• 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
    ", ans);
    	}
    	return 0;
    }
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  • 原文地址:https://www.cnblogs.com/fenice/p/5719912.html
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