C语言-操作符-典型例题多种解法(求一个数二进制中1的个数)

2021_11_18

（此做法就类似于对于一个十进制的数，如何得到它的每一位）

#include 

int count_number_if_1(unsigned int n)
{
	int c = 0;
	while (n)
	{
		if (1 == (n % 2))
		{
			c++;
		}
		n /= 2;
	}
    return c;
}

int main()
{
	int n = 0;
	scanf("%d", &n);
	int ret = count_number_of_1(n);
	printf("%dn", ret);
	return 0;
}

#include 

int count_number_of_1(int n)
{
	int c = 0;
	while (n)
	{
		if (1 == (n & 1))
		{
			c++;
		}
		n >>= 1;
	}
	return c;
}


int main()
{
	int n = 0;
	scanf("%d", &n);
	int ret = count_number_of_1(n);
	printf("%dn", ret);
	return 0;
}

#include 

int count_number_of_1(int m)
{
	int c = 0;
	while (m)
	{
		m = m & (m - 1);
		c++;
	}
	return c;
}


int main()
{
	int n = 0;
	scanf("%d", &n);
	int ret = count_number_of_1(n);
	printf("%dn", ret);
	return 0;
}