首页 > 专题系列 > Java解POJ > POJ 2453 An Easy Problem [解题报告] Java
2013
11-11

POJ 2453 An Easy Problem [解题报告] Java

An Easy Problem

问题描述 :

As we known, data stored in the computers is in binary form. The problem we discuss now is about the positive integers and its binary form.

Given a positive integer I, you task is to find out an integer J, which is the minimum integer greater than I, and the number of ’1′s in whose binary form is the same as that in the binary form of I.

For example, if “78″ is given, we can write out its binary form, “1001110″. This binary form has 4 ’1′s. The minimum integer, which is greater than “1001110″ and also contains 4 ’1′s, is “1010011″, i.e. “83″, so you should output “83″.

输入:

One integer per line, which is I (1 <= I <= 1000000).

A line containing a number “0″ terminates input, and this line need not be processed.

输出:

One integer per line, which is J.

样例输入:

1
2
3
4
78
0

样例输出:

2
4
5
8
83

解题代码:

import java.util.Scanner;    
  
public class Main {   
       
 /**  
  * �峰�涓�釜Integer瀵瑰����杩��涓���釜�� 
  * @param i  
  * @return sum  
  */  
 public static int getSum(int i) {   
  int sum = 0;   
  while (i > 0) {   
   i = i & (i - 1);   
   ++sum;   
  }   
  return sum;   
 }    
  
 public static void main(String[] args) {   
  Scanner sc = new Scanner(System.in);   
  while (sc.hasNext()) {   
   int i = sc.nextInt();   
   if (i == 0)   
    break;   
   int sum1 = getSum(i);   
   for (int j = i + 1;; ++j) {   
    int sum2 = getSum(j);   
    if (sum2 == sum1) {   
     System.out.println(j);   
     break;   
    }   
   }   
  }   
 }   
  
}

  1. 站长,你好!
    你创办的的网站非常好,为我们学习算法练习编程提供了一个很好的平台,我想给你提个小建议,就是要能把每道题目的难度标出来就好了,这样我们学习起来会有一个循序渐进的过程!