Codeforces--617B--Chocolate（规律）



Chocolate

Time Limit: 1000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit
Status

Description

Bob loves everything sweet. His favorite chocolate bar consists of pieces, each piece may contain a nut. Bob wants to break the bar of chocolate into multiple pieces so that each part would contain
exactly one nut and any break line goes between two adjacent pieces.

You are asked to calculate the number of ways he can do it. Two ways to break chocolate are considered distinct if one of them contains a break between some two adjacent pieces and the other one doesn't.

Please note, that if Bob doesn't make any breaks, all the bar will form one piece and it still has to have exactly one nut.

Input

The first line of the input contains integer n (1 ≤ n ≤ 100) — the number of pieces in the chocolate bar.

The second line contains n integers
a_i (0 ≤ a_i ≤ 1), where
0 represents a piece without the nut and
1 stands for a piece with the nut.

Output

Print the number of ways to break the chocolate into multiple parts so that each part would contain exactly one nut.

Sample Input

Input

3
0 1 0

Output

1

Input

5
1 0 1 0 1

Output

4

Sample Output

Hint

In the first sample there is exactly one nut, so the number of ways equals
1 — Bob shouldn't make any breaks.

In the second sample you can break the bar in four ways:

10|10|1

1|010|1

10|1|01

1|01|01

Source

Codeforces Round #340 (Div. 2)

两个1之间有几种选择，然后连乘，跟组合数思想是一样的

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int a[110],b[110];
int main()
{
	int n;
	while(cin>>n)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		__int64 ans=1;
		int k=0;
		for(int i=1;i<=n;i++)
		{
			cin>>a[i];
			if(a[i])
			b[k++]=i;
		}
		if(k==0)
		cout<<0<<endl;
		else if(k==1)
		cout<<1<<endl;
		else
		{
			for(int i=1;i<k;i++)
			ans=ans*(b[i]-b[i-1]);
			cout<<ans<<endl;
		}
	}
	return 0;
}