Codeforces Round #480 (Div. 2): B. Marlin

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

7 2

output

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

input

5 3

output

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

题意：

给你一个4*n的点阵，你要往其中刚好k个点上放上障碍物('#')，要求放完之后从点(1,1)到(4,n)的最短路数量和从点(4,1)到(1,n)的最短路数量仍然相同，其中n一定为奇数，障碍物不能放在最外围，如果没有满足条件的方法输出NO，否则输出任意一种

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思路：

一定可以放，当k为偶数的时候从左往右两个两个放直到放完

当k为奇数的时候，先往最中间上方放一个，之后只要从左往右对称放就行了

拿AC代码运行一下一下就明白了

#include<stdio.h>
char str[6][108];
int main(void)
{
	int n, k, i, j;
	scanf("%d%d", &n, &k);
	for(i=1;i<=4;i++)
	{
		for(j=1;j<=n;j++)
			str[i][j] = '.';
	}
	if(k%2==0)
	{
		for(i=2;i<=n-1;i++)
		{
			if(k>=2)
			{
				k -= 2;
				str[2][i] = str[3][i] = '#';
			}
		}
	}
	else
	{
		k--;
		str[2][n/2+1] = '#';
		for(i=n/2;i>=2;i--)
		{
			if(k>=2)
			{
				str[2][i] = str[2][n-i+1] = '#';
				k -= 2;
			}
		}
		for(i=2;i<=n/2;i++)
		{
			if(k>=2)
			{
				str[3][i] = str[3][n-i+1] = '#';
				k -= 2;
			}
		}
	}
	printf("YES\n");
	for(i=1;i<=4;i++)
	{
		for(j=1;j<=n;j++)
			printf("%c", str[i][j]);
		puts("");
	}
	return 0;
}