#include<stdio.h>

int numberOf1_solution1(int n)/*将一个正数以此向右移一位，与1做与运算，直到这个数为零*/
{
	int count = 0;
	while (n)
	{
		if (n&1)
			count++;
		n=n >> 1;
	}
	return count;
}

int numberOf1_solution2(int n)/*将1以此向右移动以为，与一个数（正负数均可）做与运算，直到1出现上溢为止*/
{
	int count = 0, i = 1;
	while (i)
	{
		if (n&i)
			count++;
		i = i << 1;
	}
	return count;
}

int numberOf1_solution3(int n)/*将一个数减去1后再与本身相与，便可减少一个1，利用这个原理求1的个数*/
{
	int count = 0;
	while (n)
	{
		n = (n - 1)&n;
		count++;
	}
	return count;
}

int main()
{
	printf("%d\n",numberOf1_solution3(5));
	return 0;
}