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

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int maxn = 101010;
 5 LL F[maxn] = {1},a[maxn],m[maxn],N,M,n;
 6 void init(LL mod){
 7     for(int i = 1; i < maxn; ++i)
 8         F[i] = F[i-1]*i%mod;
 9 }
10 LL quickPow(LL base,LL index,LL mod){
11     LL ret = 1;
12     base %= mod;
13     while(index){
14         if(index&1) ret = ret*base%mod;
15         index >>= 1;
16         base = base*base%mod;
17     }
18     return ret;
19 }
20 LL Inv2(LL b,LL mod){
21     return quickPow(b,mod-2,mod);
22 }
23 LL Lucas(LL n,LL m,LL mod){
24     LL ret = 1;
25     while(n && m){
26         LL a = n%mod;
27         LL b = m%mod;
28         if(a < b) return 0;
29         ret = ret*F[a]%mod*Inv2(F[b]*F[a-b]%mod,mod)%mod;
30         n /= mod;
31         m /= mod;
32     }
33     return ret;
34 }
35 LL mul(LL a,LL b,LL mod){
36     if(!a) return 0;
37     return ((a&1)*b%mod + (mul(a>>1,b,mod)<<1)%mod)%mod;
38 }
39 LL CRT(LL a[],LL m[],LL n){
40     LL M = 1,ret = 0;
41     for(int i = 0; i < n; ++i) M *= m[i];
42     for(int i = 0; i < n; ++i){
43         LL x,y,tm = M/m[i];
44         x = Inv2(tm,m[i]);
45         ret = (ret + mul(mul(tm,x,M),a[i],M))%M;
46     }
47     return ret;
48 }
49 int main(){
50     int kase;
51     scanf("%d",&kase);
52     while(kase--){
53         scanf("%I64d%I64d%I64d",&N,&M,&n);
54         for(int i = 0; i < n; ++i){
55             scanf("%I64d",m + i);
56             init(m[i]);
57             a[i] = Lucas(N,M,m[i]);
58         }
59         printf("%I64d\n",CRT(a,m,n));
60     }
61     return 0;
62 }