#define e 0.00000001
#define r 0.618
#include <math.h>
#include <iostream.h>
int p=1;
double f(double a)
{
 double y;
 y=pow(a,2)+2*a;
 return y;
}

void main()
{
 double ak;
 double a1,a2,y1,y2;
 double a=-3.0,b=5.0;
 a1=b-r*(b-a);
 a2=a+r*(b-a);
 y1=f(a1);
 y2=f(a2);
 while(fabs((b-a)/b)>=e)
 {
  if(y2<y1)
  {
   //cout<<"循環(huán)了"<<a1<<endl;
   a=a1;
   a1=a2;
   y1=y2;
   a2=a+r*(b-a);
   y2=f(a2);
  }
  else 
  {
   b=a2;
   a2=a1;
   y2=y1;
   a1=b-r*(b-a);
   y1=f(a1);
  }
  cout<<"循環(huán)了"<<p<<"次"<<endl;
  p++;
  //if(p>5)
  // break;
 }
 ak=(a+b)/2;
 cout<<"ak="<<ak<<endl;
 cout<<"f(ak)="<<f(ak)<<endl;
}

//從運行結果可以看出，其解與書上說的精確解已經一樣了