HDU 5446 Unknown Treasure (卢卡斯+CRT

代码：

 #include"bits/stdc++.h"
 #define db double
 #define ll long long
 #define vec vector<ll>
 #define Mt  vector<vec>
 #define ci(x) scanf("%d",&x)
 #define cd(x) scanf("%lf",&x)
 #define cl(x) scanf("%lld",&x)
 #define pi(x) printf("%d\n",x)
 #define pd(x) printf("%f\n",x)
 #define pl(x) printf("%lld\n",x)
 const int N   = 1e5+;
 using namespace std;
 typedef pair<ll, ll> pll;
 ll F[N], Finv[N], inv[N];//F是阶乘，Finv是逆元的阶乘
 ll mul(ll a, ll b, ll mod) {
     ll ret = ;
     while(b) {
         if(b & )    ret = (ret + a) % mod;
         a = (a + a) % mod;
         b >>= ;
     }
     return ret;
 }
 void init(int mod){
     inv[] = ;
     for(int i = ; i < N; i ++){
         inv[i] = (mod - mod / i) * 1ll * inv[mod % i] % mod;
     }
     F[] = Finv[] = ;
     for(int i = ; i < N; i ++){
         F[i] = F[i-] * 1ll * i % mod;
         Finv[i] = Finv[i-] * 1ll* inv[i] % mod;
     }
 }
 ll C(ll n, ll m,int mod){    //C(n, m)就是C(n, m)
     if(m <  || m > n)  return ;
     return F[n] * 1ll * Finv[n - m] % mod * Finv[m] % mod;
 }
 ll Lucas(ll n, ll m, int p){
     return m ? mul(Lucas(n/p, m/p, p),C(n%p, m%p, p),p) : ;
 }
 ll exgcd(ll a, ll b, ll &x, ll &y)//ax+by=gcd(a,b)的整数解
 {
     ll d;
     //if (a == 0 && b == 0) return -1;
     if (b == )
     {
         x = ,y = ;
         return a;
     }
     d = exgcd(b, a%b, y, x);
     y -= a / b * x;
     return d;
 }
 ll INV(ll a,ll p)
 {
     ll d,x,y;
     d = exgcd(a,p,x,y);
     return d==?(x%p+p)%p:-;
 }

 ll china(int n, ll *a, ll *m){
     ll M = , ret = ;
     for(int i = ; i < n; i ++) M *= m[i];
     for(int i = ; i < n; i ++){
         ll w = M / m[i];
         ret = (ret + mul(mul(w, INV(w, m[i]),M),a[i],M)) % M;
     }
     return (ret + M) % M;
 }
 ll n,m;
 int k;
 ll P[N],R[N];
 int main(){
     int T;
     scanf("%d",&T);
     while(T--){
         scanf("%lld%lld%d",&n,&m,&k);
         for(int i=;i<k;i++){
            cl(P[i]);
            init((int)P[i]);
            R[i]=Lucas(n,m,P[i]);
         }
         ll ans=china(k,R,P);
         pl(ans);
     }
     return ;
 }