Codeforces Round #340 (Div. 2) B. Chocolate 水题

B. Chocolate

题目连接：

http://www.codeforces.com/contest/617/problem/D

Descriptionww.co

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 ai (0 ≤ ai ≤ 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

5
1 0 1 0 1

Sample Output

4

Hint

题意

给你n个数字，你可以从数字间切开

然后问你有多少种切法，使得切出来的每一块恰有且仅有一个1

题解：

假如没有1，答案就是0

否则就是间隔中的0的个数+1的累乘

因为你就只能切0周围的空隙，然后根据乘法原则

代码

#include<bits/stdc++.h>
using namespace std;

int a[120];
vector<int> E;
int main()
{
    int n;scanf("%d",&n);
    int flag = 1;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        if(a[i]==1)flag = 0;
    }
    if(flag)return puts("0");
    int cnt = 0;
    for(int i=1;i<=n;i++)
    {
        if(a[i]==1)
        {
            E.push_back(cnt+1);
            cnt = 0;
        }
        else
            cnt++;
    }
    long long ans = 1;
    for(int i=1;i<E.size();i++)
        ans = ans * E[i];
    cout<<ans<<endl;
}