在這個問題中,我們給定一個值n,我們想要找零n盧比,並且我們有n個硬幣,每個硬幣的面值從1到m不等。我們需要傳回能夠組成這個總和的方式的總數。
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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