【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+15;
LL fact[maxn];
LL P[15], a[15];
LL n, m;
int k;

void init_fact(LL mod) {
    fact[0] = 1;
    rep(i, 1, maxn) fact[i] = fact[i-1] * i % mod;
}

LL Pow(LL base, LL n, LL mod) {
    LL ret = 1;
    
    while (n) {
        if (n & 1)
            ret = ret * base % mod;
        base = base * base % mod;
        n >>= 1;
    }
    
    return ret;
}

inline LL Inv(LL a, LL mod) {
    return Pow(a, mod-2, mod);
}

LL C(LL n, LL m, LL mod) {
    if (n < m)    return 0;
    #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 == 0)    return 1;
    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 = 1;
        y = 0;
    } else {
        egcd(b, a%b, d, y, x);
        y -= a/b * x;
    }
}

LL Mul(LL base, LL n, LL mod) {
    LL ret = 0;
    
    while (n) {
        if (n & 1)
            ret = (ret + base) % mod;
        base = (base + base) % mod;
        n >>= 1;
    }
    
    return ret;
}

LL china(int n, LL *a, LL *m) {
    LL M = 1, w, d, x = 0, y;
    LL tmp;
    bool sign;
    
    rep(i, 0, n) M *= m[i];
    rep(i, 0, n) {
        w = M/m[i];
        egcd(m[i], w, d, d, y);
        sign = y < 0;
        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, 0, k) {
        init_fact(P[i]);
        a[i] = Lucas(n, m, P[i]);
    }
    
    LL ans = china(k, a, P);
    printf("%I64d
", ans);
}

int main() {
    cin.tie(0);
    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, 0, k)
            scanf("%I64d", &P[i]);
        solve();
    }
    
    #ifndef ONLINE_JUDGE
        printf("time = %ldms.
", clock());
    #endif
    
    return 0;
}