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.
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.
Print the number of ways to break the chocolate into multiple parts so that each part would contain exactly one nut.
3
0 1 0
1
5
1 0 1 0 1
4
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
真是醉了，比赛时好几个题都是这样，题目必须要长整型才可以而我定义的是int型，以后不管题目什么要求都定义成长整型
题意：给一个01串，要求将这个串剪开要求每一段都有一个1问总共有多少种方法
题解：每次再两个1之间截开，则需要计算两个1之间0的个数，运用组合数学思想，求解
#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 100100
#define mod 100
#define dian 1.000000011
#define INF 0x3f3f3f
using namespace std;
LL vis[MAX];
LL s[MAX];
int main()
{
int n,m,j,i,t,k;
while(scanf("%lld",&n)!=EOF)
{
for(i=1;i<=n;i++)
scanf("%lld",&s[i]);
memset(vis,0,sizeof(vis));
int sum=0;
int ans=1;
int a[MAX];k=0;
for(i=1;i<=n;i++)
{
if(s[i]==0&&sum>0)
ans++;
else if(s[i]==1)
{
sum++;
if(sum>1)
a[k++]=ans;
ans=1;
}
}
if(sum==0)
{
printf("0
");
continue;
}
LL ant=1;
for(i=0;i<k;i++)
ant*=a[i];
printf("%lld
",ant);
}
return 0;
}