首页 > 专题系列 > Java解POJ > POJ 1546 Basically Speaking [解题报告] Java
2013
11-09

POJ 1546 Basically Speaking [解题报告] Java

Basically Speaking

问题描述 :

The Really Neato Calculator Company, Inc. has recently hired your team to help design their Super Neato Model I calculator. As a computer scientist you suggested to the company that it would be neato if this new calculator could convert among number bases. The company thought this was a stupendous idea and has asked your team to come up with the prototype program for doing base conversion. The project manager of the Super Neato Model I calculator has informed you that the calculator will have the following neato features:
  • It will have a 7-digital display.
  • Its buttons will include the capital letters A through F in addition to the digits 0 through 9.
  • It will support bases 2 through 16.

输入:

The input for your prototype program will consist of one base conversion per line. There will be three numbers per line. The first number will be the number in the base you are converting from. The second number is the base you are converting from. The third number is the base you are converting to. There will be one or more blanks surrounding (on either side of) the numbers. There are several lines of input and your program should continue to read until the end of file is reached.

输出:

The output will only be the converted number as it would appear on the display of the calculator. The number should be right justified in the 7-digit display. If the number is to large to appear on the display, then print “ERROR” (without the quotes) right justified in the display.

样例输入:

1111000  2 10
1111000  2 16
2102101  3 10
2102101  3 15
  12312  4  2
     1A 15  2
1234567 10 16
   ABCD 16 15

样例输出:

    120
     78
   1765
    7CA
  ERROR
  11001
 12D687
   D071

解题代码:

import java.io.*;
import java.util.*;
import java.math.*;
/**
 *
 * @author gongshaoqing
 */
public class Main {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {

        // TODO code application logic here
        Scanner cin=new Scanner(System.in);
        String str;
        int inbas,oubas;
        while(cin.hasNext())
        {
            str=cin.next();
            inbas=cin.nextInt();
            oubas=cin.nextInt();
            BigInteger tmp=new BigInteger(str,inbas);
      
            str=tmp.toString(oubas);
            int len=str.length();
            if(len>7)System.out.println("  ERROR");
            else
            {
                str=str.toUpperCase(Locale.FRENCH);
                for(int i=0;i< 7-len;i++)System.out.print(" ");
                System.out.println(str);
            }
        }
    }

}

  1. 代码是给出了,但是解析的也太不清晰了吧!如 13 abejkcfghid jkebfghicda
    第一步拆分为 三部分 (bejk, cfghi, d) * C(13,3),为什么要这样拆分,原则是什么?

  2. 5.1处,反了;“上一个操作符的优先级比操作符ch的优先级大,或栈是空的就入栈。”如代码所述,应为“上一个操作符的优先级比操作符ch的优先级小,或栈是空的就入栈。”