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.

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 ci, the number of balls of the i-th color (1 ≤ ci ≤ 1000).

The total number of balls doesn't exceed 1000.

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.

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：
一共有n种颜色的球，同种颜色的球视为完全相同，设置一种放置规则，即上一种颜色球的最后一个必须放在当前颜色球最后一个的前面，问放球的方案数。
题解：
把情况分开考虑，一种颜色一种颜色的往上放，那么dp[i]表示已经放了i种颜色的方案数，a[i]表示第i种颜色的球的个数，sum[i]表示前i种球一共有这么多个，所以当放入i-1种颜色的球后，要把第i种球放一个到最后面以满足题意，然后向前插球，即当前一共有sum[i-1]+a[i]-1个位置，要在这些位置中找a[i]-1个位置放剩下的第i种球，那么这种操作的方法数就是c[sum[i-1+a[i]-1]][a[i]-1]

#include<iostream>

#include<cstdio>

#include<cstring>

#define maxn 1010

#define mod 1000000007

using namespace std;

int n,a[maxn],c[maxn][maxn],dp[maxn],sum[maxn];

int main(){

    scanf("%d",&n);

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

        scanf("%d",&a[i]);

        sum[i]=sum[i-]+a[i];

    }

    c[][]=;

    for(int i=;i<=;i++){

        c[i][]=;

        for(int j=;j<=;j++)

            c[i][j]=(c[i-][j-]+c[i-][j])%mod;

    }

    dp[]=;

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

        dp[i]=1LL*dp[i-]*c[a[i]-+sum[i-]][a[i]-]%mod;

    printf("%d\n",dp[n]);

    return ;

}