Find out how many 1 different solutions in a number bits

* Returns the number of 1 in a binary representation of a number. *

#include <stdio.h>int main () {    unsigned int i;    &NBSP;SCANF ("%d", &i);     printf ("%d binary expression has  %d  1", i,count_one_bits (i));     return 0;} Int  count_one_bits (unsigned int m) {    int n,count;     //  return  1 number of digits     n = m;    count =  0;    while (n % 2 == 0)          n = n / 2;    while (n % 2 == 1)      {        count++;         do        {             if  (n != 0)                 n = n / 2;             else                 break;        }         while (n % 2 == 0);     }      return count;}

This is the most easy way to think of, and constantly in addition to 2 modulo 2 to take the remainder, the following gives another method (omit part of the statement)

... i = 32;while (i--) {if (num & 1 = = 1) count++; num = num>>1;}

This method is shifted to the right, so that the lowest bits & 1 are judged, and a more efficient method

int count_one_bit (int num) {int count = 0;    while (num) {count++; As long as a number is not 0, then there is at least one 1 num in the binary expression = num & (num-1);//The statement is not executed once, in fact, the lowest bit of the number of a 1 to 0} return count;

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How many different solutions to 1 in a number bits