Codeforces A. Kyoya and Colored Balls（分步组合）

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题目描述：

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 i before 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.

Examples

Input

Copy

3
2
2
1

Output

Copy

3

Input

Copy

4
1
2
3
4

Output

Copy

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

思路：

题目是说给一组有颜色的球，从袋子中去出球要求第i种颜色的求必须在第i+1种颜色的求取完之前取完，问这种取球方法有多少种。大致可以看出这是一道排列组合题，而且方案会很多（因为要取模）。一开始想的是整体怎么放，就是说我一下子就要先扣下每种颜色的一个球，固定住他们的顺序，然后在看其他的球的放法。但情况实际上十分复杂。然后想的是这是一种有重复元素的定序排列问题，但直接套公式好像又不可行。应该要分步考虑而不是全局考虑。考虑最后一个位子，肯定放最后一种颜色的球，之前的位置有\(sum-1\)个，剩余的最后颜色球放在这些位子上有\(C_{sum-1}^{a[last]-1}\)种放法（同种颜色的球无差别）。然后考虑倒数第二种颜色的最后一个球，这是忽略掉前面放好的球，只看空位，最后一个空位放一个球，其它空位放剩余倒数第二种颜色的球，有\(C_{sum-a[last]-1}^{a[last-1]-1}\)种放法。以此类推直到第一种颜色的球。

#include <iostream>
#define max_n 1005
#define mod 1000000007
using namespace std;
int n;
long long a[max_n];
long long ans = 1;
long long sum = 0;
long long q_mod(long long a,long long b)
{
    long long res = 1;
    while(b)
    {
        if(b&1)
        {
            res = ((res%mod)*a)%mod;
        }
        a = (a*a)%mod;
        b >>= 1;
    }
    return res;
}
long long fac[max_n];
void ini()
{
    fac[0] = 1;
    for(int i = 1;i<max_n;i++)
    {
        fac[i] = ((fac[i-1]%mod)*i)%mod;
    }
}
long long inv(long long a)
{
    return q_mod(a,mod-2);
}
long long comb(int n,int k)
{
    if(k>n) return 0;
    return (fac[n]*inv(fac[k])%mod*inv(fac[n-k])%mod)%mod;
}
int main()
{
    ini();
    //cout << comb(3,1) << endl;
    cin >> n;
    for(int i = 0;i<n;i++)
    {
        cin >> a[i];
        sum += a[i];
    }
    for(int i = n-1;i>=0;i--)
    {
        ans = (ans%mod*(comb(sum-1,a[i]-1)%mod))%mod;
        sum -= a[i];
    }
    cout << ans << endl;
    return 0;
}