金色十月线上编程比赛第二题：解密 - c++编程基础

金色十月线上 编程比赛第二题：解密

发布公司：
有效期：
- CSDN
- 2014-10-20至2015-10-20
  - 难度等级：
  - 答题时长：
  - 编程语言要求：
    - 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"ans_tip">
      
      解题思路：利用这个函数就可以计算一个数的二进制有多少个1：
```
int totalOne(int x){
    int count = 0;
    while(x){
        x = x & ( x - 1 );
        count++;
    }
    return count;
}
```
      所以我们只需遍历一边即可；
```
#include 
          
           
using namespace std;
int totalOne(int x){
    int count = 0;
    while(x){
        x = x & ( x - 1 );
        count++;
    }
    return count;
}
int main(){
    int n,a,step=1;    
    while (cin>>n){
        int max=0,num=0,min=0xffffff;
        for (int i=0;i
           
            >a; num=totalOne(a); if (num
            
             a) max=a; } cout<<"Case "<
             
              
 
              
 
               
              
  
             
            
           
          
```