codeforces 617B 水题~

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

30 1 0

Output

1

Input

51 0 1 0 1

Output

4

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

题意：给你一根牛奶棒，1为一个nut，问你有几种方法能把nut单独分开？

思路：根据样例分析，头尾的0都是无效的。所以我们只要找到第一个1和最后一个1.然后开始遍历，在每两个1之间0的个数决定多少种方案。

例如，1 0 1，有两种，1 0 0 1，有三种，所以只要将每两个1之间的(0的个数+1)乘起来就是最终答案。那如果全为0的情况呢？就特判一下输出0.

(自己wa了一发的原因，没考虑到全是0的情况)

附上AC代码：

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#include<stdio.h>#include<string.h>#include<map>using namespace std;int a[105];int main(){    int n,m,k,l,s,flag;    while(~scanf("%d",&n))    {        flag=l=s=0;        for(int i=0; i<n; i++)        {            scanf("%d",&a[i]);            if(a[i]&&!flag)            {                s=i;                flag=1;            }            if(a[i])l=i;        }       if(flag==0){printf("0\n");continue;}        long long ans=1;        int sum=0;        for(int i=s; i<=l; i++)            if(!a[i])sum++;            else            {                ans*=sum+1;                sum=0;            }        printf("%lld\n",ans);    }}