#include
     
      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。便可降低一个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;}