不使用库函数sqrt实现求一个数的平方根

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;
}