用位运算解决进制转化问题

Same binary weight

时间限制： 300  ms  |  内存限制： 65535  KB

难度： 3

描述

The binary weight of a positive  integer is the number of 1's in its binary representation.for example,the decmial number 1 has a binary weight of 1,and the decimal number 1717 (which is 11010110101 in binary) has a binary weight of 7.Give a positive integer N,return the smallest integer greater than N that has the same binary weight as N.N will be between 1 and 1000000000,inclusive,the result is guaranteed to fit in a signed 32-bit interget.

输入

The input has multicases and each case contains a integer N.

输出

For each case,output the smallest integer greater than N that has the same binary weight as N.

样例输入

1717
4
7
12
555555

样例输出

1718
8
11
17
555557

分析：这道题题意是给你一个数N把它转化成二进制以后1的个数称为The binary weight of N，求和该数有相同1的个数

的二进制中最小的那个十进制数。主要运用了c++库函数中的next_permutation函数，我之所以收藏这道题，因为用位运算就是很巧妙的把一个十进制转化成了二进制。

下面简单介绍一下这道题用到的位运算&，和<<.

1，<<的优先级高于&的优先级。

2，&运算就是对应位都是1是结果为1.

3，<<就是左移运算，2<<3=2*pow(2,3),16>>4=16/pow(2,4),pow(2,i)=1<<i;

 
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int get_binary(int n)
{  vector<int>bits(32,0); 
   for(int i=0;i<32;i++)
    if(n&1<<i)//每次让2的i次方和n进行与运算从而判断出该位存0，还是存1.
       bits[31-i]=1; 
       next_permutation(bits.begin(),bits.end());//找出比n大的最小的那个数。
       int sum=0;
       for(int i=0;i<32;i++)
             if（bits[i]）
             sum+=bits[i]<<(31-i);//计算二进制中不为0时，该位的权值。
             return sum;
}
int main()
{     int n;
        
      while(cin>>n)
     {    
        
             cout<<get_binary(n)<<endl;
     }return 0;
}