写一个函数返回参数二进制中 1 的个数（三种方法）

1.运用了除法，取余方式递推出结构 2.运用右移符(>>)运算 3.利用算术与（&）运算

三种方法效率越来越高，减少成本

#include<stdio.h>

int Number1(int n)
{
	int k;
	int count=0;
	while (n > 0)
	{
		k = n % 2;
		n /= 2;
		if (1 == k)
		{
			count++;
		}
	}
	return count;
}

int Number2(int n)
{
	int count = 0;
	while (n>0)
	{
		if ((n & 1) == 1)
		{
			count++;
		}
		n = n >>1 ;
	}
	return count;
}

int Number3(int n)
{
	int count = 0;
	while (n)
	{
		n = n&(n - 1);
		count++;
	}
	return count;
}
int main()
{
	int n;
	printf("请输入一个数:\n");
	scanf("%d",&n);
	int ret1=Number1(n);
	printf("%d\n",ret1);
	int ret2 = Number2(n);
	printf("%d\n",ret2);
	int ret3 = Number3(n);
	printf("%d\n",ret3);
	system("pause");
	return 0;
}