothrow {
immutable n = coins.length;
// Points to a cyclic buffer of length coins[i]
auto p = new Ucent*[n];
auto q = new Ucent*[n]; // iterates it.
auto buf = new Ucent[coins.sum];
p[0] = buf.ptr;
foreach (immutable i; 0 .. n) {
if (i)
p[i] = coins[i - 1] + p[i - 1];
*p[i] = Ucent.one;
q[i] = p[i];
}
Ucent prev;
foreach (immutable j; 1 .. amount + 1)
foreach (immutable i; 0 .. n) {
q[i]--;
if (q[i] < p[i])
q[i] = p[i] + coins[i] - 1;
if (i)
*q[i] += prev;
prev = *q[i];
}
return prev;
}
void main() {
immutable usCoins = [100, 50, 25, 10, 5, 1];
immutable euCoins = [200, 100, 50, 20, 10, 5, 2, 1];
foreach (immutable coins; [usCoins, euCoins]) {
countChanges( 1_00, coins[2 .. $]).writeln;
countChanges( 1000_00, coins).writeln;
countChanges( 10000_00, coins).writeln;
countChanges(100000_00, coins).writeln;
writeln;
}
}
Printing Version
This version prints all the solutions (so it can be used on the smaller input):
import std.stdio, std.conv, std.string, std.algorithm, std.range;
void printChange(in uint tot, in uint[] coins)
in {
assert(coins.isSorted);
} body {
auto freqs = new uint[coins.length];
void inner(in uint curTot, in size_t start) {
if (curTot == tot)
return writefln("%-(%s %)",
zip(coins, freqs)
.filter!(cf => cf[1] != 0)
.map!(cf => format("%u:%u", cf[])));
foreach (immutable i; start .. coins.length) {
immutable ci = coins[i];
for (auto v = (freqs[i] + 1) * ci; v <= tot; v += ci)
if (curTot + v <= tot) {
freqs[i] += v / ci;
inner(curTot + v, i + 1);
freqs[i] -= v / ci;
}
}
}
inner(0, 0);
}
void main() {
printChange(1_00, [1, 5, 10, 25]);
}
{{out}}
1:5 5:1 10:4 25:2
1:5 5:1 10:9
1:5 5:2 10:1 25:3
1:5 5:2 10:6 25:1
1:5 5:3 10:3 25:2
1:5 5:3 10:8
1:5 5:4 10:5 25:1
1:5 5:4 25:3
1:5 5:5 10:2 25:2
1:5 5:5 10:7
1:5 5:6 10:4 25:1
1:5 5:7 10:1 25:2
...
5:11 10:2 25:1
5:12 10:4
5:13 10:1 25:1
5:14 10:3
5:15 25:1
5:16 10:2
5:18 10:1
5:20
10:5 25:2
10:10
25:4
Dart
Simple recursive version plus cached version using a map.
var cache = new Map();
main() {
var stopwatch = new Stopwatch()..start();
// use the brute-force recursion for the small problem
int amount = 100;
list coinTypes = [25,10,5,1];
print (coins(amount,coinTypes).toString() + " ways for $amount using $coinTypes coins.");
// use the cache version for the big problem
amount = 100000;
coinTypes = [100,50,25,10,5,1];
print (cachedCoins(amount,coinTypes).toString() + " ways for $amount using $coinTypes coins.");
stopwatch.stop();
print ("... completed in " + (stopwatch.elapsedMilliseconds/1000).toString() + " seconds");
}
coins(int amount, list coinTypes) {
int count = 0;
if(coinTypes.length == 1) return (1); // just pennies available, so only one way to make change
for(int i=0; i<=(amount/coinTypes[0]).toInt(); i++){ // brute force recursion
count += coins(amount-(i*coinTypes[0]),coinTypes.sublist(1)); // sublist(1) is like lisp's '(rest ...)'
}
// uncomment if you want to see intermediate steps
//print("there are " + count.toString() +" ways to count change for ${amount.toString()} using ${coinTypes} coins.");
return(count);
}
cachedCoins(int amount, list coinTypes) {
int count = 0;
// this is more efficient, looks at last two coins. but not fast enough for the optional exercise.
if(coinTypes.length == 2) return ((amount/coinTypes[0]).toInt() + 1);
var key = "$amount.$coinTypes"; // lookes like "100.[25,10,5,1]"
var cacheValue = cache[key]; // check whether we have seen this before
if(cacheValue != null) return(cacheValue);
count = 0;
// same recursion as simple method, but caches all subqueries too
for(int i=0; i<=(amount/coinTypes[0]).toInt(); i++){
count += cachedCoins(amount-(i*coinTypes[0]),coinTypes.sublist(1)); // sublist(1) is like lisp's '(rest ...)'
}
cache[key] = count; // add this to the cache
return(count);
}
{{out}}
242 ways for 100 using [25, 10, 5, 1] coins.
13398445413854501 ways for 100000 using [100, 50, 25, 10, 5, 1] coins.
... completed in 3.604 seconds
Dyalect
func countCoins(coins, n) {
var xs = Array.empty(n + 1, 0)
xs[0] = 1
for c in coins {
var cj = c
while cj <= n {
xs[cj] += xs[cj - c]
cj += 1
}
}
return xs[n]
}
var coins = [1, 5, 10, 25]
print(countCoins(coins, 100))
{{out}}
242
EchoLisp
Recursive solution using memoization, adapted from CommonLisp and Racket.
(lib 'compile) ;; for (compile)
(lib 'bigint) ;; integer results > 32 bits
(lib 'hash) ;; hash table
;; h-table
(define Hcoins (make-hash))
;; the function to memoize
(define (sumways cents coins)
(+ (ways cents (cdr coins)) (ways (- cents (car coins)) coins)))
;; accelerator : ways (cents, coins) = ways ((cents - cents % 5) , coins)
(define (ways cents coins)
(cond ((null? coins) 0)
((negative? cents) 0)
((zero? cents) 1)
((eq? coins c-1) 1) ;; if coins = (1) --> 1
(else (hash-ref! Hcoins (list (- cents (modulo cents 5)) coins) sumways))))
(compile 'ways) ;; speed-up things
{{out}}
(define change '(25 10 5 1))
(define c-1 (list-tail change -1)) ;; pointer to (1)
(ways 100 change)
→ 242
(define change '(100 50 25 10 5 1))
(define c-1 (list-tail change -1))
(for ((i (in-range 0 200001 20000)))
(writeln i (time (ways i change)) (hash-count Hcoins)))
;; iterate cents = 20000, 40000, ..
;; cents ((time (msec) number-of-ways) number-of-entries-in-h-table
20000 (350 4371565890901) 9398
40000 (245 138204514221801) 18798
60000 (230 1045248220992701) 28198
80000 (255 4395748062203601) 37598
100000 (234 13398445413854501) 46998
120000 (230 33312577651945401) 56398
140000 (292 71959878152476301) 65798
160000 (736 140236576291447201) 75198
180000 (237 252625397444858101) 84598
200000 (240 427707562988709001) 93998
;; One can see that the time is linear, and the h-table size reasonably small
change
→ (100 50 25 10 5 1)
(ways 100000 change)
→ 13398445413854501
Elixir
Recursive Dynamic Programming solution in Elixir
defmodule Coins do
def find(coins,lim) do
vals = Map.new(0..lim,&{&1,0}) |> Map.put(0,1)
count(coins,lim,vals)
|> Map.values
|> Enum.max
|> IO.inspect
end
defp count([],_,vals), do: vals
defp count([coin|coins],lim,vals) do
count(coins,lim,ways(coin,coin,lim,vals))
end
defp ways(num,_coin,lim,vals) when num > lim, do: vals
defp ways(num, coin,lim,vals) do
ways(num+1,coin,lim,ad(coin,num,vals))
end
defp ad(a,b,c), do: Map.put(c,b,c[b]+c[b-a])
end
Coins.find([1,5,10,25],100)
Coins.find([1,5,10,25,50,100],100_000)
{{out}}
242
13398445413854501
Erlang
-module(coins).
-compile(export_all).
count(Amount, Coins) ->
{N,_C} = count(Amount, Coins, dict:new()),
N.
count(0,_,Cache) ->
{1,Cache};
count(N,_,Cache) when N < 0 ->
{0,Cache};
count(_N,[],Cache) ->
{0,Cache};
count(N,[C|Cs]=Coins,Cache) ->
case dict:is_key({N,length(Coins)},Cache) of
true ->
{dict:fetch({N,length(Coins)},Cache), Cache};
false ->
{N1,C1} = count(N-C,Coins,Cache),
{N2,C2} = count(N,Cs,C1),
{N1+N2,dict:store({N,length(Coins)},N1+N2,C2)}
end.
print(Amount, Coins) ->
io:format("~b ways to make change for ~b cents with ~p coins~n",[count(Amount,Coins),Amount,Coins]).
test() ->
A1 = 100, C1 = [25,10,5,1],
print(A1,C1),
A2 = 100000, C2 = [100, 50, 25, 10, 5, 1],
print(A2,C2).
