HDOJ 题目5446 Unknown Treasure（Lucas+费马小定理+CRT）

本题描述了一位数学家在洞穴中发现了一个带有组合锁的宝箱，要求通过计算从n个不同苹果中选取m个的方法数量，并对特定的质数乘积取模来找出宝箱的正确组合。

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Unknown Treasure

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1662 Accepted Submission(s): 617

Problem Description

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.

Input

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} .

Output

For each test case output the correct combination on a line.

Sample Input

Sample Output

Source

2015 ACM/ICPC Asia Regional Changchun Online

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题目大意：给n,m,k,就是求c(n,m)对下边k个素数取余

ac代码

#include<stdio.h>
#include<string.h>
#define LL __int64
LL fac[12][100005],M[12];
LL tp[12];
void get_fact(LL k)
{
	int i,j;
	for(i=0;i<k;i++)
	{
		fac[i][0]=1;
		for(j=1;j<=tp[i];j++)
			fac[i][j]=(fac[i][j-1]*j)%tp[i];
	}
}
LL qpow(LL a,LL b,LL mod)
{
	LL ans=1;
	while(b)
	{
		if(b&1)
			ans=(ans*a)%mod;
		a=(a*a)%mod;
		b>>=1;
	}
	return ans%mod;
}
LL lucas(LL n,LL m,int x)
{
	LL ans=1;
	if(n<m)
		return 0;
	while(n&&m)
	{
		LL a=n%tp[x];
		LL b=m%tp[x];
		if(a<b)
			return 0;
		ans=(ans*fac[x][a]*qpow(fac[x][b]*fac[x][a-b]%tp[x],tp[x]-2,tp[x]))%tp[x];
		n/=tp[x];
		m/=tp[x];
	}
	return ans;
}
void exgcd(LL a,LL b,LL &x,LL &y)
{
	if(b==0)
	{
		x=1;
		y=0;
		return;
	}
	exgcd(b,a%b,x,y);
	LL t=x;
	x=y;
	y=t-a/b*y;
}
LL CRT(int k)
{
	LL a1,b1,a2,b2,a,b,c,x,y;
	a1=tp[0],b1=M[0];
	for(int i=1;i<k;i++)
	{
		a2=tp[i],b2=M[i];
		a=a1;b=a2;c=b2-b1;
		exgcd(a,b,x,y);
		x=((c*x)%b+b)%b;
		b1=b1+a1*x;
		a1=a1*b;
	}
	return b1;
}
int main()
{
	LL n,g;
	int t;
	scanf("%d",&t);
	while(t--)
	{
		LL n,m,k;
		scanf("%I64d%I64d%I64d",&n,&m,&k);
		LL i;
		int j;
		memset(M,0,sizeof(M));
		for(i=0;i<k;i++)
		{
			scanf("%I64d",&tp[i]);
		//	M[j]=(M[j]+lucas(n,m,i))%tp[i];
		}
		get_fact(k);
		for(i=0;i<k;i++)
		{
			M[i]=(M[i]+lucas(n,m,i))%tp[i];
		}
		printf("%I64d\n",CRT(k));
	}
}