1 solutions
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0
C :
#include<stdio.h> int f(int n) { int i, m = 1; for(i = 1; i <= n; i++) { m *= i; } return m; } int main() { int m, n; scanf("%d%d", &m, &n); int i; i = f(n)/(f(m)*f(n-m)); printf("%d\n", i); return 0; }
C++ :
#include<stdio.h> int f(int n) { int i, m = 1; for(i = 1; i <= n; i++) { m *= i; } return m; } int main() { int m, n; scanf("%d%d", &m, &n); int i; i = f(n)/(f(m)*f(n-m)); printf("%d\n", i); return 0; }
- 1
Information
- ID
- 486
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By