C. Kyoya and Colored Balls(Codeforces Round #309 (Div. 2))

Kyoya Ootori has a bag with n colored balls that are colored with k different colors. The colors are labeled from 1 to k. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color ibefore drawing the last ball of color i + 1 for all i from 1 to k - 1. Now he wonders how many different ways this can happen.

Input

The first line of input will have one integer k (1 ≤ k ≤ 1000) the number of colors.

Then, k lines will follow. The i-th line will contain c_i, the number of balls of the i-th color (1 ≤ c_i ≤ 1000).

The total number of balls doesn't exceed 1000.

Output

A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1 000 000 007.

Sample test(s)

input
3
2
2
1

output
3

input
4
1
2
3
4

output
1680

Note

In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are:

1 2 1 2 3
1 1 2 2 3
2 1 1 2 3

题意：n种不同颜色的球。有k[n]个，要求每种颜色的球的最后一个的相对初始状态（球的种类）的位置不变；问有多少种组合

思路：定最后一个球，其它的球(同样的颜色)在前面任选,之后再定最后另外一种颜色的球<放在剩下空中离最后一个位置近期的地方>，然后剩下的任选。。

。以此类推。

题目链接：http://codeforces.com/contest/554/problem/C

转载请注明出处：寻找&星空の孩子

用了个费马小定理优化了下。也不可不优化。(a=1 mod (p-1),gcd(a,p)=1)

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define LL long long
const LL mod =  1000000007;
LL n;
LL a[1005];
LL fac[1000005];


LL ppow(LL a,LL b)
{
    LL c=1;
    while(b)
    {
        if(b&1) c=c*a%mod;
        b>>=1;
        a=a*a%mod;
    }
    return c;
}


LL work(LL m,LL i)
{
    return ((fac[m]%mod)*(ppow((fac[i]*fac[m-i])%mod,mod-2)%mod))%mod;
}

int main()
{
    LL i,j,k;
    fac[0] = 1;
    for(i = 1; i<1000005; i++)
        fac[i]=(fac[i-1]*i)%mod;
    LL ans = 1,sum = 0;
    scanf("%I64d",&n);
    for(i = 1; i<=n; i++)
    {
        scanf("%I64d",&a[i]);
        sum+=a[i];
    }
    for(i = n; i>=1; i--)
    {
        ans*=work(sum-1,a[i]-1);
        ans%=mod;
        sum-=a[i];
    }
    printf("%I64d
",ans);

    return 0;
}