
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。
例子
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}.Example
的中文翻译为:示例
#includeint coins( int S[], int m, int n ) { int i, j, x, y; int table[n+1][m]; for (i=0; i = 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; }
输出
The total number of combinations of coins that sum up to 4 is 4










