CF980B Marlin 构造 思维 二十四

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 $\mathrm{44 rows\; and\; an odd number\; of\; columns.\; It\; has\; two\; main\; villages;\; the\; first\; is\; located\; at\; the\; top-left\; cell (1,1)(1,1),\; people\; who\; stay\; there\; love\; fishing\; at\; the\; Tuna\; pond\; at\; the\; bottom-right\; cell (4,n)(4,n).\; The\; second\; village\; is\; located\; at (4,1)(4,1) and\; its\; people\; love\; the\; Salmon\; pond\; at (1,n)(1,n).}$

The mayor of Fishtopia wants to place $\mathrm{kk hotels\; in\; the\; city,\; each\; one\; occupying\; one\; cell.\; To\; allow\; people\; to\; enter\; the\; city\; from\; anywhere,\; hotels\; should\; not\; be\; placed\; on\; the\; border\; cells.}$

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, $\mathrm{nn and kk (3\le n\le 993\le n\le 99, 0\le k\le 2\times (n-2)0\le k\le 2\times (n-2)), nn is odd,\; the\; width\; of\; the\; city,\; and\; the\; number\; of\; hotels\; to\; be\; placed,\; respectively.}$

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 $\mathrm{44 lines\; that\; describe\; the\; city,\; each\; line\; should\; have nn characters,\; each\; of\; which\; is\; "\#"\; if\; that\; cell\; has\; a\; hotel\; on\; it,\; or\; "."\; if\; not.}$

Examples

input

Copy

7 2

output

Copy

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

input

Copy

5 3

output

Copy

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

 题意： 现在有一个4*n的地图，a村庄的人在（1，1），他们要去（4，n），b村庄的人在（4，1），他们要去（1，n），路途的中间有m个酒店，村民不能经过酒店。问怎么设置酒店可以使a、b村庄的人到目的地的最短路大小相同。

 题解：通过一番枚举验证，发现只要酒店能关于x轴或者y轴对称，就能满足条件。

 然后对比样例，我们不难发现，m使偶数关于x轴对称，为奇数关于y轴对称。

 接下来就是按照上面说的构造这样的一个4*n的地图，构造奇数的情况的时候小心一点就行。

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e2 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
char mapn[maxn][maxn];
int main(){
    std::ios::sync_with_stdio(false);
    ll n, m;
    cin >> n >> m;
    cout << "YES" << endl;
    for( ll i = 1; i <= 4; i ++ ) {
        for( ll j = 1; j <= n; j ++ ) {
            mapn[i][j] = '.';
        }
    }
    if( m % 2 == 0 ) {
        for( ll i = 2; i <= 3; i ++ ) {
            for( ll j = 2; j <= 1+m/2; j ++ ) {
                mapn[i][j] = '#';
            }
        }
    } else {
        ll t = ( n+1 ) / 2, num = m; //奇数对称的中心点纵坐标是t，num不大于1时不用再填充了，大于1时关于t对称填充
        for( ll i = 2; i <= ( (m-1) / (n-2) ) + 2 && num > 1; i ++ ) {
            for( ll j = 1; j <= t-2 && num > 1; j ++ ) {
                mapn[i][t-j] = mapn[i][t+j] = '#';
                num -= 2;
            }
        }
        mapn[2][t] = '#';
    }
    for( ll i = 1; i <= 4; i ++ ) {
        for( ll j = 1; j <= n; j ++ ) {
            cout << mapn[i][j];
        }
        cout << endl;
    }
    return 0;
}