HDU 5446 Unknown Treasure Lucas+中国剩余定理

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5446

Unknown Treasure

#### 问题描述
> On the way to the next secret treasure hiding place, the mathematician discovered a cave unknown to the map. The mathematician entered the cave because it is there. Somewhere deep in the cave, she found a treasure chest with a combination lock and some numbers on it. After quite a research, the mathematician found out that the correct combination to the lock would be obtained by calculating how many ways are there to pick m different apples among n of them and modulo it with M. M is the product of several different primes.
#### 输入
> On the first line there is an integer T(T≤20) representing the number of test cases.
>
> Each test case starts with three integers n,m,k(1≤m≤n≤1018,1≤k≤10) on a line where k is the number of primes. Following on the next line are k different primes p1,...,pk. It is guaranteed that M=p1⋅p2⋅⋅⋅pk≤1018 and pi≤105 for every i∈{1,...,k}.
#### 输出
> For each test case output the correct combination on a line.
#### 样例
> **sample input**
> 1
> 9 5 2
> 3 5
>
> **sample output**
> 6

题意

求C[n][m]%(P1 * P2 * P3 * ... * pk)

题解

由于n,m都特别大，所以我们用卢卡斯定理对C[n][m]进行pi进制的拆项得到结果ai，用卢卡斯定理的时候p不能太大，否则就没有意义了，所以我们不能直接用M=P1 * P2 * P3 * ... * pk（而且这个不是质数！！！）进行拆项。

然后对所有的ai用中国剩余定理求出C[n][m]%(P1 * P2 * P3 * ... * pk)。

代码

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

typedef long long LL;

LL pi[22], a[22];
LL mul(LL a, LL n, LL mod) {
	LL ret = 0;
	LL t1 = a, t2 = n;
	while (n) {
		//puts("mul");
		if (n & 1) ret = (ret + a) % mod;
		a = (a + a) % mod;
		n >>= 1;
	}
	return ret;
}

void gcd(LL a, LL b, LL& d, LL& x, LL& y) {
	if (!b) { d = a; x = 1; y = 0; }
	else { gcd(b, a%b, d, y, x); y -= x*(a / b); }
}

LL inv(LL a, LL mod) {
	LL d, x, y;
	gcd(a, mod, d, x, y);
	return d == 1 ? (x + mod) % mod : -1;
}

LL get_C(LL n, LL m, LL mod) {
	if (n < m) return 0;
	LL ret = 1;
	for (int i = 0; i < m; i++) ret = ret*(n - i) % mod;
	LL fac_m = 1;
	for (int i = 1; i <= m; i++) fac_m = fac_m*i%mod;
	return ret*inv(fac_m, mod) % mod;
}

LL lucas(LL n, LL m, LL mod) {
	if (m == 0) return 1LL;
	return get_C(n%mod, m%mod, mod)*lucas(n / mod, m / mod, mod) % mod;
}

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

int main() {
	int tc;
	scanf("%d", &tc);
	while (tc--) {
		LL n, m; int k;
		scanf("%lld%lld%d", &n, &m, &k);
		for (int i = 0; i < k; i++) {
			scanf("%lld", &pi[i]);
			a[i] = lucas(n, m, pi[i]);
		}
		LL ans = china(k);
		printf("%lld\n", ans);
	}
	return 0;
}