{{out}} 42> coins:test(). 242 ways to make change for 100 cents with [25,10,5,1] coins 13398445413854501 ways to make change for 100000 cents with [100,50,25,10,5,1] coins ok
=={{header|F_Sharp|F#}}== {{trans|OCaml}}
Forward iteration, which can also be seen in Scala.
let changes amount coins =
let ways = Array.zeroCreate (amount + 1)
ways.[0] <- 1L
List.iter (fun coin ->
for j = coin to amount do ways.[j] <- ways.[j] + ways.[j - coin]
) coins
ways.[amount]
[<EntryPoint>]
let main argv =
printfn "%d" (changes 100 [25; 10; 5; 1]);
printfn "%d" (changes 100000 [100; 50; 25; 10; 5; 1]);
0
{{out}}
242
13398445413854501
Factor
USING: combinators kernel locals math math.ranges sequences sets sorting ;
IN: rosetta.coins
<PRIVATE
! recursive-count uses memoization and local variables.
! coins must be a sequence.
MEMO:: recursive-count ( cents coins -- ways )
coins length :> types
{
! End condition: 1 way to make 0 cents.
{ [ cents zero? ] [ 1 ] }
! End condition: 0 ways to make money without any coins.
{ [ types zero? ] [ 0 ] }
! Optimization: At most 1 way to use 1 type of coin.
{ [ types 1 number= ] [
cents coins first mod zero? [ 1 ] [ 0 ] if
] }
! Find all ways to use the first type of coin.
[
! f = first type, r = other types of coins.
coins unclip-slice :> f :> r
! Loop for 0, f, 2*f, 3*f, ..., cents.
0 cents f <range> [
! Recursively count how many ways to make remaining cents
! with other types of coins.
cents swap - r recursive-count
] [ + ] map-reduce ! Sum the counts.
]
} cond ;
PRIVATE>
! How many ways can we make the given amount of cents
! with the given set of coins?
: make-change ( cents coins -- ways )
members [ ] inv-sort-with ! Sort coins in descending order.
recursive-count ;
From the listener:
'''USE: rosetta.coins''' ( scratchpad ) '''100 { 25 10 5 1 } make-change .''' 242 ( scratchpad ) '''100000 { 100 50 25 10 5 1 } make-change .''' 13398445413854501
This algorithm is '''slow'''. A test machine needed '''1 minute''' to run ''100000 { 100 50 25 10 5 1 } make-change .'' and get 13398445413854501. The same machine needed less than 1 second to run the Common Lisp ([[SBCL]]), Ruby ([[MRI]]) or Tcl ([[tclsh]]) programs and get the same answer.
One might make use of the rosetta-code.count-the-coins vocabulary as shown:
For reference, the implementation is shown next.
<lang>
USING: arrays locals math math.ranges sequences sets sorting ;
IN: rosetta-code.count-the-coins
<PRIVATE
:: (make-change) ( cents coins -- ways )
cents 1 + 0 <array> :> ways
1 ways set-first
coins [| coin |
coin cents [a,b] [| j |
j coin - ways nth j ways [ + ] change-nth
] each
] each ways last ;
PRIVATE>
! How many ways can we make the given amount of cents
! with the given set of coins?
: make-change ( cents coins -- ways )
members [ ] inv-sort-with (make-change) ;
Or one could implement the algorithm like described in http://www.cdn.geeksforgeeks.org/dynamic-programming-set-7-coin-change.
USE: math.ranges
:: exchange-count ( seq val -- cnt )
val 1 + 0 <array> :> tab
0 :> old!
1 0 tab set-nth
seq length iota [
seq nth old!
old val [a,b] [| j |
j old - tab nth
j tab nth +
j tab set-nth
] each
] each
val tab nth
;
[ { 1 5 10 25 50 100 } 100000 exchange-count . ] time
13398445413854501
Running time: 0.029163549 seconds
Forth
\ counting change (SICP section 1.2.2)
: table create does> swap cells + @ ;
table coin-value 0 , 1 , 5 , 10 , 25 , 50 ,
: count-change ( total coin -- n )
over 0= if
2drop 1
else over 0< over 0= or if
2drop 0
else
2dup coin-value - over recurse
>r 1- recurse r> +
then then ;
100 5 count-change .
FreeBASIC
Translation from "Dynamic Programming Solution: Python version" on this webside [http://www.geeksforgeeks.org/dynamic-programming-set-7-coin-change/]
' version 09-10-2016
' compile with: fbc -s console
Function count(S() As UInteger, n As UInteger) As ULongInt
Dim As Integer i, j
' calculate m from array S()
Dim As UInteger m = UBound(S) - LBound(S) +1
Dim As ULongInt x, y
'' We need n+1 rows as the table is consturcted in bottom up manner using
'' the base case 0 value case (n = 0)
Dim As ULongInt table(n +1, m)
'' Fill the enteries for 0 value case (n = 0)
For i = 0 To m -1
table(0, i) = 1
Next
'' Fill rest of the table enteries in bottom up manner
For i = 1 To n
For j = 0 To m -1
'' Count of solutions including S[j]
x = IIf (i >= S(j), table(i - S(j), j), 0)
'' Count of solutions excluding S[j]
y = IIf (j >= 1, table(i, j -1), 0)
''total count
table(i, j) = x + y
Next
Next
Return table(n, m -1)
End Function
' ------=< MAIN >=------
Dim As UInteger n
Dim As UInteger value()
ReDim value(3)
value(0) = 1 : value(1) = 5 : value(2) = 10 : value(3) = 25
n = 100
print
Print " There are "; count(value(), n); " ways to make change for $";n/100;" with 4 coins"
Print
n = 100000
Print " There are "; count(value(), n); " ways to make change for $";n/100;" with 4 coins"
Print
ReDim value(5)
value(0) = 1 : value(1) = 5 : value(2) = 10
value(3) = 25 : value(4) = 50 : value(5) = 100
n = 100000
Print " There are "; count(value(), n); " ways to make change for $";n/100;" with 6 coins"
Print
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
{{out}}
There are 242 ways to make change for $ 1 with 4 coins
There are 133423351001 ways to make change for $ 1000 with 4 coins
There are 13398445413854501 ways to make change for $ 1000 with 6 coins
FutureBasic
include "ConsoleWindow"
dim as long penny, nickel, dime, quarter , count
penny = 1 : nickel = 1
dime = 1 : quarter = 1
for penny = 0 to 100
for nickel = 0 to 20
for dime = 0 to 10
for quarter = 0 to 4
if penny + nickel * 5 + dime * 10 + quarter * 25 == 100
print penny; " pennies "; nickel;" nickels "; dime; " dimes "; quarter; " quarters"
count++
end if
next quarter
next dime
next nickel
next penny
print count;" ways to make a dollar"
Output:
0 pennies 0 nickels 0 dimes 4 quarters
0 pennies 0 nickels 5 dimes 2 quarters
0 pennies 0 nickels 10 dimes 0 quarters
0 pennies 1 nickels 2 dimes 3 quarters
......
65 pennies 5 nickels 1 dimes 0 quarters
65 pennies 7 nickels 0 dimes 0 quarters
70 pennies 0 nickels 3 dimes 0 quarters
70 pennies 1 nickels 0 dimes 1 quarters
242 ways to make a dollar
Go
A translation of the Lisp code referenced by the task description:
package main
import "fmt"
func main() {
amount := 100
fmt.Println("amount, ways to make change:", amount, countChange(amount))
}
func countChange(amount int) int64 {
return cc(amount, 4)
}
func cc(amount, kindsOfCoins int) int64 {
switch {
case amount == 0:
return 1
case amount < 0 || kindsOfCoins == 0:
return 0
}
return cc(amount, kindsOfCoins-1) +
cc(amount - firstDenomination(kindsOfCoins), kindsOfCoins)
}
func firstDenomination(kindsOfCoins int) int {
switch kindsOfCoins {
case 1:
return 1
case 2:
return 5
case 3:
return 10
case 4:
return 25
}
panic(kindsOfCoins)
}
Output:
amount, ways to make change: 100 242
Alternative algorithm, practical for the optional task.
