Codeforces 980B Marlin （构造）

B. Marlin

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The city of Fishtopia can be imagined as a grid of $4$ rows and an odd number of columns. It has two main villages; the first is located at the top-left cell $(1,1)$, people who stay there love fishing at the Tuna pond at the bottom-right cell $(4,n)$. The second village is located at $(4,1)$and its people love the Salmon pond at $(1,n)$.

The mayor of Fishtopia wants to place $k$ 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, $n$ and $k$ ($3\le n\le 99$, $0\le k\le 2\times (n-2)$), $n$ 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 $4$ lines that describe the city, each line should have $n$ characters, each of which is "#" if that cell has a hotel on it, or "." if not.

Examples

input

Copy

7 2

output

Copy

YES
.......
.#.....
.#.....
.......

input

Copy

5 3

output

Copy

YES
.....
.###.
.....
.....

【题目链接】 http://codeforces.com/contest/980/problem/B

【题意】有一个4*n的格子，有a和b两个人，a初始位于（1,1）处，要走到（n,n）处；b初始位于（4,1）处，要走到（1，n）处，可以在除了边界处的格子放置k个酒店（障碍物），人不能够通过酒店，输入一种放置方式使得a和b到达目的地的最短路的路径数目相等。

【思路】打个表发现，当酒店放置的位置关于x轴或者y轴对称的时候，最短路径的路径数目相等。

【代码如下】

#include <bits/stdc++.h>
using namespace std;

int n,k;
char str[210][210];

int main(){
    scanf("%d%d",&n,&k);
    for(int i = 1; i <= 4; i ++) for(int j = 1; j <= n; j ++) str[i][j] = '.';
    if(k%2==0){
        int c=2;
        while(k){
            str[2][c] = str[3][c] = '#';
            c ++; k -= 2;
        }
    }
    else{
        str[2][n/2+1] = '#'; k --;
        int c1=n/2,c2=(n+1)/2+1;
        while(k){
            str[2][c1] = str[2][c2] = '#';
            c1 --; c2 ++; k -= 2;
            if(c1 == 1) break;
        }
        c1=n/2,c2=(n+1)/2+1;
        while(k){
            str[3][c1] = str[3][c2] = '#';
            c1 --; c2 ++; k -= 2;
            if(c1 == 1) break;
        }
    }
    puts("YES");
    for(int i = 1; i <= 4; i ++){
        for(int j = 1; j <= n; j ++) printf("%c",str[i][j]);
        puts("");
    }
    return 0;
}