Codeforces 617B Chocolate【思维】

B. Chocolate

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 containexactly 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
ai (0 ≤ ai ≤ 1), where0 represents a piece without the nut and1stands 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.

Examples

Input

3
0 1 0

Output

1

Input

5
1 0 1 0 1

Output

4

Note
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
题目大意：

给你一个长度为n的01序列，问有多少种切法，使得切割开的每个序列中都有1，并且1可以相连。

思路：

对序列去掉前导0之后，统计每两个1之间的0的个数，累乘即可，有cont个0，就累乘一个（cont+1）【因为在连续的0的序列中的每一个0左边都可以切一下，在最后一个0的右边也可以切一下，所以是cont+1】。

注意特判一下全是0的情况，需要直接输出0.

Ac代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
using namespace std;
#define ll __int64int a[1000];
int main()
{
int n;
while(~scanf("%d",&n))
{
int flag=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
if(a[i]==1)flag=1;
}
if(flag==0)
{
printf("0\n");
continue;
}
flag=0;
int cont=0;
ll output=1;
for(int i=1;i<=n;i++)
{
if(a[i]==1)
{
if(flag==0)
{
flag=1;continue;
}
else
{
output*=(cont+1);
cont=0;
}
}
if(a[i]==0&&flag==1)
{
cont++;
}
}
printf("%I64d\n",output);
}
}