Hdu 5446 Unknown Treasure (2015 ACM/ICPC Asia Regional Changchun Online Lucas定理 + 中国剩余定理)

题目链接：

　　Hdu 5446 Unknown Treasure

题目描述：

　　就是有n个苹果，要选出来m个，问有多少种选法？还有k个素数，p1,p2,p3,...pk，结果对lcm(p1,p2,p3.....,pk)取余。

解题思路：

　　Lucas + 中国剩余定理，注意的是中国剩余定理的时候有可能会爆long long。然后用一个快速加法就好辣。

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef __int64 LL;
 const int maxn = ;

 LL quick_mul (LL a, LL b, LL mod)
 {
     LL res = ;
     while (b)
     {
         if (b % )
             res = (res * a) % mod;
         a = (a * a) % mod;
         b /= ;
     }
     return res;
 }

 LL quick_add (LL a, LL b, LL mod)
 {
     LL res = ;
     while (b)
     {
         if (b % )
             res =(res + a) % mod;
         a = (a + a) % mod;
         b /= ;
     }
     return res;
 }

 LL C (LL n, LL m, LL mod)
 {
     if (n < m)
         return ;
     LL ans = ;
     for (int i=; i<=m; i++)
     {
         LL a = (n - m + i) % mod;
         LL b = i % mod;
         ans = ans * (a * quick_mul(b, mod - , mod) % mod) % mod;
     }
     return ans;
 }

 LL Extended_Euclid (LL a, LL b, LL &x, LL &y)
 {//a, b只能是正数
     if (b == )
     {
         x = ;
         y = ;
         return a;
     }

     LL r = Extended_Euclid (b, a%b, x, y), t;
     t = x;
     x = y;
     y = t - a / b * y;
     return r;
 }

 LL CRT (LL a[], LL b[], LL n)
 {
     LL M = , ans = ;
     LL Mi, x, y;

     for (int i=; i<n; i++)
         M *= a[i];

     for (int i=; i<n; i++)
     {
         Mi = M / a[i];
         LL d = Extended_Euclid (Mi, a[i], x, y);
         x = (x % a[i] + a[i]) % a[i];
         //注意，这里的x有可能是负数要注意取mod，变成正的

         //或者quick_add 的第二个参数传Mi
         LL res = quick_add (x, Mi, M);
         res = quick_add (res, b[i], M);
         ans = (ans + res) % M;
     }
     return (ans + M) % M;
 }
 LL Lucas (LL n, LL m, LL mod)
 {
     if (m == )
         return ;
     return (C(n%mod, m%mod, mod) * Lucas(n/mod, m/mod, mod)%mod);
 }

 int main ()
 {
     LL t, n, m, k, a[maxn], b[maxn];
     scanf ("%I64d", &t);
     while (t --)
     {
         scanf ("%I64d %I64d %I64d", &n, &m, &k);
         for (int i=; i<k; i++)
         {
             scanf ("%I64d", &a[i]);
             b[i] = Lucas (n, m, a[i]);
         }
         printf ("%I64d\n", CRT (a, b, k));
     }
     return ;
 }