The city of Fishtopia can be imagined as a grid of 44 rows and an odd number of columns. It has two main villages; the first is located at the top-left cell (1,1)(1,1), people who stay there love fishing at the Tuna pond at the bottom-right cell (4,n)(4,n). The second village is located at (4,1)(4,1) and its people love the Salmon pond at (1,n)(1,n).
The mayor of Fishtopia wants to place kk hotels in the city, each one occupying one cell. To allow people to enter the city from anywhere, hotels should not be placed on the border cells.
A person can move from one cell to another if those cells are not occupied by hotels and share a side.
Can you help the mayor place the hotels in a way such that there are equal number of shortest paths from each village to its preferred pond?
Input
The first line of input contain two integers, nn and kk (3≤n≤993≤n≤99, 0≤k≤2×(n−2)0≤k≤2×(n−2)), nn is odd, the width of the city, and the number of hotels to be placed, respectively.
Output
Print "YES", if it is possible to place all the hotels in a way that satisfies the problem statement, otherwise print "NO".
If it is possible, print an extra 44 lines that describe the city, each line should have nn characters, each of which is "#" if that cell has a hotel on it, or "." if not.
Examples
Input
7 2
Output
YES ....... .#..... .#..... .......
Input
5 3
Output
YES ..... .###. ..... .....
如何去看这个题，可以把他看成关于对称的图形，上下对称和左右对称
代码：
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>
const int maxn=3e5+5;
typedef long long ll;
using namespace std;
char Map[5][105];
int n,k;
void init()
{
for(int t=0;t<4;t++)
{
for(int j=0;j<n;j++)
{
Map[t][j]='.';
}
}
}
int main()
{
cin>>n>>k;
init();
if(k%2==0)
{
int s=k/2;
for(int t=1;t<s+1;t++)
{
Map[1][t]='#';
}
for(int t=1;t<s+1;t++)
{
Map[2][t]='#';
}
cout<<"YES"<<endl;
for(int t=0;t<4;t++)
{
puts(Map[t]);
}
}
else
{
int s=n/2;
Map[1][s]='#';
k--;
for(int t=1;t<n/2;t++)
{
if(k==0)
{
break;
}
else
{
Map[1][t]='#';
Map[1][n-1-t]='#';
k-=2;
}
}
for(int t=1;t<n/2;t++)
{
if(k==0)
{
break;
}
else
{
Map[2][t]='#';
Map[2][n-1-t]='#';
k-=2;
}
}
cout<<"YES"<<endl;
for(int t=0;t<4;t++)
{
puts(Map[t]);
}
}
return 0;
}作者：李斌
个性签名：独学而无友，则孤陋而寡闻。做一个灵魂有趣的人！
