codeforces 617B 水題~

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

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

題意：給你一根牛奶棒，1爲一個nut，問你有幾種方法能把nut單獨分開？

思路：根據樣例分析，頭尾的0都是無效的。所以我們只要找到第一個1和最後一個1.然後開始遍歷，在每兩個1之間0的個數決定多少種方案。

例如，1 0 1，有兩種，1 0 0 1，有三種，所以只要將每兩個1之間的(0的個數+1)乘起來就是最終答案。那如果全爲0的情況呢？就特判一下輸出0.

(自己wa了一發的原因，沒考慮到全是0的情況)

附上AC代碼：

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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);
    }
}