HDU 4706 Children's Day

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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(); } } } }  

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