【HDOJ】5446 Unknown Treasure

1. 题目描述
题目很简单，就是求$C(n,m) % M$。

2. 基本思路
这是一道应用了众多初等数论定理的题目，因为数据范围较大因此使用Lucas求$C(n,m) % P$。
而M较大，因此通过$a[i] = C(n,m)%P_i$再综合中国剩余定理可解。由于数据可能为$10^{18}$，再进行乘法可能超long long。
因此，还需要模拟乘法。

3. 代码

 /* 5446 */
 #include <iostream>
 #include <sstream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <bitset>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 typedef long long LL;
 const int maxn = 1e5+;
 LL fact[maxn];
 LL P[], a[];
 LL n, m;
 int k;

 void init_fact(LL mod) {
     fact[] = ;
     rep(i, , maxn) fact[i] = fact[i-] * i % mod;
 }

 LL Pow(LL base, LL n, LL mod) {
     LL ret = ;

     while (n) {
         if (n & )
             ret = ret * base % mod;
         base = base * base % mod;
         n >>= ;
     }

     return ret;
 }

 inline LL Inv(LL a, LL mod) {
     return Pow(a, mod-, mod);
 }

 LL C(LL n, LL m, LL mod) {
     if (n < m)    return ;
     #ifndef ONLINE_JUDGE
     assert(n >= m);
     #endif
     return fact[n] * Inv(fact[n-m]*fact[m]%mod, mod) % mod;
 }

 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;
 }

 void egcd(LL a, LL b, LL& d, LL& x, LL& y) {
     if (!b) {
         d = a;
         x = ;
         y = ;
     } else {
         egcd(b, a%b, d, y, x);
         y -= a/b * x;
     }
 }

 LL Mul(LL base, LL n, LL mod) {
     LL ret = ;

     while (n) {
         if (n & )
             ret = (ret + base) % mod;
         base = (base + base) % mod;
         n >>= ;
     }

     return ret;
 }

 LL china(int n, LL *a, LL *m) {
     LL M = , w, d, x = , y;
     LL tmp;
     bool sign;

     rep(i, , n) M *= m[i];
     rep(i, , n) {
         w = M/m[i];
         egcd(m[i], w, d, d, y);
         sign = y < ;
         tmp = Mul(w, abs(y), M);
         tmp = Mul(tmp, a[i], M);
         if (sign) tmp = -tmp;
         x = (x + tmp) % M;
     }

     return (x + M) % M;
 }

 void solve() {
     rep(i, , k) {
         init_fact(P[i]);
         a[i] = Lucas(n, m, P[i]);
     }

     LL ans = china(k, a, P);
     printf("%I64d\n", ans);
 }

 int main() {
     cin.tie();
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     int t;

     scanf("%d", &t);
     while (t--) {
         scanf("%I64d%I64d%d", &n,&m,&k);
         rep(i, , k)
             scanf("%I64d", &P[i]);
         solve();
     }

     #ifndef ONLINE_JUDGE
         printf("time = %ldms.\n", clock());
     #endif

     return ;
 }