LightOJ 1226 - One Unit Machine Lucas/组合数取模

**题意：**按要求完成n个任务，每个任务必须进行a[i]次才算完成，且按要求，第i个任务必须在大于i任务完成之前完成，问有多少种完成顺序的组合。（n

/** @Date    : 2016-11-21-18.26

  * @Author  : Lweleth (SoungEarlf@gmail.com)

  * @Link    : https://github.com/

  * @Version :

  */

#include <stdio.h>

#include <iostream>

#include <string.h>

#include <algorithm>

#include <utility>

#include <vector>

#include <map>

#include <set>

#include <string>

#include <stack>

#include <queue>

//#include<bits/stdc++.h>

#define LL long long

#define MMF(x) memset((x),0,sizeof(x))

#define MMI(x) memset((x), INF, sizeof(x))

using namespace std;



const int INF = 0x3f3f3f3f;

const int N = 1e5+20;

const LL mod = 1000000007;



LL inv[N*10], fa[N*10];



LL fpow(LL a, int n)

{

    LL r = 1;

    while(n > 0)

    {

        if(n & 1)

            r = r * a % mod;

        a = a * a % mod;

        n >>= 1;

    }

    return r;

}

void init()

{

    fa[0] = 1;

    inv[0] = 1;

    for(LL i = 1; i <= 1000100; i++)

    {

        fa[i] = fa[i-1] * i % mod;

        inv[i] = fpow(fa[i], mod - 2);

    }

}



LL C(LL n, LL m)

{

    if(m > n)

        return 0;

    LL ans = 0;

    ans = ((fa[n] * inv[m] % mod)* inv[n-m]) % mod;

    return ans;

}



LL lucas(LL n, LL m)

{

    if(m == 0)

        return 1;

    return C(n % mod, m % mod) * lucas(n / mod, m / mod) % mod;

}

LL a[1100];

int main()

{

    int T;

    cin >> T;

    init();

    int cnt = 0;

    while(T--)

    {

        LL n;

        scanf("%lld", &n);

        for(int i = 1; i <= n; i++)

            scanf("%lld", a + i);

        LL ans = 1;

        LL ct = 0;

        for(int i = 1; i <= n; i++)

        {

            ct += a[i];

            ans = (ans*lucas(ct-1, a[i]-1) % mod);

        }

        printf("Case %d: %lld\n", ++cnt, ans);

    }

    return 0;

}