package main
import "fmt"
func main() {
amount := 1000 * 100
fmt.Println("amount, ways to make change:", amount, countChange(amount))
}
func countChange(amount int) int64 {
ways := make([]int64, amount+1)
ways[0] = 1
for _, coin := range []int{100, 50, 25, 10, 5, 1} {
for j := coin; j <= amount; j++ {
ways[j] += ways[j-coin]
}
}
return ways[amount]
}
Output:
amount, ways to make change: 100000 13398445413854501
Groovy
{{trans|Go}} Intuitive Recursive Solution:
def ccR
ccR = { BigInteger tot, List<BigInteger> coins ->
BigInteger n = coins.size()
switch ([tot:tot, coins:coins]) {
case { it.tot == 0 } :
return 1g
case { it.tot < 0 || coins == [] } :
return 0g
default:
return ccR(tot, coins[1..<n]) +
ccR(tot - coins[0], coins)
}
}
Fast Iterative Solution:
def ccI = { BigInteger tot, List<BigInteger> coins ->
List<BigInteger> ways = [0g] * (tot+1)
ways[0] = 1g
coins.each { BigInteger coin ->
(coin..tot).each { j ->
ways[j] += ways[j-coin]
}
}
ways[tot]
}
Test:
println '\nBase:'
[iterative: ccI, recursive: ccR].each { label, cc ->
print "${label} "
def start = System.currentTimeMillis()
def ways = cc(100g, [25g, 10g, 5g, 1g])
def elapsed = System.currentTimeMillis() - start
println ("answer: ${ways} elapsed: ${elapsed}ms")
}
print '\nExtra Credit:\niterative '
def start = System.currentTimeMillis()
def ways = ccI(1000g * 100, [100g, 50g, 25g, 10g, 5g, 1g])
def elapsed = System.currentTimeMillis() - start
println ("answer: ${ways} elapsed: ${elapsed}ms")
Output:
Base:
iterative answer: 242 elapsed: 5ms
recursive answer: 242 elapsed: 220ms
Extra Credit:
iterative answer: 13398445413854501 elapsed: 1077ms
Haskell
Naive implementation:
count :: (Integral t, Integral a) => t -> [t] -> a
count 0 _ = 1
count _ [] = 0
count x (c:coins) =
sum
[ count (x - (n * c)) coins
| n <- [0 .. (quot x c)] ]
main :: IO ()
main = print (count 100 [1, 5, 10, 25])
Much faster, probably harder to read, is to update results from bottom up:
[Int] -> [a]
count = foldr addCoin (1 : repeat 0)
where
addCoin c oldlist = newlist
where
newlist = take c oldlist ++ zipWith (+) newlist (drop c oldlist)
main :: IO ()
main = do
print (count [25, 10, 5, 1] !! 100)
print (count [100, 50, 25, 10, 5, 1] !! 10000)
Or equivalently, (reformulating slightly, and adding a further test):
import Data.Function (fix)
count :: Integral a => [Int] -> [a]
count =
foldr
(\x a ->
let (l, r) = splitAt x a
in fix (mappend l . flip (zipWith (+)) r))
(1 : repeat 0)
-- TEST -----------------------------------------------------------------------
main :: IO ()
main =
mapM_
(print . uncurry ((!!) . count))
[ ([25, 10, 5, 1], 100)
, ([100, 50, 25, 10, 5, 1], 10000)
, ([100, 50, 25, 10, 5, 1], 1000000)
]
{{Out}}
242
139946140451
1333983445341383545001
=={{header|Icon}} and {{header|Unicon}}==
procedure main()
US_coins := [1, 5, 10, 25]
US_allcoins := [1,5,10,25,50,100]
EU_coins := [1, 2, 5, 10, 20, 50, 100, 200]
CDN_coins := [1,5,10,25,100,200]
CDN_allcoins := [1,5,10,25,50,100,200]
every trans := ![ [15,US_coins],
[100,US_coins],
[1000*100,US_allcoins]
] do
printf("There are %i ways to count change for %i using %s coins.\n",CountCoins!trans,trans[1],ShowList(trans[2]))
end
procedure ShowList(L) # helper list to string
every (s := "[ ") ||:= !L || " "
return s || "]"
end
This is a naive implementation and very slow. {{improve|Icon|Needs a better algorithm.}}
procedure CountCoins(amt,coins) # very slow, recurse by coin value
local count
static S
if type(coins) == "list" then {
S := sort(set(coins))
if *S < 1 then runerr(205,coins)
return CountCoins(amt)
}
else {
/coins := 1
if value := S[coins] then {
every (count := 0) +:= CountCoins(amt - (0 to amt by value), coins + 1)
return count
}
else
return (amt ~= 0) | 1
}
end
{{libheader|Icon Programming Library}} [http://www.cs.arizona.edu/icon/library/src/procs/printf.icn printf.icn provides formatting]
Output:
There are 6 ways to count change for 15 using [ 1 5 10 25 ] coins.
There are 242 ways to count change for 100 using [ 1 5 10 25 ] coins.
^c
Another one:
# coin.icn
# usage: coin value
procedure count(coinlist, value)
if value = 0 then return 1
if value < 0 then return 0
if (*coinlist <= 0) & (value >= 1) then return 0
return count(coinlist[1:*coinlist], value) + count(coinlist, value - coinlist[*coinlist])
end
procedure main(params)
money := params[1]
coins := [1,5,10,25]
writes("Value of ", money, " can be changed by using a set of ")
every writes(coins[1 to *coins], " ")
write(" coins in ", count(coins, money), " different ways.")
end
Output:
Value of 15 can be changed by using a set of 1 5 10 25 coins in 6 different ways.
Value of 100 can be changed by using a set of 1 5 10 25 coins in 242 different ways.
J
In this draft intermediate results are a two column array. The first column is tallies -- the number of ways we have for reaching the total represented in the second column, which is unallocated value (which we will assume are pennies). We will have one row for each different in-range value which can be represented using only nickles (0, 5, 10, ... 95, 100).
merge=: ({:"1 (+/@:({."1),{:@{:)/. ])@;
count=: {.@] <@,. {:@] - [ * [ i.@>:@<.@%~ {:@]
init=: (1 ,. ,.)^:(0=#@$)
nsplits=: 0 { [: +/ [: (merge@:(count"1) init)/ }.@/:~@~.@,
This implementation special cases the handling of pennies and assumes that the lowest coin value in the argument is 1. If I needed additional performance, I would next special case the handling of nickles/penny combinations...
Thus:
100 nsplits 1 5 10 25
242
And, on a 64 bit machine with sufficient memory:
100000 nsplits 1 5 10 25 50 100
13398445413854501
Warning: the above version can miss one when the largest coin is equal to the total value.
For British viewers change from £10 using £10 £5 £2 £1 50p 20p 10p 5p 2p and 1p
init =: 4 : '(1+x)$1'
length1 =: 4 : '1=#y'
f =: 4 : ',/ +/\ (-x) ]\ y'
1000 { f ` init @. length1 / 1000 500 200 100 50 20 10 5 2 , 1000 0
327631322
NB. this is a foldLeft once initialised the intermediate right arguments are arrays
1000 f 500 f 200 f 100 f 50 f 20 f 10 f 5 f 2 f (1000 init 0)
Java
{{trans|D}} {{works with|Java|1.5+}}
import java.util.Arrays;
import java.math.BigInteger;
class CountTheCoins {
private static BigInteger countChanges(int amount, int[] coins){
final int n = coins.length;
int cycle = 0;
for (int c : coins)
if (c <= amount && c >= cycle)
cycle = c + 1;
cycle *= n;
BigInteger[] table = new BigInteger[cycle];
Arrays.fill(table, 0, n, BigInteger.ONE);
Arrays.fill(table, n, cycle, BigInteger.ZERO);
int pos = n;
for (int s = 1; s <= amount; s++) {
for (int i = 0; i < n; i++) {
if (i == 0 && pos >= cycle)
pos = 0;
if (coins[i] <= s) {
final int q = pos - (coins[i] * n);
table[pos] = (q >= 0) ? table[q] : table[q + cycle];
}
if (i != 0)
table[pos] = table[pos].add(table[pos - 1]);
pos++;
}
}
return table[pos - 1];
}
public static void main(String[] args) {
final int[][] coinsUsEu = {{100, 50, 25, 10, 5, 1},
{200, 100, 50, 20, 10, 5, 2, 1}};
for (int[] coins : coinsUsEu) {
System.out.println(countChanges( 100,
Arrays.copyOfRange(coins, 2, coins.length)));
System.out.println(countChanges( 100000, coins));
System.out.println(countChanges( 1000000, coins));
System.out.println(countChanges(10000000, coins) + "\n");
}
}
}
Output:
242
13398445413854501
1333983445341383545001
133339833445334138335450001
4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001
JavaScript
Iterative
Efficient iterative algorithm (cleverly calculates number of combinations without permuting them)
function countcoins(t, o) {
'use strict';
var targetsLength = t + 1;
var operandsLength = o.length;
t = [1];
for (var a = 0; a < operandsLength; a++) {
for (var b = 1; b < targetsLength; b++) {
// initialise undefined target
t[b] = t[b] ? t[b] : 0;
// accumulate target + operand ways
t[b] += (b < o[a]) ? 0 : t[b - o[a]];
}
}
return t[targetsLength - 1];
}
{{out}} JavaScript hits integer limit for optional task
countcoins(100, [1,5,10,25]);
242
Recursive
Inefficient recursive algorithm (naively calculates number of combinations by actually permuting them)
function countcoins(t, o) {
'use strict';
var operandsLength = o.length;
var solutions = 0;
function permutate(a, x) {
// base case
if (a === t) {
solutions++;
}
// recursive case
else if (a < t) {
for (var i = 0; i < operandsLength; i++) {
if (i >= x) {
permutate(o[i] + a, i);
}
}
}
}
permutate(0, 0);
return solutions;
}
{{Out}} Too slow for optional task
countcoins(100, [1,5,10,25]);
242
Iterative again
{{Trans|C#}}
var amount = 100,
coin = [1, 5, 10, 25]
var t = [1];
for (t[amount] = 0, a = 1; a < amount; a++) t[a] = 0 // initialise t[0..amount]=[1,0,...,0]
for (var i = 0, e = coin.length; i < e; i++)
for (var ci = coin[i], a = ci; a <= amount; a++)
t[a] += t[a - ci]
document.write(t[amount])
{{Out}}
242
jq
Currently jq uses IEEE 754 64-bit numbers. Large integers are approximated by floats, and therefore the answer that the following program provides for the optional task is only correct for the first 15 digits.
