不使用库函数sqrt实现求一个数的平方根

二分法：

double mysqrt(double a)
{
    if(a == 0 ) return 0;
    double precision = 1.0e-7, start = 0, end = a;
    if(a < 1) end = 1;
    while(end - start > precision)
    {
        double mid = (start + end) / 2;
        if( mid == a / mid) return mid;
        else if(mid > a/mid) end = mid;
        else start = mid;
    }
    return (start + end)/2;
}

 

牛顿迭代：

/*  牛顿法求解 */
#define E 0.0000001
double newton(double number, int *count)  //*count 可统计出迭代次数
{
    double x0 = number;
    double x1;
    if(x0<E)return 0;
    while(true)
    {
        (*count)++;
        x1 = -(x0*x0 - number) / (2 * x0) + x0;
        //x1 = (x0*x0 + number) / (2 * x0)
        if(x1 * x1 - number <= E && x1 * x1 - number >= -E)
            return x1;
        x0 = x1;
    }
    return 0;
}