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.
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
题意：分割所给数，使得每一部分都只有1个1，问有几种分割方案。
思路：最一开始感觉和划分数问题很像，那样就得搜索了，不过后面猜了几组样例，感觉是只要算出每两个1中间0的个数加1再相乘就可以了，然后A了。。。真的就是猜题意的赶脚，具体肯定是有证明的。
注意：全是0的情况要考虑，因为中间有相乘的情况，所以要考虑变量的范围。
#include <bits/stdc++.h>
using namespace std;
unsigned long long ans = 1;
int main()
{
int n;
int a[105];
int cnt = 0;
int flag = 0;
cin>>n;
for(int i = 0;i < n;i++)
{
cin>>a[i];
if(a[i] == 1)
{
flag = 1;
}
}
for(int i = 0;i < n;i++)
{
if(a[i] == 1)
{
for(int j = i + 1;j < n;j++)
{
if(a[j] == 1)
{
ans *= (j - i);
i = j - 1;
break;
}
}
}
}
if(flag == 0)
{
printf("0\n");
return 0;
}
printf("%llu\n",ans);
return 0;
}