hdu5446 盧卡斯＋中國剩餘定理

一道題集合了盧卡斯，中國剩餘定理，費馬小定理，擴展歐幾里德  也是挺厲害的。

盧卡斯求得摸線性方程組，在用中國剩餘定理解，會爆long long，用擴展加法

#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <cmath>
#define N 100005
#define P 105

using namespace std;

long long pow_mod(long long a, long long b, long long p) {
    long long re = 1;
    while( b ) {
        if( b & 1) {
            re = (a * re) % p;
        }
        a *= a;
        a %= p;
        b >>= 1;
    }
    return re;
}

long long fac[N];
void f(long long p){
    fac[1] = 1;
    fac[0] = 1;
    for( long long i=2; i<N; ++i) {
        fac[i] = (fac[i-1] * i) % p;
    }
}
long long lukas(long long a, long long b, long long p) {
    long long re = 1;
    while( a && b) {
        long long aa = a % p;
        long long bb = b % p;
        if( aa < bb ){
            return 0;
        }
        re = ((re * fac[aa]) % p) * pow_mod(fac[bb]*fac[aa-bb]%p, p-2, p) % p;
        a /= p;
        b /= p;
    }
    return re;
}

long long ext_gcd(long long a, long long b, long long &x, long long &y) {
    if(b==0) {
        x = 1;
        y = 0;
        return a;
    }
    long long r = ext_gcd(b, a%b, x, y);
    long long t = x;
    x = y;
    y = t - a/b*y;
    return r;
}

long long mul_mod(long long a, long long b, long long p) {
    long long re = 0;
    while(b) {
        if(b & 1){
            re = (re + a) % p;
        }
        a += a;
        a %= p;
        b >>= 1;
    }
    return re;
}

int main(int argc, char* argv[]){
    int t;
    long long m, n;
    long long prime[P];
    int p;
    scanf("%d", &t);
    while(t-- && cin >> m >> n >> p) {
        long long pro = 1;
        for(int i=0; i<p; ++i) {
            cin >> prime[i];
            pro *= prime[i];
        }
        long long re = 0;
        for(int i=0; i<p; ++i) {
            f(prime[i]);
            long long luka = lukas(m, n, prime[i]);
            long long x, y;
            ext_gcd(pro/prime[i], prime[i], x, y);
            x = (x + prime[i]) % prime[i];
            re += mul_mod(mul_mod(luka, x, pro), pro / prime[i], pro) % pro;
            re %= pro;
        }
        cout << re << endl;
    }
    return 0;
}