用递归求一个正数的开平方，精确到小数点后3位，要求使用递归实现

Question

用递归求一个正数的开平方，精确到小数点后3位，要求使用递归实现

Answer 1

 #include <stdio.h>
double sq(double d, double r1, double r2)
{
    double diff = ((r1 + r2) * (r1 + r2)) / 4.0 - d;
    if (diff >= -0.000001 && diff <= 0.000001) return (r1 + r2) / 2;
    if (diff < -0.000001)
        return sq(d, (r1 + r2) / 2.0, r2);
    else
        return sq(d, r1, (r1 + r2) / 2.0);
}
int main()
{
    double x = 2.0;
    double r = sq(x, 0, x);
    printf("%lf", r);
}