# How many ways are there to make "target" cents, given a list of coin
# denominations as input.
# The strategy is to record at total[n] the number of ways to make n cents.
def countcoins(target):
. as $coin
| reduce range(0; length) as $a
( [1]; # there is 1 way to make 0 cents
reduce range(1; target + 1) as $b
(.; # total[]
if $b < $coin[$a] then .
else .[$b - $coin[$a]] as $count
| if $count == 0 then .
else .[$b] += $count
end
end ) )
| .[target] ;
'''Example''': [1,5,10,25] | countcoins(100) {{Out}} 242
Julia
{{trans|Python}}
function changes(amount::Int, coins::Array{Int})::Int128
ways = zeros(Int128, amount + 1)
ways[1] = 1
for coin in coins, j in coin+1:amount+1
ways[j] += ways[j - coin]
end
return ways[amount + 1]
end
@show changes(100, [1, 5, 10, 25])
@show changes(100000, [1, 5, 10, 25, 50, 100])
{{out}}
changes(100, [1, 5, 10, 25]) = 242
changes(100000, [1, 5, 10, 25, 50, 100]) = 13398445413854501
Kotlin
{{trans|C#}}
// version 1.0.6
fun countCoins(c: IntArray, m: Int, n: Int): Long {
val table = LongArray(n + 1)
table[0] = 1
for (i in 0 until m)
for (j in c[i]..n) table[j] += table[j - c[i]]
return table[n]
}
fun main(args: Array<String>) {
val c = intArrayOf(1, 5, 10, 25, 50, 100)
println(countCoins(c, 4, 100))
println(countCoins(c, 6, 1000 * 100))
}
{{out}}
242
13398445413854501
Lasso
Inspired by the javascript iterative example for the same task
define cointcoins(
target::integer,
operands::array
) => {
local(
targetlength = #target + 1,
operandlength = #operands -> size,
output = staticarray_join(#targetlength,0),
outerloopcount
)
#output -> get(1) = 1
loop(#operandlength) => {
#outerloopcount = loop_count
loop(#targetlength) => {
if(loop_count >= #operands -> get(#outerloopcount) and loop_count - #operands -> get(#outerloopcount) > 0) => {
#output -> get(loop_count) += #output -> get(loop_count - #operands -> get(#outerloopcount))
}
}
}
return #output -> get(#targetlength)
}
cointcoins(100, array(1,5,10,25,))
'<br />'
cointcoins(100000, array(1, 5, 10, 25, 50, 100))
Output:
242
13398445413854501
Lua
Lua uses one-based indexes but table keys can be any value so you can define an element 0 just as easily as you can define an element "foo"...
function countSums (amount, values)
local t = {}
for i = 1, amount do t[i] = 0 end
t[0] = 1
for k, val in pairs(values) do
for i = val, amount do t[i] = t[i] + t[i - val] end
end
return t[amount]
end
print(countSums(100, {1, 5, 10, 25}))
print(countSums(100000, {1, 5, 10, 25, 50, 100}))
{{out}}
242
1.3398445413855e+16
M2000 Interpreter
===Fast O(n*m)=== Works with decimals in table()
Module FindCoins {
Function count(c(), n) {
dim table(n+1)=0@ : table(0)=1@
for c=0 to len(c())-1 {
if c(c)>n then exit
}
if c else exit
for i=0 to c-1 {for j=c(i) to n {table(j)+=table(j-c(i))}}
=table(n)
}
Print "For 1$ ways to change:";count((1,5,10,25),100)
Print "For 100$ (optional task ways to change):";count((1,5,10,25,50,100),100000)
}
FindCoins
{{out}}
For 1$ ways to change:242
For 100$ (optional task) ways to change:13398445413854501
With Recursion with saving partial results
Using an inventory (a kind of vector) to save first search (but is slower than previous one)
Module CheckThisToo {
inventory c=" 0 0":=1@
make_change=lambda c (amount, coins()) ->{
m=lambda c,coins() (n,m)->{if n<0 or m<0 then =0@:exit
if exist(c,str$(n)+str$(m)) then =eval(c):exit
append c,str$(n)+str$(m):=lambda(n-coins(m), m)+lambda(n, m-1):=c(str$(n)+str$(m))}
=m(amount,len(coins())-1)
}
Print make_change(100, (1,5,10,25,50,100))=293
Print make_change(100, (1,5,10,25))=242
Print make_change(15, (1,5,10,25))=6
Print make_change(5, (1,5,10,25))=2
}
CheckThisToo
=={{header|Mathematica}} / {{header|Wolfram Language}}== {{trans|Go}}
CountCoins[amount_, coinlist_] := ( ways = ConstantArray[1, amount];
Do[For[j = coin, j <= amount, j++,
If[ j - coin == 0,
ways[[j]] ++,
ways[[j]] += ways[[j - coin]]
]]
, {coin, coinlist}];
ways[[amount]])
Example usage:
CountCoins[100, {25, 10, 5}]
-> 242
CountCoins[100000, {100, 50, 25, 10, 5}]
-> 13398445413854501
=={{header|MATLAB}} / {{header|Octave}}==
%% Count_The_Coins
clear;close all;clc;
tic
for i = 1:2 % 1st loop is main challenge 2nd loop is optional challenge
if (i == 1)
amount = 100; % Matlab indexes from 1 not 0, so we need to add 1 to our target value
amount = amount + 1;
coins = [1 5 10 25]; % Value of coins we can use
else
amount = 100*1000; % Matlab indexes from 1 not 0, so we need to add 1 to our target value
amount = amount + 1;
coins = [1 5 10 25 50 100]; % Value of coins we can use
end % End if
ways = zeros(1,amount); % Preallocating for speed
ways(1) = 1; % First solution is 1
% Solves from smallest sub problem to largest (bottom up approach of dynamic programming).
for j = 1:length(coins)
for K = coins(j)+1:amount
ways(K) = ways(K) + ways(K-coins(j));
end % End for
end % End for
if (i == 1)
fprintf(‘Main Challenge: %d \n', ways(amount));
else
fprintf(‘Bonus Challenge: %d \n', ways(amount));
end % End if
end % End for
toc
Example Output:
Main Challenge: 242
Bonus Challenge: 13398445413854501
Mercury
:- module coins.
:- interface.
:- import_module int, io.
:- type coin ---> quarter; dime; nickel; penny.
:- type purse ---> purse(int, int, int, int).
:- pred sum_to(int::in, purse::out) is nondet.
:- pred main(io::di, io::uo) is det.
:- implementation.
:- import_module solutions, list, string.
:- func value(coin) = int.
value(quarter) = 25.
value(dime) = 10.
value(nickel) = 5.
value(penny) = 1.
:- pred supply(coin::in, int::in, int::out) is multi.
supply(C, Target, N) :- upto(Target div value(C), N).
:- pred upto(int::in, int::out) is multi.
upto(N, R) :- ( nondet_int_in_range(0, N, R0) -> R = R0 ; R = 0 ).
sum_to(To, Purse) :-
Purse = purse(Q, D, N, P),
sum(Purse) = To,
supply(quarter, To, Q),
supply(dime, To, D),
supply(nickel, To, N),
supply(penny, To, P).
:- func sum(purse) = int.
sum(purse(Q, D, N, P)) =
value(quarter) * Q + value(dime) * D +
value(nickel) * N + value(penny) * P.
main(!IO) :-
solutions(sum_to(100), L),
show(L, !IO),
io.format("There are %d ways to make change for a dollar.\n",
[i(length(L))], !IO).
:- pred show(list(purse)::in, io::di, io::uo) is det.
show([], !IO).
show([P|T], !IO) :-
io.write(P, !IO), io.nl(!IO),
show(T, !IO).
