CodeForces - 554C Kyoya and Colored Balls （组合数学&逆元模板）

CodeForces - 554C

Kyoya and Colored Balls

Time Limit: 2000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

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.

Sample Input

Input

3
2
2
1

Output

3

Input

4
1
2
3
4

Output

1680

Hint

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

Source

Codeforces Round #309 (Div. 2)

//题意：

一个包里有k种颜色的球，每种颜色的球有a[i]个，现在要求在取完第i种颜色的球之前得先将第i-1种颜色的球取完，问总共有几种取法。

//思路：

从后往前推，先确定第k种颜色的球的取法，此时中共有sum个位置，因为最后一个位置是确定的，所以在剩下的位置中取a[k]-1个球的去取法为C（sum-1，a[k]-1），接着去确定第k-1种颜色的球的方法数，此时一共有sum=sum-a[k]个位置，排法和上面的一样这种颜色的球的方法数为C（sum-1,a[k-1]-1）,以此类推，直到取到第一种球为止。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define ll __int64
#define N 2010
#define M 1000000007
using namespace std;
ll a[1010];
ll C[N][N]; 
void GetC()
{
    C[0][0] =C[1][0] = C[1][1] = 1;
    for(int i=2;i<N;i++)
	{
        C[i][0] = 1;C[i][i] = 1;
        for(int j=1;j<i;j++)
		{
            C[i][j] = (C[i-1][j-1] + C[i-1][j]) % M;
        }
    }
}
int main()
{
	int n,i,j,k;
	GetC();
	while(scanf("%d",&k)!=EOF)
	{
		ll sum=0;
		memset(a,0,sizeof(a));
		for(i=1;i<=k;i++)
		{
			scanf("%I64d",&a[i]);
			sum+=a[i];
		}
		ll num=1;
		for(i=k;i>=2;i--)
		{
			num=(num%M*C[sum-1][a[i]-1])%M;
			sum-=a[i];
		}
		printf("%I64d\n",num%M);
	}
	return 0;
}

//逆元求解：（模板）

#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\n",ans);

    return 0;
}