HDU 5446 Unknown Treasure ACM/ICPC 2015 Changchun Online（Lucass+CRT）

Unknown Treasure

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2901    Accepted Submission(s): 1059

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 pickm 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

19 5 23 5

Sample Output

6

Source

2015 ACM/ICPC Asia Regional Changchun Online

        可以说是一道裸题，求C(n,m)对p1*p2*……pn取模，这个n、m都可以到1e18级别，然后pi是1e5级别。

        如果说只对一个质数p取模，而且质数不大，那么我们直接可以用Lucass定理拆分求组合数，这个我之前讲过了，我就不多说了。关键是，本题取模是对很多质数的乘积取模，故不能直接用Lucass定理。如果我们设最后答案为X，那么X=C(n,m)%(p1p2……pn)。根据同余的性质，我们可以有X%pi=C(n,m)%pi，这个很容易理解，于是我们X对于所有的pi的方程都写出来。则有：

                                                                      

        发现了什么吗？这个就相当于是一系列的同余方程，然后我们要做的就是解这个同余方程。如此，我们就可以用中国剩余定理CRT了。下面，再大致的讲讲中国剩余定理吧：

                                              

        定理知道之后，我们就只需要对应的求出每一个质数pi对应的用Lucass定理计算之后的组合数的余数，然后把把余数和除数代入，用中国剩余定理（CRT）解方程即可。具体的话要注意中间乘的时候LL*LL，可能会爆，所以可以用按位乘来解决。当然了，HDU比较高级，可以直接用__int128，但是现场可能是没有的，所以我还是写了一遍按位乘。具体见代码：

#include<bits/stdc++.h>#define LL long long#define N 101000using namespace std;LL f[N],a[11],Mod[11],n,m,k;LL multiply(LL a,LL b,LL mod){    LL res=0;    while(b)    {        if(b&1) res=(res+a)%mod;        b>>=1;a=(a+a)%mod;    }    return res;}LL qpow(LL x,int n,int p){    LL ret=1;    while (n)    {        if (n&1) ret=ret*x%p;        x=x*x%p; n>>=1;    }    return ret;}LL ex_gcd(LL a,LL b,LL &x,LL &y){    if(b==0){x=1,y=0;return a;}    else    {        LL r=ex_gcd(b,a%b,y,x);        y-=x*(a/b); return r;    }}void init(int p){    f[0]=1;    for(int i=1;i<=p;i++)        f[i]=f[i-1]*i%p;}LL lucass(LL n,LL m,int p){    init(p);    LL ret=1;    while (n&&m)    {        int a=n%p,b=m%p;        if (a<b) return 0;        ret=ret*f[a]*qpow(f[b]*f[a-b]%p,p-2,p)%p;        n/=p,m/=p;    }    return ret;}LL CRT(LL a[],LL m[],int n){    LL M=1,res=0;    for(int i=1;i<=n;i++) M*=m[i];    for(int i=1;i<=n;i++)    {        LL x,y,Mi=M/m[i];        ex_gcd(m[i],Mi,x,y);//ex_gcd求逆元        res=(res+multiply(multiply(y,Mi,M),a[i],M))%M;    }    if(res<0) res+=M;    return res;}int main(){    int T_T;    cin>>T_T;    while(T_T--)    {        scanf("%I64d%I64d%I64d",&n,&m,&k);        for(int i=1;i<=k;i++)        {            scanf("%I64d",&a[i]);            Mod[i]=lucass(n,m,a[i]);        }        printf("%I64d\n",CRT(Mod,a,k));    }    return 0;}

X=Cmn%(∏pi)