HDU 3037 saving beans (direct application of the Lucas theorem)

Solution:

Evaluate C (n + M, m) % P directly. Because n, m, and P are both very large, we need to use the Lucas theorem to solve the modulo problem of the number of large groups.

#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 (direct application of the Lucas theorem)