CodeForces 510 A. Fox And Snake（模拟啊 ）

题目链接：http://codeforces.com/problemset/problem/510/A

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.

Sample test(s)

input

3 3

output

###..####

input

3 4

output

####...#####

input

5 3

output

###..#####..###

input

9 9

output

#########........###########........#########........###########........#########

代码如下：

#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>using namespace std;int main(){    int t, n, m;    while(~scanf("%d%d",&n,&m))    {        int k = 0;        for(int i = 1; i <= n; i++)        {            for(int j = 1; j <= m; j++)            {                if(i%2 == 1)                {                    printf("#");                }                else if(i%2 == 0 && k == 0)                {                    if(j == m)                        printf("#");                    else                        printf(".");                    if(j == m)                        k = 1;                }                else if(i%2 == 0 && k == 1)                {                    if(j == 1)                        printf("#");                    else                        printf(".");                    if(j == m)                        k = 0;                }            }            printf("\n");        }    }    return 0;}