HDU 5446 Unknown Treasure Lucas+中国剩余定理+按位乘

HDU 5446 Unknown Treasure

题意：求C(n, m) %(p[1] * p[2] ··· p[k])     0< n,m < 10¹⁸

思路：这题基本上算是模版题了，Lucas定理求C(n,m),再用中国剩余定理合并模方程，因为LL相乘会越界，所以用到按位乘。

 #include <iostream>
 #include <cstdio>
 #include <fstream>
 #include <algorithm>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <string>
 #include <cstring>
 #include <map>
 #include <stack>
 #include <set>
 #define LL long long
 #define eps 1e-8
 #define INF 0x3f3f3f3f
 #define MAXN 10005
 #define MAXK 15
 using namespace std;
 LL p[MAXK], mod[MAXK];
 LL quick_mod(LL x, LL y, LL mod){
     if (y == ) return ;
     LL res = quick_mod(x, y >> , mod);
     res = res * res % mod;
     if (y & ){
         res = res * x % mod;
     }
     return res;
 }
 LL comb(LL n, LL m, LL p){
     LL res = ;
     for (int i = ; i <= m; i++){
         LL x = (n + i - m) % p;
         LL y = i % p;
         res = res * (x * quick_mod(y, p - , p) % p) % p;
     }
     return res;
 }
 LL lucas(LL n, LL m, LL p){
     if (m == ) return ;
     return lucas(n / p, m / p, p) * comb(n % p, m % p, p) % p;
 }

 LL exgcd(LL  a, LL  b, LL & x, LL & y)
 {
     if (b == ){
         x = ;
         y = ;
         return a;
     }
     else{
         LL temp = exgcd(b, a % b, x, y);
         LL t = y;
         y = x - y * (a / b);
         x = t;
         return temp;
     }
 }
 LL multi(LL a, LL b, LL m){
     LL res = ;
     while (b){
         if (b & ){
             res = (res + a) % m;
         }
         a = (a + a) % m;
         b >>= ;
     }
     return (res + m) % m;
 }
 LL china(LL p[], LL mod[], LL m){
     LL M = ;
     for (int i = ; i <= m; i++){
         M *= p[i];
     }
     LL x, y, res = ;
     LL d;
     for (int i = ; i <= m; i++){
         LL s = M / p[i];
         d = exgcd(p[i], s, d, y);
         res = (res + multi(multi(y, s, M), mod[i], M)) % M;
     }
     return res;
 }
 int main()
 {
 #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin);
     //freopen("out.txt", "w", stdout);
 #endif // OPEN_FILE
     int T;
     scanf("%d", &T);
     LL n, m, k;
     while (T--){
         scanf("%I64d%I64d%I64d", &n, &m, &k);
         for (int i = ; i <= k; i++){
             scanf("%I64d", &p[i]);
             mod[i] = lucas(n, m, p[i]);
         }
         LL ans = china(p, mod, k);
         printf("%I64d\n", ans);
     }
 }