Sqrt

牛顿迭代法

 1 double NewtonMethod(double fToBeSqrted)
 2 {
 3     double x = 1.0;
 4     while(abs(x*x-fToBeSqrted) > 1e-5)
 5     {
 6         x = (x+fToBeSqrted/x)/2;
 7     }
 8 
 9     return x;
10 }

二分法

 1 float get_sqrt(float x)
 2 {
 3     float low=0, up=x, mid, now;
 4     mid=(low+up)/2;
 5     do
 6     {
 7         now=mid;        //**now保存上一次计算的值
 8         if(mid*mid<x)   //**mid偏小,右移
 9         {
10             low=mid;
11         }
12         else       //**mid偏大,左移
13         {
14             up=mid;
15         }
16         mid=(low+up)/2;
17     }while(abs(mid-now)>eps); //**两次计算的误差小于eps,mid即为所求值
18     return mid;
19 }
原文地址:https://www.cnblogs.com/Juntaran/p/5903010.html