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题目链接 : http://acm.hdu.edu.cn/showproblem.php?pid=4828
Catalan数的公式为 C[n+1] = C[n] * (4 * n + 2) / (n + 2)
题目要求对M = 1e9+7 取模
利用乘法逆元将原式中除以(n+2)取模变为对(n+2)逆元的乘法取模
C[n+1] = C[n] * (4 * n + 2) * Pow(n+2, MOD-2) % MOD
其中Pow用快速幂解决
#include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; typedef long long LL; const int MAXN = 1e6+10; const int MOD = 1e9+7; LL C[MAXN]; LL QuickPow(LL x, LL n) { LL ans = 1; while(n) { if(n & 1) ans = (ans * x) % MOD; x = (x * x) % MOD; n /= 2; } return ans; } void Pre() { C[1] = 1; for(int i = 2; i <= MAXN; i++) { C[i] = C[i-1] * (4 * i - 2) % MOD * QuickPow(i + 1, MOD-2) % MOD; } } int main() { Pre(); int t; int n; scanf("%d", &t); for(int cas = 1; cas <= t; cas++) { scanf("%d", &n); printf("Case #%d:\n%I64d\n", cas, C[n]); } return 0; }
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原文地址:http://www.cnblogs.com/Quinte/p/4969874.html