hdu5446 Unknown Treasure（数论综合题：大组合数取大合数模:Lucas+CRT）

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

Unknown Treasure

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 1590 Accepted Submission(s): 585



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. Mis
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 kdifferent
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

1
9 5 2
3 5

Sample Output

6

Source

2015 ACM/ICPC Asia Regional Changchun Online



题意：求C(n,m)%M,其中M=(m0*m2*…*m(k-1)),mi为素数。

AC code：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<map>
#include<stack>
#include<vector>
#define LL long long
#define MAXN 1000010
using namespace std;
const  int N=20;//模方程数 
LL a
,mod
;
LL mul(LL a,LL b,LL mod)//a*b%mod
{
	LL ans=0;
	while(b){
		if(b&1)
			ans=(ans+a)%mod;
		b>>=1;
		a=(a+a)%mod;
	}
	return ans;
}

LL quick_mod(LL a,LL b,LL m)//a^b%m 
{
	LL ans=1;
	a%=m;
	while(b)
	{
		if(b&1)
		{
			ans=ans*a%m;
		}			
		b>>=1;
		a=a*a%m;
	}
	return ans;
}

LL getC(LL n,LL m,int cur)//C(n,m)%mod[cur]
{
	LL p=mod[cur];
	if(m>n)
		return 0;
	if(m>n-m)
		m=n-m;
	LL ans=1;
	for(int i=1;i<=m;i++)
	{
		LL a=(n+i-m)%p;
		LL b=i%p;
		//ans=mul(ans,mul(a,quick_mod(b,p-2,p),p),p);//p为素数，i对p的逆元可以不用扩张欧几里得进行求解  re=i^(P-2) 
		ans = ans * (a * quick_mod(b, p-2,p) % p) % p;  
	}
	return ans; 
}

LL Lucas(LL n,LL k,int cur)//求C(n,k)%mod[cur] 
{
	LL p=mod[cur];
	if(k==0)
		return 1%p;
	//return getC(n%p,k%p,cur)*Lucas(n/p,k/p,cur)%p;
	return getC(n % p, k % p,cur) * Lucas(n / p, k / p,cur) % p;  
}

void extend_Euclid(LL a,LL b,LL &x,LL &y)
{
	if(b==0)
	{
		x=1;
		y=0;
		return;
	}
	extend_Euclid(b,a%b,x,y);
	LL tmp=x;
	x=y;
	y=tmp-a/b*y;
}

LL CRT(LL a[],LL m[],int k)//求C(n,m)%M,其中M=(m0*m2*…*m(k-1)),mi为素数，则先用a[i]存储模方程C(n,m)%mi,
{                           //m[]存储所有素数因子mi，k表示总共有k个模方程，返回C(n,m)%M的值 
	LL M=1;
	LL ans=0;
	for(int i=0;i<k;i++)
		M*=mod[i];
	for(int i=0;i<k;i++)
	{
		LL x,y,tmp;
		LL Mi=M/m[i];
		extend_Euclid(Mi,m[i],x,y);
		if(x<0)
		{
			x=-x;
			tmp=mul(Mi,x,M);
			tmp=mul(tmp,a[i],M);
			tmp=-tmp;
		}
		else
		{
			tmp=mul(Mi,x,M);
			tmp=mul(tmp,a[i],M);
		}
		ans=(ans+tmp)%M;
	}
	while(ans<0)
		ans+=M;
	return ans;
}

int main()
{
	//freopen("D:\\in.txt","r",stdin);
	int T;
	scanf("%d",&T);
	while(T--)
	{
		LL n,m;
		int k;
		scanf("%lld%lld%d",&n,&m,&k);
		//k=1;
		for(int i=0;i<k;i++)
			scanf("%lld",&mod[i]);
		for(int i=0;i<k;i++)
			a[i]=Lucas(n,m,i)%mod[i];
		//printf("%I64d\n",a[0]);
		printf("%lld\n",CRT(a,mod,k)); 
	}
  	return 0;
}