1 条题解
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0
#include <bits/stdc++.h> using namespace std; typedef unsigned long long LL; int n, m, a[110], b[110]; LL gcd(LL x, LL y) { if (y == 0) return x; else return gcd(y, x % y); } LL cal(LL n, LL m) { m = min(m, n - m); LL s = 1; for (int i = 1; i <= m; i++) a[i] = n - i + 1, b[i] = i; for (int i = 1; i <= m; i++) { for (int j = 1; j <= m; j++) { LL d = gcd(a[i], b[j]); a[i] /= d; b[j] /= d; } s *= a[i]; } return s; } int main() { int n; scanf("%d", &n); LL ans = cal(2 * n, n) / (n + 1); printf("%lu", ans); return 0; }
- 1
信息
- ID
- 401
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- 8
- 标签
- 递交数
- 372
- 已通过
- 66
- 上传者