Nim
{{trans|Python}}
proc changes(amount: int, coins: openArray[int]): int =
var ways = @[1]
ways.setLen(amount+1)
for coin in coins:
for j in coin..amount:
ways[j] += ways[j-coin]
ways[amount]
echo changes(100, [1, 5, 10, 25])
echo changes(100000, [1, 5, 10, 25, 50, 100])
Output:
242
13398445413854501
OCaml
Translation of the D minimal version:
let changes amount coins =
let ways = Array.make (amount + 1) 0L in
ways.(0) <- 1L;
List.iter (fun coin ->
for j = coin to amount do
ways.(j) <- Int64.add ways.(j) ways.(j - coin)
done
) coins;
ways.(amount)
let () =
Printf.printf "%Ld\n" (changes 1_00 [25; 10; 5; 1]);
Printf.printf "%Ld\n" (changes 1000_00 [100; 50; 25; 10; 5; 1]);
;;
Output:
$ ocaml coins.ml
242
13398445413854501
PARI/GP
coins(v)=prod(i=1,#v,1/(1-'x^v[i]));
ways(v,n)=polcoeff(coins(v)+O('x^(n+1)),n);
ways([1,5,10,25],100)
ways([1,5,10,25,50,100],100000)
Output:
%1 = 242
%2 = 13398445413854501
Perl
use 5.01;
use Memoize;
sub cc {
my $amount = shift;
return 0 if !@_ || $amount < 0;
return 1 if $amount == 0;
my $first = shift;
cc( $amount, @_ ) + cc( $amount - $first, $first, @_ );
}
memoize 'cc';
# Make recursive algorithm run faster by sorting coins descending by value:
sub cc_optimized {
my $amount = shift;
cc( $amount, sort { $b <=> $a } @_ );
}
say 'Ways to change $ 1 with common coins: ',
cc_optimized( 100, 1, 5, 10, 25 );
say 'Ways to change $ 1000 with addition of less common coins: ',
cc_optimized( 1000 * 100, 1, 5, 10, 25, 50, 100 );
{{out}} Ways to change $ 1 with common coins: 242 Ways to change $ 1000 with addition of less common coins: 13398445413854501
Perl 6
{{works with|rakudo|2018.10}} {{trans|Ruby}}
# Recursive (cached)
sub change-r($amount, @coins) {
my @cache = [1 xx @coins], |([] xx $amount);
multi ways($n where $n >= 0, @now [$coin,*@later]) {
@cache[$n;+@later] //= ways($n - $coin, @now) + ways($n, @later);
}
multi ways($,@) { 0 }
# more efficient to start with coins sorted in descending order
ways($amount, @coins.sort(-*).list);
}
# Iterative
sub change-i(\n, @coins) {
my @table = [1 xx @coins], [0 xx @coins] xx n;
(1..n).map: -> \i {
for ^@coins -> \j {
my \c = @coins[j];
@table[i;j] = [+]
@table[i - c;j] // 0,
@table[i;j - 1] // 0;
}
}
@table[*-1][*-1];
}
say "Iterative:";
say change-i 1_00, [1,5,10,25];
say change-i 1000_00, [1,5,10,25,50,100];
say "\nRecursive:";
say change-r 1_00, [1,5,10,25];
say change-r 1000_00, [1,5,10,25,50,100];
{{out}}
Iterative:
242
13398445413854501
Recursive:
242
13398445413854501
Phix
Very fast, from http://www.geeksforgeeks.org/dynamic-programming-set-7-coin-change
function coin_count(sequence coins, integer amount)
sequence s = repeat(0,amount+1)
s[1] = 1
for c=1 to length(coins) do
for n=coins[c] to amount do
s[n+1] += s[n-coins[c]+1]
end for
end for
return s[amount+1]
end function
An attempt to explain this algorithm further seems worthwhile:
function coin_count(sequence coins, integer amount)
-- start with 1 known way to achieve 0 (being no coins)
-- (nb: s[1] holds the solution for 0, s[n+1] for n)
sequence s = repeat(0,amount+1)
s[1] = 1
-- then for every coin that we can use, increase number of
-- solutions by that previously found for the remainder.
for c=1 to length(coins) do
-- this inner loop is essentially behaving as if we had
-- called this routine with 1..amount, but skipping any
-- less than the coin's value, hence coins[c]..amount.
for n=coins[c] to amount do
s[n+1] += s[n-coins[c]+1]
end for
end for
return s[amount+1]
end function
-- The key to understanding the above is to try a dry run of this:
printf(1,"%d\n",coin_count({2,3},5)) -- (prints 1)
-- You'll need 4 2p coins, 3 3p coins, and 5 spaces marked 1..5.
-- Place 2p wherever it fits: 1:0 2:1 3:1 4:1 5:1
-- Add previously found solns: +0 +1 +0 +1 +0 [1]
-- Place 3p wherever it fits: 1:0 2:0 3:1 4:1 5:1
-- Add previously found solns: +0 +0 +1 +0 +1 [2]
-- [1] obviously at 2: we added the base soln for amount=0,
-- and at 4: we added the previously found soln for 2.
-- also note that we added nothing for 2p+3p, yet, that
-- fact is central to understanding why this works. [3]
-- [2] obviously at 3: we added the base soln for amount=0,
-- at 4: we added the zero solutions yet found for 1p,
-- and at 5: we added the previously found soln for 2.
-- you can imagine at 6,9,12 etc all add in soln for 3,
-- albeit by adding that as just added to the precessor.
-- [3] since we add no 3p solns when processing 2p, we do
-- not count 2p+3p and 3p+2p as two solutions.
--For N = 4 and S = {1,2,3}, there are four solutions: {1,1,1,1},{1,1,2},{2,2},{1,3}.
printf(1,"%d\n",coin_count({1,2,3},4))
--For N = 10 and S = {2, 5, 3, 6}, there are five solutions: {2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5} and {5,5}.
printf(1,"%d\n\n",coin_count({2,3,5,6},10))
printf(1,"%d\n",coin_count({25, 10, 5, 1},1_00))
printf(1,"%,d\n",coin_count({100, 50, 25, 10, 5, 1},1000_00))
{{out}}
1
4
5
242
13,398,445,413,854,501
Note that a slightly wrong value is printed when running this on 32 bits:
13,398,445,413,854,501 -- 64 bit (exact)
13,398,445,413,854,496 -- 32 bit (5 out)
9,007,199,254,740,992 -- max precision (53 bits) of a 64-bit float
PicoLisp
{{trans|C}}
(de coins (Sum Coins)
(let (Buf (mapcar '((N) (cons 1 (need (dec N) 0))) Coins) Prev)
(do Sum
(zero Prev)
(for L Buf
(inc (rot L) Prev)
(setq Prev (car L)) ) )
Prev ) )
Test:
(for Coins '((100 50 25 10 5 1) (200 100 50 20 10 5 2 1))
(println (coins 100 (cddr Coins)))
(println (coins (* 1000 100) Coins))
(println (coins (* 10000 100) Coins))
(println (coins (* 100000 100) Coins))
(prinl) )
Output:
242
13398445413854501
1333983445341383545001
133339833445334138335450001
4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001
Python
Simple version
{{trans|Go}}
def changes(amount, coins):
ways = [0] * (amount + 1)
ways[0] = 1
for coin in coins:
for j in xrange(coin, amount + 1):
ways[j] += ways[j - coin]
return ways[amount]
print changes(100, [1, 5, 10, 25])
print changes(100000, [1, 5, 10, 25, 50, 100])
Output:
242
13398445413854501
Fast version
{{trans|C}}
try:
import psyco
psyco.full()
except ImportError:
pass
def count_changes(amount_cents, coins):
n = len(coins)
# max([]) instead of max() for Psyco
cycle = max([c+1 for c in coins if c <= amount_cents]) * n
table = [0] * cycle
for i in xrange(n):
table[i] = 1
pos = n
for s in xrange(1, amount_cents + 1):
for i in xrange(n):
if i == 0 and pos >= cycle:
pos = 0
if coins[i] <= s:
q = pos - coins[i] * n
table[pos]= table[q] if (q >= 0) else table[q + cycle]
if i:
table[pos] += table[pos - 1]
pos += 1
return table[pos - 1]
def main():
us_coins = [100, 50, 25, 10, 5, 1]
eu_coins = [200, 100, 50, 20, 10, 5, 2, 1]
for coins in (us_coins, eu_coins):
print count_changes( 100, coins[2:])
print count_changes( 100000, coins)
print count_changes( 1000000, coins)
print count_changes(10000000, coins), "\n"
main()
Output:
242
13398445413854501
1333983445341383545001
133339833445334138335450001
4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001
Racket
This is the basic recursive way:
#lang racket
(define (ways-to-make-change cents coins)
(cond ((null? coins) 0)
((negative? cents) 0)
((zero? cents) 1)
(else
(+ (ways-to-make-change cents (cdr coins))
(ways-to-make-change (- cents (car coins)) coins)))))
(ways-to-make-change 100 '(25 10 5 1)) ; -> 242
This works for the small numbers, but the optional task is just too slow with this solution, so with little change to the code we can use memoization:
#lang racket
(define memos (make-hash))
(define (ways-to-make-change cents coins)
(cond [(or (empty? coins) (negative? cents)) 0]
[(zero? cents) 1]
[else (define (answerer-for-new-arguments)
(+ (ways-to-make-change cents (rest coins))
(ways-to-make-change (- cents (first coins)) coins)))
(hash-ref! memos (cons cents coins) answerer-for-new-arguments)]))
(time (ways-to-make-change 100 '(25 10 5 1)))
(time (ways-to-make-change 100000 '(100 50 25 10 5 1)))
(time (ways-to-make-change 1000000 '(200 100 50 20 10 5 2 1)))
#| Times in milliseconds, and results:
cpu time: 1 real time: 1 gc time: 0
242
cpu time: 524 real time: 553 gc time: 163
13398445413854501
cpu time: 20223 real time: 20673 gc time: 10233
99341140660285639188927260001 |#
REXX
recursive
The recursive calls to the subroutine have been unrolled somewhat, this reduces the number of recursive calls substantially.
