hdu 5446 Unknown Treasure （Lucas定理+中国剩余定理）

最新推荐文章于 2018-08-21 21:58:53 发布

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最新推荐文章于 2018-08-21 21:58:53 发布

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分类专栏：数论组合数取模

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本文描述了一个数学家在未知洞穴中寻找宝藏的故事，并通过计算特定的组合数来解开宝藏箱上的锁。采用Lucas定理与中国剩余定理解决大规模组合数计算的问题。

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Unknown Treasure

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2461 Accepted Submission(s): 905

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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题解：Lucas定理+中国剩余定理

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 20
#define LL long long
#define pa pair<LL,LL>
using namespace std;
LL n,m,k,p[N],a[N],mod,jc[20][200003];
LL init()
{
	for (int i=1;i<=k;i++) {
		jc[i][0]=jc[i][1]=1;
		for (int j=2;j<=p[i];j++)
		 jc[i][j]=jc[i][j-1]*j%p[i];
	}
}
LL mul(LL a,LL b,LL mod)
{
	LL res=0;
	while (b) {
		if (b&1) res=(res+a)%mod;
		a=(a+a)%mod;
		b>>=1;
	}
	return res;
}
LL quickpow(LL num,LL x,LL p)
{
	LL base=num%p; LL ans=1;
	while (x) {
		if (x&1) ans=(ans*base)%p;
		x>>=1;
		base=(base*base)%p;
	}
	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=y; 
	y=x-(a/b)*y;
	x=t;
}
LL inverse(LL a,LL b)
{
	LL x,y;
	exgcd(a,b,x,y);
	return (x%b+b)%b;
}
/*pa solve(int k,LL n)
{
	if (n==0) return make_pair(0,1);
	LL x=n/p[k],y=n/p[k];
	LL ans=1;
	if (y) {
		for (LL i=2;i<p[k];i++) {
			if (i%p[k]!=0) ans=(ans*(LL)i)%p[k];
		}
		ans=quickpow(ans,y,p[k]);
	}
	for (LL i=y*p[k]+1;i<=n;i++)
	 if (i%p[k]!=0) ans=(ans*(LL)i)%p[k];
	pa t=solve(k,x);
	return make_pair(t.first+x,ans*t.second%p[k]);
}*/
/*LL calc(int k,LL n,LL m)
{
	if (n<m) return 0;
	pa a=solve(k,n),b=solve(k,m),c=solve(k,n-m);
	LL t1=quickpow(p[k],a.first-b.first-c.first,p[k])*a.second%p[k];
	LL t2=inverse(b.second,p[k])%p[k];
	LL t3=inverse(c.second,p[k])%p[k];
	return t1*t2*t3%p[k];
}*/
LL calc(int i,LL n,LL m)
{
	if (m>n) return 0;
	return jc[i][n]*quickpow(jc[i][m]*jc[i][n-m]%p[i],p[i]-2,p[i])%p[i];
}
LL lucas(int i,LL n,LL m)
{
	if (m==0) return 1;
	return calc(i,n%p[i],m%p[i])*lucas(i,n/p[i],m/p[i])%p[i]; 
}
LL china()
{
	LL x,y,ans=0;
    for (int i=1;i<=k;i++){
    	LL r=mod/p[i];
    	//exgcd(r,p[i]);
    	ans=(ans+mul(mul(r,inverse(r,p[i]),mod),a[i],mod))%mod;
	}
	return ans;	
}
int main()
{
	freopen("a.in","r",stdin);
	freopen("my.out","w",stdout);
	int T; 
	scanf("%d",&T);
	while (T--) {
		cin>>n>>m>>k; mod=1;
		for (int i=1;i<=k;i++) 
		 scanf("%I64d",&p[i]),mod*=p[i];
		init();
		for (int i=1;i<=k;i++)
		 a[i]=lucas(i,n,m);
		//for (int i=1;i<=k;i++) cout<<a[i]<<" ";
	//	cout<<endl;
		printf("%I64d\n",china());
	}
}