CF 553A 组合DP

http://codeforces.com/problemset/problem/553/A

A. Kyoya and Colored Balls

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

/**
CF 553A  组合DP
题目大意：在一个口袋里有n种颜色的小球，每个小球有ai个，现在从口袋里依次取出小球，要求第i种颜色的最后一个球要在第i+1种颜色的最后一个小球的前面被拿出
          问有多少种取法？
解题思路：考虑第i种球最后一个球取出来之前前i-1种球必须已经安排好了，那么对于第i种球的ai-1个球只需插入sum[i]之间即可，故dp[i]=dp[i-1]*c[sum[i]-1][a[i01]];
*/
#include 
#include 
#include 
#include 
using namespace std;
typedef long long LL;
const LL mod=1e9+7;
const int maxn=1200;
LL c[maxn][maxn],dp[maxn],a[maxn];
int n;
void init()
{
    c[0][0]=1;
    for(int i=1;i<1110;i++)
    {
        c[i][0]=c[i][1]=1;
        for(int j=1;j<=i;j++)
        {
            c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod;
        }
    }
    /*for(int i=1;i<=10;i++)
    {
        for(int j=0;j<=i;j++)
        {
            printf("%d ",c[i][j]);
        }
        printf("\n");
    }*/
}
int main()
{
    init();
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
        }
        int sum=0;
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for(int i=1;i<=n;i++)
        {
            sum+=a[i];
            dp[i]=(dp[i-1]*c[sum-1][a[i]-1])%mod;
        }
        printf("%lld\n",dp[n]);
    }
    return 0;
}