HDU 5446 中国剩余定理+lucas

Posted 2020-08-21 半根毛线code

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

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdio>
  4 #include<algorithm>
  5 #define ll __int64
  6 #define mod 10000000007
  7 using namespace std;
  8 ll n,m,k;
  9 int t;
 10 ll exm;
 11 ll f[1000010];
 12 void init(int p) {                 //f[n] = n!
 13     f[0] = 1;
 14     for (int i=1; i<=p; ++i) f[i] = f[i-1] * i % p;
 15 }
 16 ll mulmod(ll x,ll y,ll m)
 17 {
 18     ll ans=0;
 19     while(y)
 20     {
 21         if(y%2)
 22         {
 23             ans+=x;
 24             ans%=m;
 25         }
 26         x+=x;
 27         x%=m;
 28         y/=2;
 29     }
 30     ans=(ans+m)%m;
 31     return ans;
 32 }
 33 
 34 void exgcd(ll a, ll b, ll &x, ll &y)
 35 {
 36     if(b == 0)
 37     {
 38         x = 1;
 39         y = 0;
 40         return;
 41     }
 42     exgcd(b, a % b, x, y);
 43     ll tmp = x;
 44     x = y;
 45     y = tmp - (a / b) * y;
 46 }
 47 
 48 ll pow_mod(ll a, ll x, int p)   {
 49     ll ret = 1;
 50     while (x)   {
 51         if (x & 1)  ret = ret * a % p;
 52         a = a * a % p;
 53         x >>= 1;
 54     }
 55     return ret;
 56 }
 57 ll CRT(ll a[],ll m[],ll n)
 58 {
 59     ll M = 1;
 60     ll ans = 0;
 61     for(ll i=0; i<n; i++)
 62         M *= m[i];
 63     for(ll i=0; i<n; i++)
 64     {
 65         ll x, y;
 66         ll Mi = M / m[i];
 67         exgcd(Mi, m[i], x, y);
 68         ans = (ans +mulmod( mulmod( x , Mi ,M ), a[i] , M ) ) % M;
 69     }
 70     ans=(ans + M )% M;
 71     return ans;
 72 }
 73 
 74 ll Lucas(ll n, ll k, ll p) {       //C (n, k) % p
 75      init(p);
 76      ll ret = 1;
 77      while (n && k) {
 78         ll nn = n % p, kk = k % p;
 79         if (nn < kk) return 0;                   //inv (f[kk]) = f[kk] ^ (p - 2) % p
 80         ret = ret * f[nn] * pow_mod (f[kk] * f[nn-kk] % p, p - 2, p) % p;
 81         n /= p, k /= p;
 82      }
 83      return ret;
 84 }
 85 int main ()
 86 {
 87     scanf("%d",&t);
 88     {
 89         for(int i=1;i<=t;i++)
 90         {
 91             ll ee[15];
 92             ll gg[15];
 93             scanf("%I64d %I64d %I64d",&n,&m,&k);
 94             for(ll j=0;j<k;j++)
 95             {
 96                 scanf("%I64d",&exm);
 97                 gg[j]=exm;;
 98                 ee[j]=Lucas(n,m,exm);
 99             }
100             printf("%I64d\n",CRT(ee,gg,k));
101         }
102     }
103     return 0;
104 }