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

Chocolate

题目链接https://codeforces.com/problemset/problem/617/B

time limit per test 1 second
memory limit per test 256 megabytes

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.

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

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题目大意：给你一个数列只包含0和1，问你有多少种分法使得每个区间有且只有一个1.
我们可以直接手动模拟分割线：如果1后面有0，我们从将分割线从1后开始移动，其余的分割线不动，一直移动到碰到1为止，设1后面的0的个数为xi，那么我们很容易发现，分割的总方法为x1 *x2 *x3 …… xn。当然我们还要特判一下最后一位是否为0。
以下是详细代码：

#include <cstdio>
#define ll long long
int a[120];
ll b[120];
int main()
{
	int n,mark=0;
	ll sum=1;
	scanf ("%d",&n);
	for (int i=1; i<=n; i++){
		scanf ("%d",&a[i]);
	}
	int flag;
	for (int i=1; i<=n; i++){
		flag=i;
		if (a[i]) {
			mark=1;
			b[flag]++;i++;
			while (!a[i] && i<=n){
				b[flag]++;i++;
			}
			i--;
		}
	}
	if (!mark) {
		printf ("0\n");
		return 0;
	}
	if (!a[n]) b[flag]=1;
	for (int i=1; i<=n; i++)
	 if (b[i]>1) sum*=b[i];
	printf ("%lld\n",sum);
	return 0;
}