Huawei interview question: divide candy

subject

Xiao Ming grabs a handful of candy at random from the candy box
Every time Xiao Ming takes out half of the candy and gives it to the students
When candy is not evenly distributed
Xiao Ming can take the candy box ( Suppose there are enough candy in the box ) Take out or put back a candy
How many times does Xiao Ming need at least ( Take it out, put it back and distribute it evenly ) Can divide the candy in your hand to only one

 Input description ：
   Number of candy grabs ( Less than 1000000)：15
 Output description ：
   The number of times to get at least one candy ：5

  Example 1：
    Input 

15

    Output 

  5

    remarks 
       explain ：(1) 15+1=16;
           (2) 16/2=8;
           (3) 8/2=4;
           (4) 4/2=2;
           (5) 2/2=1;

Code

Be careful ： Is to count the number of times to retrieve and put in ！！

#include<iostream>
using namespace std;
int main(){
    
    int n;
    cin >> n;
    int count = 0;
    for(int i = n; i != 1;i /= 2,count++){
    
        if(i == 3){
    
            cout << count+2 << endl;
            return 0;
        }
        // If it can be 2 to be divisible by , Never take it out or put it back ;
        // If it can't be 2 to be divisible by , Unify +1 and -1 Who has the least ！; Local optima are used here ！！
        if(i % 2 != 0){
    
            if((i + 1)/2 % 2 == 0) i+=1;
            else --i;
            ++count;
        }
    }
    cout << count << endl;
    return 0;
}