These REXX versions also support fractional cents (as in a ½-cent and ¼-cent coins). Any fractional coin can be
specified as a decimal fraction (.5, .25, ···).
Support was included to allow specification of half-cent and quarter-cent coins as '''1/2''' and '''1/4'''.
The amount can be specified in cents (as a number), or in dollars (as for instance, $1000).
/*REXX program counts the number of ways to make change with coins from an given amount.*/
numeric digits 20 /*be able to handle large amounts of $.*/
parse arg N $ /*obtain optional arguments from the CL*/
if N='' | N="," then N=100 /*Not specified? Then Use $1 (≡100¢).*/
if $='' | $="," then $=1 5 10 25 /*Use penny/nickel/dime/quarter default*/
if left(N,1)=='$' then N=100*substr(N,2) /*the amount was specified in dollars.*/
coins=words($) /*the number of coins specified. */
NN=N; do j=1 for coins /*create a fast way of accessing specie*/
_=word($,j) /*define an array element for the coin.*/
if _=='1/2' then _=.5 /*an alternate spelling of a half-cent.*/
if _=='1/4' then _=.25 /* " " " " " quarter-¢.*/
$.j=_ /*assign the value to a particular coin*/
end /*j*/
_=n//100; cnt=' cents' /* [↓] is the amount in whole dollars?*/
if _=0 then do; NN='$' || (NN%100); cnt=; end /*show the amount in dollars, not cents*/
say 'with an amount of ' commas(NN)cnt", there are " commas( MKchg(N, coins) )
say 'ways to make change with coins of the following denominations: ' $
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: procedure; parse arg _; n=_'.9'; #=123456789; b=verify(n,#,"M")
e=verify(n,#'0',,verify(n,#"0.",'M'))-4
do j=e to b by -3; _=insert(',',_,j); end /*j*/; return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
MKchg: procedure expose $.; parse arg a,k /*this function is invoked recursively.*/
if a==0 then return 1 /*unroll for a special case of zero. */
if k==1 then return 1 /* " " " " " " unity. */
if k==2 then f=1 /*handle this special case of two. */
else f=MKchg(a, k-1) /*count, and then recurse the amount. */
if a==$.k then return f+1 /*handle this special case of A=a coin.*/
if a <$.k then return f /* " " " " " A<a coin.*/
return f+MKchg(a-$.k,k) /*use diminished amount ($) for change.*/
'''output''' when using the default input:
with an amount of $1, there are 242
ways to make change with coins of the following denominations: 1 5 10 25
'''output''' when using the following input: $1 1/4 1/2 1 2 3 5 10 20 25 50 100
with an amount of $1, there are 29,034,171
ways to make change with coins of the following denominations: 1/4 1/2 1 2 3 5 10 20 25 50 100
with memoization
This REXX version is more than a couple of orders of magnitude faster than the 1st version when using larger amounts.
/*REXX program counts the number of ways to make change with coins from an given amount.*/
numeric digits 20 /*be able to handle large amounts of $.*/
parse arg N $ /*obtain optional arguments from the CL*/
if N='' | N="," then N=100 /*Not specified? Then Use $1 (≡100¢).*/
if $='' | $="," then $=1 5 10 25 /*Use penny/nickel/dime/quarter default*/
if left(N,1)=='$' then N=100*substr(N,2) /*the amount was specified in dollars.*/
coins=words($) /*the number of coins specified. */
!.=.; NN=N; do j=1 for coins /*create a fast way of accessing specie*/
_=word($,j); ?=_ ' coin' /*define an array element for the coin.*/
if _=='1/2' then _=.5 /*an alternate spelling of a half-cent.*/
if _=='1/4' then _=.25 /* " " " " " quarter-¢.*/
$.j=_ /*assign the value to a particular coin*/
end /*j*/
_=n//100; cnt=' cents' /* [↓] is the amount in whole dollars?*/
if _=0 then do; NN='$' || (NN%100); cnt=; end /*show the amount in dollars, not cents*/
say 'with an amount of ' commas(NN)cnt", there are " commas( MKchg(N, coins) )
say 'ways to make change with coins of the following denominations: ' $
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: procedure; parse arg _; n=_'.9'; #=123456789; b=verify(n,#,"M")
e=verify(n,#'0',,verify(n,#"0.",'M'))-4
do j=e to b by -3; _=insert(',',_,j); end /*j*/; return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
MKchg: procedure expose $. !.; parse arg a,k /*function is recursive. */
if !.a.k\==. then return !.a.k /*found this A & K before? */
if a==0 then return 1 /*unroll for a special case*/
if k==1 then return 1 /* " " " " " */
if k==2 then f=1 /*handle this special case.*/
else f=MKchg(a, k-1) /*count, recurse the amount*/
if a==$.k then do; !.a.k=f+1; return !.a.k; end /*handle this special case.*/
if a <$.k then do; !.a.k=f ; return f ; end /* " " " " */
!.a.k=f + MKchg(a-$.k, k); return !.a.k /*compute, define, return. */
'''output''' when using the following input for the optional test case: $1000 1 5 10 25 50 100
with an amount of $1,000, there are 13,398,445,413,854,501
ways to make change with coins of the following denominations: 1 5 10 25 50 100
with error checking
This REXX version is identical to the previous REXX version, but has error checking for the amount and the coins specified.
/*REXX program counts the number of ways to make change with coins from an given amount.*/
numeric digits 20 /*be able to handle large amounts of $.*/
parse arg N $ /*obtain optional arguments from the CL*/
if N='' | N="," then N=100 /*Not specified? Then Use $1 (≡100¢).*/
if $='' | $="," then $=1 5 10 25 /*Use penny/nickel/dime/quarter default*/
X=N /*save original for possible error msgs*/
if left(N,1)=='$' then do /*the amount has a leading dollar sign.*/
_=substr(N,2) /*the amount was specified in dollars.*/
if \isNum(_) then call ser "amount isn't numeric: " N
N=100*_ /*change amount (in $) ───► cents (¢).*/
end
max$=10**digits() /*the maximum amount this pgm can have.*/
if \isNum(N) then call ser X " amount isn't numeric."
if N=0 then call ser X " amount can't be zero."
if N<0 then call ser X " amount can't be negative."
if N>max$ then call ser X " amount can't be greater than " max$'.'
coins=words($); !.=.; NN=N; p=0 /*#coins specified; coins; amount; prev*/
@.=0 /*verify a coin was only specified once*/
do j=1 for coins /*create a fast way of accessing specie*/
_=word($,j); ?=_ ' coin' /*define an array element for the coin.*/
if _=='1/2' then _=.5 /*an alternate spelling of a half-cent.*/
if _=='1/4' then _=.25 /* " " " " " quarter-¢.*/
if \isNum(_) then call ser ? "coin value isn't numeric."
if _<0 then call ser ? "coin value can't be negative."
if _<=0 then call ser ? "coin value can't be zero."
if @._ then call ser ? "coin was already specified."
if _<p then call ser ? "coin must be greater than previous:" p
if _>N then call ser ? "coin must be less or equal to amount:" X
@._=1; p=_ /*signify coin was specified; set prev.*/
$.j=_ /*assign the value to a particular coin*/
end /*j*/
_=n//100; cnt=' cents' /* [↓] is the amount in whole dollars?*/
if _=0 then do; NN='$' || (NN%100); cnt=; end /*show the amount in dollars, not cents*/
say 'with an amount of ' commas(NN)cnt", there are " commas( MKchg(N, coins) )
say 'ways to make change with coins of the following denominations: ' $
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
isNum: return datatype(arg(1), 'N') /*return 1 if arg is numeric, 0 if not.*/
ser: say; say '***error***'; say; say arg(1); say; exit 13 /*error msg.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: procedure; parse arg _; n=_'.9'; #=123456789; b=verify(n,#,"M")
e=verify(n,#'0',,verify(n,#"0.",'M'))-4
do j=e to b by -3; _=insert(',',_,j); end /*j*/; return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
MKchg: procedure expose $. !.; parse arg a,k /*function is recursive. */
if !.a.k\==. then return !.a.k /*found this A & K before? */
if a==0 then return 1 /*unroll for a special case*/
if k==1 then return 1 /* " " " " " */
if k==2 then f=1 /*handle this special case.*/
else f=MKchg(a, k-1) /*count, recurse the amount*/
if a==$.k then do; !.a.k=f+1; return !.a.k; end /*handle this special case.*/
if a <$.k then do; !.a.k=f ; return f ; end /* " " " " */
!.a.k=f + MKchg(a-$.k, k); return !.a.k /*compute, define, return. */
'''output''' is the same as the previous REXX versions.
