HDU 5446 Unknown Treasure

Unknown Treasure

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 721    Accepted Submission(s): 251

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 $T(T\leq 20)$ representing the number of test cases.

Each test case starts with three integers $n,m,k(1\leq m\leq n\leq 10^{18},1\leq k\leq 10)$ 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 $M=p_1⋅p_2\cdots p_k\leq 10^{18}\, and\, p_i\leq 10^5 for\, every\, i\in\{1,\dots,k\}.$

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

解题：中国剩余定理+Lucas定理

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int maxn = ;
 LL F[maxn] = {},a[maxn],m[maxn],N,M,n;
 void init(LL mod){
     for(int i = ; i < maxn; ++i)
         F[i] = F[i-]*i%mod;
 }
 LL quickPow(LL base,LL index,LL mod){
     LL ret = ;
     base %= mod;
     while(index){
         if(index&) ret = ret*base%mod;
         index >>= ;
         base = base*base%mod;
     }
     return ret;
 }
 LL Inv2(LL b,LL mod){
     return quickPow(b,mod-,mod);
 }
 LL Lucas(LL n,LL m,LL mod){
     LL ret = ;
     while(n && m){
         LL a = n%mod;
         LL b = m%mod;
         if(a < b) return ;
         ret = ret*F[a]%mod*Inv2(F[b]*F[a-b]%mod,mod)%mod;
         n /= mod;
         m /= mod;
     }
     return ret;
 }
 LL mul(LL a,LL b,LL mod){
     if(!a) return ;
     return ((a&)*b%mod + (mul(a>>,b,mod)<<)%mod)%mod;
 }
 LL CRT(LL a[],LL m[],LL n){
     LL M = ,ret = ;
     for(int i = ; i < n; ++i) M *= m[i];
     for(int i = ; i < n; ++i){
         LL x,y,tm = M/m[i];
         x = Inv2(tm,m[i]);
         ret = (ret + mul(mul(tm,x,M),a[i],M))%M;
     }
     return ret;
 }
 int main(){
     int kase;
     scanf("%d",&kase);
     while(kase--){
         scanf("%I64d%I64d%I64d",&N,&M,&n);
         for(int i = ; i < n; ++i){
             scanf("%I64d",m + i);
             init(m[i]);
             a[i] = Lucas(N,M,m[i]);
         }
         printf("%I64d\n",CRT(a,m,n));
     }
     return ;
 }