首页 > 专题系列 > Java解POJ > POJ 1565 Skew Binary [解题报告] Java
2013
11-09

POJ 1565 Skew Binary [解题报告] Java

Skew Binary

问题描述 :

When a number is expressed in decimal, the kth digit represents a multiple of 10k. (Digits are numbered from right to left, where the least significant digit is number 0.) For example,

81307(10) = 8 * 10^4 + 1 * 10 ^3 + 3 * 10^2 + 0 * 10^1 + 7 * 10^0

= 80000 + 1000 + 300 + 0 + 7

= 81307.

When a number is expressed in binary, the kth digit represents a multiple of 2^k . For example,

10011(2) = 1 * 2^4 + 0 * 2^3 + 0 * 2^2 + 1 * 2^1 + 1 * 2^0

= 16 + 0 + 0 + 2 + 1

= 19.

In skew binary, the kth digit represents a multiple of 2^(k+1)-1. The only possible digits are 0 and 1, except that the least-significant nonzero digit can be a 2. For example,

10120(skew) = 1 * (2^5-1) + 0 * (2^4-1) + 1 * (2^3-1) + 2 * (2^2-1) + 0 * (2^1-1)

= 31 + 0 + 7 + 6 + 0

= 44.

The first 10 numbers in skew binary are 0, 1, 2, 10, 11, 12, 20, 100, 101, and 102. (Skew binary is useful in some applications because it is possible to add 1 with at most one carry. However, this has nothing to do with the current problem.)

输入:

The input contains one or more lines, each of which contains an integer n. If n = 0 it signals the end of the input, and otherwise n is a nonnegative integer in skew binary.

输出:

For each number, output the decimal equivalent. The decimal value of n will be at most 2^31-1 = 2147483647.

样例输入:

10120
200000000000000000000000000000
10
1000000000000000000000000000000
11
100
11111000001110000101101102000
0

样例输出:

44
2147483646
3
2147483647
4
7
1041110737

解题代码:

import java.util.*;   
  
public class Main {   
  
    public static void main(String[] args) {   
        Scanner cin = new Scanner(System.in);   
           
        while(true)   
        {   
            String skew = cin.nextLine();   
            if(skew.equals("0"))   
                break;   
               
            int decimal = trans(skew);   
            System.out.println(decimal);   
        }   
  
    }   
       
    private static int trans(String skew)   
    {   
        int num = 0;   
        int index = 0;   
        for(int i = skew.length()-1; i>=0; i--)   
        {   
            int cnum = Integer.valueOf(skew.substring(i, i+1));   
            num += (Math.pow(2, index+1) - 1) * cnum;   
            index++;   
        }   
           
        return num;   
    }   
  
}

  1. 思路二可以用一个长度为k的队列来实现,入队后判断下队尾元素的next指针是否为空,若为空,则出队指针即为所求。