Ring
penny = 1
nickel = 1
dime = 1
quarter = 1
count = 0
for penny = 0 to 100
for nickel = 0 to 20
for dime = 0 to 10
for quarter = 0 to 4
if (penny + nickel * 5 + dime * 10 + quarter * 25) = 100
see "" + penny + " pennies " + nickel + " nickels " + dime + " dimes " + quarter + " quarters" + nl
count = count + 1
ok
next
next
next
next
see count + " ways to make a dollar" + nl
Output:
0 pennies 0 nickels 0 dimes 4 quarters
0 pennies 0 nickels 5 dimes 2 quarters
0 pennies 0 nickels 10 dimes 0 quarters
0 pennies 1 nickels 2 dimes 3 quarters
......
65 pennies 5 nickels 1 dimes 0 quarters
65 pennies 7 nickels 0 dimes 0 quarters
70 pennies 0 nickels 3 dimes 0 quarters
70 pennies 1 nickels 0 dimes 1 quarters
242 ways to make a dollar
Ruby
The algorithm also appears [http://www.algorithmist.com/index.php/Coin_Change here]
'''Recursive, with caching'''
def make_change(amount, coins)
@cache = Array.new(amount+1){|i| Array.new(coins.size, i.zero? ? 1 : nil)}
@coins = coins
do_count(amount, @coins.length - 1)
end
def do_count(n, m)
if n < 0 || m < 0
0
elsif @cache[n][m]
@cache[n][m]
else
@cache[n][m] = do_count(n-@coins[m], m) + do_count(n, m-1)
end
end
p make_change( 1_00, [1,5,10,25])
p make_change(1000_00, [1,5,10,25,50,100])
outputs
242
13398445413854501
'''Iterative'''
def make_change2(amount, coins)
n, m = amount, coins.size
table = Array.new(n+1){|i| Array.new(m, i.zero? ? 1 : nil)}
for i in 1..n
for j in 0...m
table[i][j] = (i<coins[j] ? 0 : table[i-coins[j]][j]) +
(j<1 ? 0 : table[i][j-1])
end
end
table[-1][-1]
end
p make_change2( 1_00, [1,5,10,25])
p make_change2(1000_00, [1,5,10,25,50,100])
outputs
242
13398445413854501
Run BASIC
for penny = 0 to 100
for nickel = 0 to 20
for dime = 0 to 10
for quarter = 0 to 4
if penny + nickel * 5 + dime * 10 + quarter * 25 = 100 then
print penny;" pennies ";nickel;" nickels "; dime;" dimes ";quarter;" quarters"
count = count + 1
end if
next quarter
next dime
next nickel
next penny
print count;" ways to make a buck"
Output:
0 pennies 0 nickels 0 dimes 4 quarters
0 pennies 0 nickels 5 dimes 2 quarters
0 pennies 0 nickels 10 dimes 0 quarters
0 pennies 1 nickels 2 dimes 3 quarters
......
65 pennies 5 nickels 1 dimes 0 quarters
65 pennies 7 nickels 0 dimes 0 quarters
70 pennies 0 nickels 3 dimes 0 quarters
70 pennies 1 nickels 0 dimes 1 quarters
.....
242 ways to make a buck
Rust
fn make_change(coins: &[usize], cents: usize) -> usize {
let size = cents + 1;
let mut ways = vec![0; size];
ways[0] = 1;
for &coin in coins {
for amount in coin..size {
ways[amount] += ways[amount - coin];
}
}
ways[cents]
}
fn main() {
println!("{}", make_change(&[1,5,10,25], 100));
println!("{}", make_change(&[1,5,10,25,50,100], 100_000));
}
{{output}}
242
13398445413854501
SAS
Generate the solutions using CLP solver in SAS/OR:
/* call OPTMODEL procedure in SAS/OR */
proc optmodel;
/* declare set and names of coins */
set COINS = {1,5,10,25};
str name {COINS} = ['penny','nickel','dime','quarter'];
/* declare variables and constraint */
var NumCoins {COINS} >= 0 integer;
con Dollar:
sum {i in COINS} i * NumCoins[i] = 100;
/* call CLP solver */
solve with CLP / findallsolns;
/* write solutions to SAS data set */
create data sols(drop=s) from [s]=(1.._NSOL_) {i in COINS} <col(name[i])=NumCoins[i].sol[s]>;
quit;
/* print all solutions */
proc print data=sols;
run;
Output:
Obs penny nickel dime quarter
1 100 0 0 0
2 95 1 0 0
3 90 2 0 0
4 85 3 0 0
5 80 4 0 0
...
238 5 2 1 3
239 0 3 1 3
240 5 0 2 3
241 0 1 2 3
242 0 0 0 4
Scala
def countChange(amount: Int, coins:List[Int]) = {
val ways = Array.fill(amount + 1)(0)
ways(0) = 1
coins.foreach (coin =>
for (j<-coin to amount)
ways(j) = ways(j) + ways(j - coin)
)
ways(amount)
}
countChange (15, List(1, 5, 10, 25))
Output:
res0: Int = 6
Recursive implementation:
def count(target: Int, coins: List[Int]): Int = {
if (target == 0) 1
else if (coins.isEmpty || target < 0) 0
else count(target, coins.tail) + count(target - coins.head, coins)
}
count(100, List(25, 10, 5, 1))
Scheme
A simple recursive implementation:
(define ways-to-make-change
(lambda (x coins)
(cond
[(null? coins) 0]
[(< x 0) 0]
[(zero? x) 1]
[else (+ (ways-to-make-change x (cdr coins)) (ways-to-make-change (- x (car coins)) coins))])))
(ways-to-make-change 100)
Output:
242
Scilab
Straightforward solution
Fairly simple solution for the task. Expanding it to the optional task is not recommend, for Scilab will spend a lot of time processing the nested for loops.
for a=0:4 for b=0:10 for c=0:20 for d=0:100 n_coins=[a b c d]; change=sum(n_coins.*coins); if change==amount then ways=ways+1; elseif change>amount break end end end end end
disp(ways);
{{out}}
```txt
242.
