HDU 5446 中国剩余定理+lucas

Unknown Treasure

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2389    Accepted Submission(s): 885

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

1
9 5 2
3 5

 

Sample Output

6

 

Source

2015 ACM/ICPC Asia Regional Changchun Online

 

题意：C(n,m)%p1*p2*p3..pk

题解：中国剩余定理+lucas

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 #define ll __int64
 #define mod 10000000007
 using namespace std;
 ll n,m,k;
 int t;
 ll exm;
 ll f[];
 void init(int p) {                 //f[n] = n!
     f[] = ;
     for (int i=; i<=p; ++i) f[i] = f[i-] * i % p;
 }
 ll mulmod(ll x,ll y,ll m)
 {
     ll ans=;
     while(y)
     {
         if(y%)
         {
             ans+=x;
             ans%=m;
         }
         x+=x;
         x%=m;
         y/=;
     }
     ans=(ans+m)%m;
     return ans;
 }

 void exgcd(ll a, ll b, ll &x, ll &y)
 {
     if(b == )
     {
         x = ;
         y = ;
         return;
     }
     exgcd(b, a % b, x, y);
     ll tmp = x;
     x = y;
     y = tmp - (a / b) * y;
 }

 ll pow_mod(ll a, ll x, int p)   {
     ll ret = ;
     while (x)   {
         if (x & )  ret = ret * a % p;
         a = a * a % p;
         x >>= ;
     }
     return ret;
 }
 ll CRT(ll a[],ll m[],ll n)
 {
     ll M = ;
     ll ans = ;
     for(ll i=; i<n; i++)
         M *= m[i];
     for(ll i=; i<n; i++)
     {
         ll x, y;
         ll Mi = M / m[i];
         exgcd(Mi, m[i], x, y);
         ans = (ans +mulmod( mulmod( x , Mi ,M ), a[i] , M ) ) % M;
     }
     ans=(ans + M )% M;
     return ans;
 }

 ll Lucas(ll n, ll k, ll p) {       //C (n, k) % p
      init(p);
      ll ret = ;
      while (n && k) {
         ll nn = n % p, kk = k % p;
         if (nn < kk) return ;                   //inv (f[kk]) = f[kk] ^ (p - 2) % p
         ret = ret * f[nn] * pow_mod (f[kk] * f[nn-kk] % p, p - , p) % p;
         n /= p, k /= p;
      }
      return ret;
 }
 int main ()
 {
     scanf("%d",&t);
     {
         for(int i=;i<=t;i++)
         {
             ll ee[];
             ll gg[];
             scanf("%I64d %I64d %I64d",&n,&m,&k);
             for(ll j=;j<k;j++)
             {
                 scanf("%I64d",&exm);
                 gg[j]=exm;;
                 ee[j]=Lucas(n,m,exm);
             }
             printf("%I64d\n",CRT(ee,gg,k));
         }
     }
     return ;
 }