POJ2136 Vertical Histogram

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POJ2136 Vertical Histogram

 

 

题目:http://acm.pku.edu.cn/JudgeOnline/problem?id=2136

很喜欢这种的画图题啊,特别是做完后很有成就感。。。

看看结果先:

                            *

                            *

        *                   *

        *                   *     *   *

        *                   *     *   *

*       *     *             *     *   *

*       *     * *     * *   *     * * *

*       *   * * *     * *   * *   * * * *

*     * * * * * *     * * * * *   * * * *     * *

* * * * * * * * * * * * * * * * * * * * * * * * * *

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

题目的意思是求一个字符出现的频率,画出来很有意思的,有兴趣试试。

 

下面是代码:

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.Arrays;

 

public class Main {

 

    public static void main(String[] args) throws NumberFormatException,

           IOException {

       BufferedReader read = new BufferedReader(new InputStreamReader(

              System.in));

       int[] a = new int[26];

       int tt;

       for (int i = 0; i < 4; i++) {

           String str = read.readLine();

           for (int j = 0; j < str.length(); j++) {

              tt = str.charAt(j) - 'A';

              if (tt >= 0 && tt <= 25) {

                  a[tt]++;

              }

           }

       }

       int max = 0;

       for (int i = 0; i < 26; i++) {

           if (a[i] > max) {

              max = a[i];

           }

       }

       char[][] c = new char[max + 1][51];

       for (int i = 0; i < max + 1; i++) {

           Arrays.fill(c[i], ' ');

       }

       for (int i = 0; i < 26; i++) {

           int h = a[i];

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

              c[max - 1 - j][i * 2] = '*';

           }

       }

       for (int i = 0; i < 26; i++) {

           c[max][i * 2] = (char) ('A' + i);

       }

       for (int i = 0; i < max + 1; i++) {

           String str = String.valueOf(c[i]);

           while (str.endsWith(" ")) {

              str = str.substring(0, str.length() - 1);

           }

           System.out.println(str);

       }

    }

}