Faster approach
{{trans|Python}}
varargout=list(ways(length(ways)))
endfunction
a=changes(100, [1, 5, 10, 25]); b=changes(100000, [1, 5, 10, 25, 50, 100]); mprintf("%.0f, %.0f", a, b);
{{out}}
```txt
242, 13398445413854540
Seed7
$ include "seed7_05.s7i";
include "bigint.s7i";
const func bigInteger: changeCount (in integer: amountCents, in array integer: coins) is func
result
var bigInteger: waysToChange is 0_;
local
var array bigInteger: t is 0 times 0_;
var integer: pos is 0;
var integer: s is 0;
var integer: i is 0;
begin
t := length(coins) times 1_ & (length(coins) * amountCents) times 0_;
pos := length(coins) + 1;
for s range 1 to amountCents do
if coins[1] <= s then
t[pos] := t[pos - (length(coins) * coins[1])];
end if;
incr(pos);
for i range 2 to length(coins) do
if coins[i] <= s then
t[pos] := t[pos - (length(coins) * coins[i])];
end if;
t[pos] +:= t[pos - 1];
incr(pos);
end for;
end for;
waysToChange := t[pos - 1];
end func;
const proc: main is func
local
const array integer: usCoins is [] (1, 5, 10, 25, 50, 100);
const array integer: euCoins is [] (1, 2, 5, 10, 20, 50, 100, 200);
begin
writeln(changeCount( 100, usCoins[.. 4]));
writeln(changeCount( 100000, usCoins));
writeln(changeCount(1000000, usCoins));
writeln(changeCount( 100000, euCoins));
writeln(changeCount(1000000, euCoins));
end func;
Output:
242
13398445413854501
1333983445341383545001
10056050940818192726001
99341140660285639188927260001
Sidef
{{trans|Perl}}
func cc(_) { 0 }
func cc({ .is_neg }, *_) { 0 }
func cc({ .is_zero }, *_) { 1 }
func cc(amount, first, *rest) is cached {
cc(amount, rest...) + cc(amount - first, first, rest...);
}
func cc_optimized(amount, *rest) {
cc(amount, rest.sort_by{|v| -v }...);
}
var x = cc_optimized(100, 1, 5, 10, 25);
say "Ways to change $1 with common coins: #{x}";
var y = cc_optimized(1000 * 100, 1, 5, 10, 25, 50, 100);
say "Ways to change $1000 with addition of less common coins: #{y}";
{{out}}
Ways to change $1 with common coins: 242
Ways to change $1000 with addition of less common coins: 13398445413854501
Swift
{{trans|Python}} {{libheader|Attaswift BigInt}}
import BigInt
func countCoins(amountCents cents: Int, coins: [Int]) -> BigInt {
let cycle = coins.filter({ $0 <= cents }).map({ $0 + 1 }).max()! * coins.count
var table = [BigInt](repeating: 0, count: cycle)
for x in 0..<coins.count {
table[x] = 1
}
var pos = coins.count
for s in 1..<cents+1 {
for i in 0..<coins.count {
if i == 0 && pos >= cycle {
pos = 0
}
if coins[i] <= s {
let q = pos - coins[i] * coins.count
table[pos] = q >= 0 ? table[q] : table[q + cycle]
}
if i != 0 {
table[pos] += table[pos - 1]
}
pos += 1
}
}
return table[pos - 1]
}
let usCoins = [100, 50, 25, 10, 5, 1]
let euCoins = [200, 100, 50, 20, 10, 5, 2, 1]
for set in [usCoins, euCoins] {
print(countCoins(amountCents: 100, coins: Array(set.dropFirst(2))))
print(countCoins(amountCents: 100000, coins: set))
print(countCoins(amountCents: 1000000, coins: set))
print(countCoins(amountCents: 10000000, coins: set))
print()
}
{{out}}
242
13398445413854501
1333983445341383545001
133339833445334138335450001
4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001
Tcl
{{trans|Ruby}}
package require Tcl 8.5
proc makeChange {amount coins} {
set table [lrepeat [expr {$amount+1}] [lrepeat [llength $coins] {}]]
lset table 0 [lrepeat [llength $coins] 1]
for {set i 1} {$i <= $amount} {incr i} {
for {set j 0} {$j < [llength $coins]} {incr j} {
set k [expr {$i - [lindex $coins $j]}]
lset table $i $j [expr {
($k < 0 ? 0 : [lindex $table $k $j]) +
($j < 1 ? 0 : [lindex $table $i [expr {$j-1}]])
}]
}
}
return [lindex $table end end]
}
puts [makeChange 100 {1 5 10 25}]
puts [makeChange 100000 {1 5 10 25 50 100}]
# Making change with the EU coin set:
puts [makeChange 100 {1 2 5 10 20 50 100 200}]
puts [makeChange 100000 {1 2 5 10 20 50 100 200}]
Output:
242
13398445413854501
4563
10056050940818192726001
uBasic/4tH
{{trans|Run BASIC}}
{{out}}
```txt
0 pennies 0 nickels 0 dimes 4 quarters
0 pennies 0 nickels 5 dimes 2 quarters
0 pennies 0 nickels 10 dimes 0 quarters
...
90 pennies 2 nickels 0 dimes 0 quarters
95 pennies 1 nickels 0 dimes 0 quarters
100 pennies 0 nickels 0 dimes 0 quarters
242 ways to make a buck
0 OK, 0:312
UNIX Shell
{{trans|Common Lisp}} {{works with|bash}}
function count_change {
local -i amount=$1 coin j
local ways=(1)
shift
for coin; do
for (( j=coin; j <= amount; j++ )); do
let ways[j]=${ways[j]:-0}+${ways[j-coin]:-0}
done
done
echo "${ways[amount]}"
}
count_change 100 25 10 5 1
count_change 100000 100 50 25 10 5 1
{{works with|ksh|93}}
function count_change {
typeset -i amount=$1 coin j
typeset ways
set -A ways 1
shift
for coin; do
for (( j=coin; j <= amount; j++ )); do
let ways[j]=${ways[j]:-0}+${ways[j-coin]:-0}
done
done
echo "${ways[amount]}"
}
count_change 100 25 10 5 1
count_change 100000 100 50 25 10 5 1
{{works with|ksh|88}}
function count_change {
typeset -i amount=$1 coin j
typeset ways
set -A ways 1
shift
for coin; do
let j=coin
while (( j <= amount )); do
let ways[j]=${ways[j]:-0}+${ways[j-coin]:-0}
let j+=1
done
done
echo "${ways[amount]}"
}
count_change 100 25 10 5 1
# (optional task exceeds a subscript limit in ksh88)
And just for fun, here's one that works even with the original V7 shell:
{{works with|sh|v7}}
if [ $# -lt 2 ]; then
set ${1-100} 25 10 5 1
fi
amount=$1
shift
ways_0=1
for coin in "$@"; do
j=$coin
while [ $j -le $amount ]; do
d=`expr $j - $coin`
eval "ways_$j=\`expr \${ways_$j-0} + \${ways_$d-0}\`"
j=`expr $j + 1`
done
done
eval "echo \$ways_$amount"
{{Out}}
242
13398445413854501
VBA
{{trans|Phix}}
Private Function coin_count(coins As Variant, amount As Long) As Variant 'return type will be Decimal
'sequence s = Repeat(0, amount + 1)
Dim s As Variant
ReDim s(amount + 1)
Dim c As Integer
s(1) = CDec(1)
For c = 1 To UBound(coins)
For n = coins(c) To amount
s(n + 1) = CDec(s(n + 1) + s(n - coins(c) + 1))
Next n
Next c
coin_count = s(amount + 1)
End Function
Public Sub main2()
Dim us_commons_coins As Variant
'The next line creates a base 1 array
us_common_coins = [{25, 10, 5, 1}]
Debug.Print coin_count(us_common_coins, 100)
Dim us_coins As Variant
us_coins = [{100,50,25, 10, 5, 1}]
Debug.Print coin_count(us_coins, 100000)
End Sub
{{out}}
242
13398445413854501
VBScript
{{trans|C#}}
Function count(coins,m,n)
ReDim table(n+1)
table(0) = 1
i = 0
Do While i < m
j = coins(i)
Do While j <= n
table(j) = table(j) + table(j - coins(i))
j = j + 1
Loop
i = i + 1
Loop
count = table(n)
End Function
'testing
arr = Array(1,5,10,25)
m = UBound(arr) + 1
n = 100
WScript.StdOut.WriteLine count(arr,m,n)
{{Out}}
242
Visual Basic
{{trans|VBA}} {{works with|Visual Basic|6}}
Option Explicit
'----------------------------------------------------------------------
Private Function coin_count(coins As Variant, amount As Long) As Variant
'return type will be Decimal
Dim s() As Variant
Dim n As Long, c As Long
ReDim s(amount + 1)
s(1) = CDec(1)
For c = LBound(coins) To UBound(coins)
For n = coins(c) To amount
s(n + 1) = CDec(s(n + 1) + s(n - coins(c) + 1))
Next n
Next c
coin_count = s(amount + 1)
End Function
'----------------------------------------------------------------------
Sub Main()
Dim us_common_coins As Variant
Dim us_coins As Variant
'The next line creates 0-based array
us_common_coins = Array(25, 10, 5, 1)
Debug.Print coin_count(us_common_coins, 100)
us_coins = Array(100, 50, 25, 10, 5, 1)
Debug.Print coin_count(us_coins, 100000)
End Sub
{{out}}
242
13398445413854501
zkl
{{trans|Scheme}}
fcn ways_to_make_change(x, coins=T(25,10,5,1)){
if(not coins) return(0);
if(x<0) return(0);
if(x==0) return(1);
ways_to_make_change(x, coins[1,*]) + ways_to_make_change(x - coins[0], coins)
}
ways_to_make_change(100).println();
{{out}}
242
Blows the stack on the optional part, so try this:
{{trans|Ruby}}
fcn make_change2(amount, coins){
n, m := amount, coins.len();
table := (0).pump(n+1,List, (0).pump(m,List().write,1).copy);
foreach i,j in ([1..n],[0..m-1]){
table[i][j] = (if(i<coins[j]) 0 else table[i-coins[j]][j]) +
(if(j<1) 0 else table[i][j-1])
}
table[-1][-1]
}
println(make_change2( 100, T(1,5,10,25)));
make_change2(0d1000_00, T(1,5,10,25,50,100)) : "%,d".fmt(_).println();
{{out}}
242
13,398,445,413,854,501
ZX Spectrum Basic
{{trans|AWK}} Test with emulator at full speed for reasonable performance.
10 LET amount=100
20 GO SUB 1000
30 STOP
1000 LET nPennies=amount
1010 LET nNickles=INT (amount/5)
1020 LET nDimes=INT (amount/10)
1030 LET nQuarters=INT (amount/25)
1040 LET count=0
1050 FOR p=0 TO nPennies
1060 FOR n=0 TO nNickles
1070 FOR d=0 TO nDimes
1080 FOR q=0 TO nQuarters
1090 LET s=p+n*5+d*10+q*25
1100 IF s=100 THEN LET count=count+1
1110 NEXT q
1120 NEXT d
1130 NEXT n
1140 NEXT p
1150 PRINT count
1160 RETURN