A binary integer number of 1

method one:

Ideas: If a number with 1 for the calculation can be found in the number of the last digit is 1, 0 otherwise.
According to this thinking can have the following code:

 1 #include <iostream>
 2   using namespace std;
 3 
 4   int main()
 5   {
 6 
 7       int n = 0;
 8       cout << "输入一个数";
 9       cin >> n;
10       int count = 0;
11       while (n)
12  {
13 
14       if (n & 1)
15       {
16           count++;
17       }
 18 is        n->> = 1 ;
 . 19    }
 20 is        COUT << " number of 1s " << COUNT << endl;
 21 is        return  0 ;
 22 is    }

Write pictures described here

But there are a few flaws in this method if the input is negative loop will be infinite die

Method Two:

N 1 and the first operation to do with determining the lowest bit of n is not 1. Then the left one obtained 1 2, n, and then do with the calculation, can determine the time n is not 1 ... low. Thus the left repeatedly, each time n can be determined which is not a solution of the 1 binary integer number of cycles equal to the number of bits, an integer of 32 to 32 cycles need

 1 #include <iostream>
 2   using namespace std;
 3 
 4   int main()
 5   {
 6 
 7       int n = 0;
 8       int key = 1;
 9       cout << "输入一个数";
10       cin >> n;
11       int count = 0;
12       while (key)
13  {
14 
15       if (n & key)
16       {
17           count++;
18       }
19       key <<= 1;
20   }
21       cout << "1的个数为"<<count<<endl;
22       return 0;
23   }

Method Three

Ideas: integer to a minus 1, and then to do with the original integer arithmetic, the integer will rightmost a 1 to a 0, then the binary representation of an integer has
the number 1, on how many times such an operation can be performed.

 1 #include <iostream>
 2   using namespace std;
 3 
 4   int main()
 5   {
 6 
 7       int n = 0;
 8       cout << "输入一个数";
 9       cin >> n;
10       int count = 0;
11       while (n)
12  {
13           n = ((n - 1)& n);
14           count++;
15 
16   }
17       cout << "1的个数为"<<count<<endl;
18 }

End of this article.

[Reserved] https://blog.csdn.net/weibo_dm/article/details/80531465