transfer LCT Run away , Brush the water and adjust it .

``` #include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int a[],n,m,f[],ans;
bool cal()
{
for (int i=;i<=n;i++) {f[i+]=a[i]-f[i]-f[i-];}
if (a[n]-f[n]-f[n-]) return ;
else return ;
}
int main()
{
scanf("%d",&n);
for (int i=;i<=n;i++) scanf("%d",&a[i]);
if (a[]==) ans+=cal();
else if (a[]==)
{
f[]=;
ans+=cal();
memset(f,,sizeof(f));
f[]=;
ans+=cal();
}
else
{
f[]=f[]=;
ans+=cal();
}
printf("%d\n",ans);
return ;
}```

## Description

I believe everyone has played the game of minesweeping . It was in a city n*m There's some ray in the matrix , I want you to find ray based on some information . It's Halloween ,“ more than ” A simple minesweeping game has become popular in China , The rules of the game are the same as mine sweeping , If there's no thunder in a grid , So the number in it and it 8 The number of rays in a connected lattice . Now the chessboard is n×2 Of , Some of the lattices in the first column are ray , And there's no thunder in the second column , Here's the picture ： Because the first column of mine may have a variety of programs to meet the second column of the number of restrictions , Your task is to determine the number of placement options for the first row of mines according to the information in the second row .

## Input

First act N, The second line has N Number , In turn, the numbers in the second column of the lattice .（1<= N <= 10000）

## Output

a number , That is, the number of placement schemes of mines in the first column .

2
1 1

2

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