Fox And Snake

文章目录

一、Fox And Snake

总结

一、Fox And Snake

本题链接：

题目：

A. Fox And Snake

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.

A snake is a pattern on a n by m table. Denote c-th cell of r-th row as (r, c). The tail of the snake is located at (1, 1), then it’s body extends to (1, m), then goes down 2 rows to (3, m), then goes left to (3, 1) and so on.

Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters (’.’) and the snake cells should be filled with number signs (’#’).

Consider sample tests in order to understand the snake pattern.

Input

The only line contains two integers: n and m (3 ≤ n, m ≤ 50).

n is an odd number.

Output

Output n lines. Each line should contain a string consisting of m characters. Do not output spaces.

Examples

input

3 3

output

###

..#

###

input

3 4

output

####

...#

####

input

5 3

output

###

..#

###

#..

###

input

9 9

output

#########

........#

#########

#........

#########

........#

#########

#........

#########

本博客给出本题截图：

题意：按照如图所示的样子打印出来结即可

AC代码

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 55;
char g[N][N];
int main()
{
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ )
        {
            if (i % 2) g[i][j] = '#';
            else
            {
                if (j == 1 && !(i % 4))
                {
                    g[i][j]=  '#';
                    continue;
                }
                else if (j == m && !(i % 2) && g[i][1] != '#')
                {
                    g[i][j]=  '#';
                    continue;
                }
                g[i][j] = '.';
            }
        }
    for (int i = 1; i <= n; i ++ )
    {
        for (int j = 1; j <= m; j ++ )
            cout << g[i][j];
        puts("");
    }
    return 0;
}

总结

水题，不解释