c语言| |二进制中1的个数

二进制中1的个数


//第一种:短除法
int count_one(unsigned num)
{
    int count = 0;
    while (num != 0)
    {
        if (num % 2 == 1)
        {
            count++;
        }
        num /= 2;
    }
    return count;
}

//第二种:移位按位与1法
int count_one(int num)
{
    int count = 0;
    int i = 0;
    for (i = 0; i < 32; i++)
    {
        count += (num >> i) & 1;
    }
    return count;
}

//第三种:num与num-1进行按位与
int count_one(int num)
{
    int count = 0;
    while (num != 0)
    {
        num = num & (num - 1);
        count++;
    }
    return count;
}

int main()
{
    int num = 0;
    int count = 0;
    printf("请输入一个数字>:");
    scanf("%d", &num);
    count = count_one(num);
    printf("1的个数为:%d\n", count);
    return 0;
}

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转载自blog.csdn.net/qq_40399012/article/details/80879519