Lucas定理+中国剩余定理 hdu5446 Unknown Treasure

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题意：求C(n,m)%p,其中n<=1e18,m<=1e18,p1*p2*....*pn<=1e18,pi<1e5

思路：这题出的非常好，也揭示了如何处理p不是质数的方案。可以用lucas求C(n,m) n<=1e18,m<=1e18,p<=1e5，然后得到n个模方程，再通过中国剩余定理合并模方程就行了，但是要注意一个地方，就是在使用中国剩余定理的时候，最后那个M可能会很大，所以乘法的时候可能会爆LL，要用快速乘去处理

#include<map>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck printf("fuck")
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;

const int MX = 10 + 5;

LL multi(LL a, LL b, LL p) {
    LL ret = 0;
    while(b) {
        if(b & 1) ret = (ret + a) % p;
        a = (a + a) % p;
        b >>= 1;
    }
    return ret;
}

LL power(LL a, LL b, LL p) {
    LL res = 1;
    while(b != 0) {
        if(b & 1) res = (res * a) % p;
        a = (a * a) % p;
        b >>= 1;
    }
    return res;
}

LL Comb(LL a, LL b, LL p) {
    if(a < b)   return 0;
    if(a == b)  return 1;
    if(b > a - b)   b = a - b;

    LL ans = 1, ca = 1, cb = 1;
    for(LL i = 0; i < b; ++i) {
        ca = (ca * (a - i)) % p;
        cb = (cb * (b - i)) % p;
    }
    ans = (ca * power(cb, p - 2, p)) % p;
    return ans;
}

LL Lucas(LL n, LL m, LL p) {
    LL ans = 1;

    while(n && m && ans) {
        ans = (ans * Comb(n % p, m % p, p)) % p;
        n /= p;
        m /= p;
    }
    return ans;
}

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

/*x % m = a*/
LL china(int n, int *m, int *a) {
    LL M = 1, d, y, x = 0;
    for(int i = 0; i < n; i++) M *= m[i];
    for(int i = 0; i < n; i++) {
        LL w = M / m[i];
        d = exgcd(m[i], w, d, y);
        x = (x + multi(multi(y, w, M), a[i], M)) % M;
    }
    return (x + M) % M;
}

int A[MX], B[MX];

int main() {
    int T; //FIN;
    scanf("%d", &T);
    while(T--) {
        LL m, n, k;
        scanf("%I64d%I64d%I64d", &n, &m, &k);
        for(int i = 0; i < k; i++) {
            scanf("%d", &A[i]);
            B[i] = Lucas(n, m, A[i]);
        }
        printf("%I64d\n", china(k, A, B));
    }
    return 0;
}