在这个问题中，我们给定一个值n，我们想要找零n卢比，并且我们有n个硬币，每个硬币的面值从1到m不等。我们需要返回能够组成这个总和的方式的总数。
Input : N = 6 ; coins = {1,2,4}.
Output : 6
Explanation : The total combination that make the sum of 6
is :
{1,1,1,1,1,1} ; {1,1,1,1,2}; {1,1,2,2}; {1,1,4}; {2,2,2} ; {2,4}.#include <stdio.h>
int coins( int S[], int m, int n ) {
int i, j, x, y;
int table[n+1][m];
for (i=0; i<m; i++)
table[0][i] = 1;
for (i = 1; i < n+1; i++) {
for (j = 0; j < m; j++) {
x = (i-S[j] >= 0)? table[i - S[j]][j]: 0;
y = (j >= 1)? table[i][j-1]: 0;
table[i][j] = x + y;
}
}
return table[n][m-1];
}
int main() {
int arr[] = {1, 2, 3};
int m = sizeof(arr)/sizeof(arr[0]);
int n = 4;
printf("The total number of combinations of coins that sum up to %d",n);
printf(" is %d ", coins(arr, m, n));
return 0;
}The total number of combinations of coins that sum up to 4 is 4
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