Remainders Game

Problem:

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

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

Solution:

組合數學。題意是求保證\(i\)種顏色氣球的最後一個之後必須是\(i+1\)種顏色的氣球的情況下的排列數。對於每一種顏色計算當前所剩的總位數\(sum\)，當前顏色的個數\(a\)，\(C[sum-1][a-1]\),將所有情況相乘即可。

預先處理組合數，遞推方法計算：

\[c[i][j]=c[i-1][j]+c[i-1][j-1] \\(楊輝三角求法) \]

Code:

#include <bits/stdc++.h>
#define lowbit(x) (x&(-x))
#define CSE(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define Abs(x) (x>=0?x:(-x))
#define FAST ios::sync_with_stdio(false);cin.tie(0);
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int maxn = 1111;
const ll mod = 1000000007;
ll c[maxn][maxn];
int a[maxn];

void Ini()
{
	c[0][0] = 1; c[1][0] = 1; c[1][1] = 1;
	for (int i = 2; i < maxn; i++) {
		c[i][0] = c[i][i] = 1;
		for (int j = 1; j < maxn; j++) {
			c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
			c[i][j] %= mod;
		}
	}
	return;
}

int main()
{
	FAST;
	Ini();
	int k, sum = 0;
	cin >> k;
	for (int i = 1; i <= k; i++) {
		cin >> a[i];
		sum += a[i];
	}
	ll ans = 1;
	for (int i = k; i >= 1; i--) {
		ans *= c[sum - 1][a[i] - 1];
		ans %= mod;
		sum -= a[i];
	}
	cout << ans << endl;
	return 0;
}