hdu 5446 Lucas定理，中國剩餘定理，處理爆long long 的乘法取模運算

Unknown Treasure

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3056    Accepted Submission(s): 1128

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

1 9 5 2 3 5

 

Sample Output

6

 

題意：

問組合數 C(n,m) % M 等於多少

其中 n  m  M 都是 小於等於 10^18 次方的比較大的數字

並且題目給出 M 的唯一分解的所有質因子 p ，求解決上述問題

題解：

首先介紹中國剩餘定理：

逆元詳解：

本題中求逆元 inv數組 還利用了費馬小定理，得到了一個遞推性質，然後快速的求了出來

lucas定理：

C（n , m）= C（n%mod , m%mod）* lucas( n/mod , m/mod )

爆longlong的乘法取模處理：

LL mul(LL a,LL b,LL mod){///二進制，按位相乘
    a=(a%mod+mod)%mod;
    b=(b%mod+mod)%mod;
    LL ans=0;
    while(b){
        if(b&1) ans+=a,ans=(ans>=mod?ans-mod:ans);
        b>>=1; a<<=1;
        a=(a>=mod?a-mod:a);
    }
    return ans;
}

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define maxn 100005
#define LL long long
LL fac[maxn],inv[maxn];

LL powmod(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;
}

void init(int n){
    fac[0]=1;
    for(int i=1;i<n;i++) fac[i]=fac[i-1]*i%n;
    inv[n-1]=powmod(fac[n-1],n-2,n);
    for(int i=n-2;i>=0;i--) inv[i]=inv[i+1]*(i+1)%n;
}

LL lucas(LL n,LL m,LL mod){
    LL ans=1;
    while(n&&m){
        LL a=n%mod,b=m%mod;
        if(a<b) return 0;
        ans=ans*fac[a]%mod*inv[b]%mod*inv[a-b]%mod;
        n/=mod,m/=mod;
    }
    return ans;
}

void ex_gcd(LL a,LL b,LL gcd,LL& x,LL& y){
    if(!b) x=1,y=0,gcd=a;
    else ex_gcd(b,a%b,gcd,y,x),y-=x*(a/b);
}

LL mul(LL a,LL b,LL mod){
    a=(a%mod+mod)%mod;
    b=(b%mod+mod)%mod;
    LL ans=0;
    while(b){
        if(b&1) ans+=a,ans=(ans>=mod?ans-mod:ans);
        b>>=1; a<<=1;
        a=(a>=mod?a-mod:a);
    }
    return ans;
}

LL china(LL n,LL* a,LL* m){
    LL M=1,w,gcd,x,y,ans=0;
    for(int i=0;i<n;i++) M*=m[i];
    for(int i=0;i<n;i++){
        w=M/m[i];
        ex_gcd(m[i],w,gcd,x,y);
        ans=(ans+mul(mul(y,w,M),a[i],M)+M)%M;
    }
    return ans;
}

int main()
{
    LL k,n,m,a[20],p[20],T;
    //freopen("in.txt","r",stdin);
    scanf("%lld",&T);
    while(T--)
    {
        scanf("%lld%lld%lld",&n,&m,&k);
        for(int i=0;i<k;i++){
            scanf("%lld",&p[i]);
            init(p[i]);
            a[i]=lucas(n,m,p[i]);
        }
        printf("%lld\n",china(k,a,p));
    }
    return 0;
}