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解题思路:
直接求C(n+m , m) % p , 由于n , m ,p都非常大,所以要用Lucas定理来解决大组合数取模的问题。
#include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <cstdio> #include <stdlib.h> #include <time.h> #include <assert.h> #define LL long long #define FOR(i, x, y) for(int i=x;i<=y;i++) using namespace std; const int maxn = 100000 + 10; LL Pow_mod(LL a, LL b, LL mod) { LL ret = 1; while(b) { if(b & 1) ret = ret * a % mod; a = a * a % mod; b >>= 1; } return ret; } LL fac[maxn]; void init(LL p) { fac[0] = 1; FOR(i, 1, p) fac[i] = (fac[i-1] * i) % p; } LL Lucas(LL n, LL m, LL p) { LL ret = 1; while(n && m) { LL a = n % p, b = m % p; if(a < b) return 0; ret = (ret * fac[a] * Pow_mod(fac[b] * fac[a-b] % p , p - 2 , p)) % p; n /= p, m /= p; } return ret; } int main() { int T; scanf("%d",&T); while(T--) { int n, m, p; scanf("%d%d%d", &n, &m, &p); init(p); printf("%d\n", (int)Lucas(n + m, m, p)); } return 0; }
HDU 3037 Saving Beans(Lucas定理的直接应用)
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原文地址:http://www.cnblogs.com/mengfanrong/p/5207555.html