The beauty of programming a number of binary representations of 1 in number

Here are 3 solutions to be introduced

The first kind; (Conventional solution)

The number is represented in the computer by binary, so it can be continuously use the number n to divide by 2

Code Listing 1:

#include <iostream>using namespace Std;int main (void) {int n,m;    m=0;    cin>>n;        while (n) {if (n%2)//If n cannot be divisible by 2, indicating that the end number of the current n is 1 m++;    n>>=1;    n Shift right 1 bits, i.e. N/2} cout<< m <<endl; return 0;}

The second type: using bit manipulation

&: If A is 1 and B is 1, then a&b is 1, otherwise, A&b is 0

Here initializes the i=0x1;

If the n&i is 1, then the end of N is 1, otherwise 0

Code Listing 2:

#include <iostream>using namespace Std;int main (void) {//Use bit operation, 1 and binary the last one to be performed with (&) operation Int N,I,M;M=0;I=0X1;CIN&G T;>n;while (n) {m+= (n&i); n>>=1;}    cout << M <<endl; return 0;}

Third: Use a different bit operation

If the binary representation of n is 0010 0000, then the second algorithm obviously needs to be improved

Allows N and n-1 to perform a & operation, if n& (n-1) ==0, the above case (the binary representation of N, only one is 0).

Code Listing 3:

#include <iostream>using namespace Std;int main (void) {int n,m;m=0;cin>>n;while (n) {n&= (n-1); m++;} cout << m <<endl;return 0;}

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The beauty of programming a number of binary representations of 1 in number