HDU 4706 Children's Day

Children's Day

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 315 Accepted Submission(s): 196

Problem Description

Today is Children's Day. Some children ask you to output a big letter 'N'. 'N' is constituted by two vertical linesand one diagonal. Each pixel of this letter is a character orderly. No tail blank is allowed.
For example, this is a big 'N' start with 'a' and it's size is 3.

a e

bdf

c g

Your task is to write different 'N' from size 3 to size 10. The pixel character used is from 'a' to 'z' continuously and periodic('a' is reused after 'z').

 

 

Input

This problem has no input.

 

 

Output

Output different 'N' from size 3 to size 10. There is no blank line among output.

 

 

Sample Output

[pre] a e bdf c g h n i mo jl p k q ......... r j [/pre]

Hint

Not all the resultsare listed in the sample. There are just some lines. The ellipsis expresseswhat you should write.

 

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup

 

 

读题很重要.... 我看了半天， 输出倒数第二行虚线是什么意思...  读了半天，参反应过来，那只是省略了 一些字符...

哎... 看来还需要做做练习....

 总之水题一道....

 

import java.io.*;

import java.util.*;



public class Main {



	public static void main(String[] args) {

		new Main().work();

	}



	void work() {

		char c = 'a';

		//求第一列数据

		for (int i = 3; i <=10; i++) {

			char[][] ch = new char[i][i];

			for (int j = 0; j < i; j++) {

				

				ch[j][0] = c;

				c=(char) ((c + 1));

				if (c > 'z')

					c = 'a';

			}

			//求 第二列 到 倒数第二列

			for (int k = i - 2, j = 1; k > 0 && j < i - 1; k--, j++) {

				

				ch[k][j] = c;

				c=(char) ((c + 1));

				if (c > 'z')

					c = 'a';

			}

			//求最后一列

			for (int j = 0; j < i; j++) {

				

				ch[j][i - 1] = c;

				c=(char) ((c + 1));

				if (c > 'z')

					c = 'a';

			}

			//输出

			for (int j = 0; j < i; j++) {

				for (int k = 0; k < i; k++) {

					if (ch[j][k] >= 'a' && ch[j][k] <= 'z')

						System.out.print(ch[j][k]);

					else

						System.out.print(" ");

				}

				System.out.println();

			}

		}



	}

}