求一个正整数N的开方，要求不能用库函数sqrt()，结果的精度在0.001

 #include<iostream>
 using namespace std;
 int main()
 {
     int N;
     cout<<"输入N的值：";
     cin>>N

     double x1 = 1;//初值
     double x2 = x1/2.0+N/2.0;
     while( fabs(x2-x1)>0.001)
     {
         x1 = x2;
         x2 = x1/2.0+N/2.0;
     }
     cout<<x1<<endl;

     return 0;
 }