980B 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

Copy

7 2

output

Copy

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

input

Copy

5 3

output

Copy

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

题意：输入n,k，一个4行n列的城市（矩阵），让你放置k个宾馆，有两个村庄，第一个村庄（1，1）要去（4，n）捕鱼，第二个村庄（4，1）要去（1，n）捕鱼，问怎么放置宾馆，才能让两个村庄的最短路线的个数相同，然后输出YES，打印出，4行n列的放置情况宾馆不能放置在边界上，' .'表示路，#表示宾馆，如果不可能就输出NO。然而我想说对的是这个题，还是不好懂的，昨晚比赛的时候，群里讨论都在那里跟题意较真，然后过了的大佬说是构造题。我们开始构造，构造了就wa 1，难适应我就想改了，电脑也没电了，就没做了，我能说CF开题，能不开在23点么，寝室断电23点，电脑只能坚持一个多小时点....刚开始我是从非边界第一个开始放置宾馆的，结果wa 1，群里问其他人，有几个和我情况一样，同样再问。。。这道题当时还是卡了很多人的...

题解：构造   对k的奇偶性讨论，奇数的时候就一行一行的放置宾馆，并且要从中间开始放，k为偶数的时候就一列一列的放置宾馆...

#include<bits/stdc++.h>
using namespace std;
char s[10][110];
int main()
{
    int n,k;
    cin>>n>>k;
    for(int i=0; i<4; i++)
        for(int j=0; j<n; j++)
            s[i][j]='.';
    if(k&1)
    {
        s[1][n/2]='#',k--;///先在中间放置一个宾馆
        for(int i=0; i<4; i++)
            for(int j=0; j<n/2; j++)
                if(i&&j&&k)
                    s[i][j]='#',s[i][n-j-1]='#',k-=2;然后一左一右放置
    }
    else
    {
        for(int i=0; i<4; i++)///三层循环一列一列的放置。
            for(int j=0; j<n; j++)
                for(int l=0; l<3; l++)
                    if(i&&j&&l&&k)
                        s[l][j]='#',k--;
    }
    puts("YES");
    for(int i=0; i<4; i++)
        cout<<s[i]<<endl;
    return 0;
}