hdu 2899 Strange fuction
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3212    Accepted Submission(s): 2351




Problem Description
Now, here is a fuction:
  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
Can you find the minimum value when x is between 0 and 100.
 


Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has only one real numbers Y.(0 < Y <1e10)
 


Output
Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.
 


Sample Input
2
100
200
 


Sample Output
-74.4291
-178.8534


/*题解：
二分法求解，精度1e-7 ，原理简单，不知道最好在纸上比划比划 

*/ 

#include<cstdio>
#include<cmath> 
double ys(double x,double y)
{
  return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y; 
}
double ans(double x,double y)
{
  return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x; 
}
int main()
{
  int T;
  double y,fx;
  scanf("%d",&T);
  while(T--)
  {
    scanf("%lf",&y);
    double f1,f2;
    f1=0,f2=100;
    while(f2-f1>1e-7)
    {
      fx=(f2+f1)/2;
      if(ys(fx,y)>0)
      {
        
        f2=fx-1e-9;
        
      }
      else
      {
        f1=fx+1e-9;
      }
    }
    printf("%.4lf\n",ans(fx,y));
  }
  return 0;
}

/*


三分法可直接求解


定义了L和R，m = (L + R) / 2，mm = (m + R) / 2; 如果m靠近极值点，则R = mm；否则就是mm靠近极值点，则L = m ，和二分法原理基本一样。


*/ 


#include<cstdio>
#include<cmath>
int y;
double fun(double x)
{
  return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-x*y;
} 
double search()
{
  double L,R,mid,midmid;
  L=0,R=100;
  while(R-L>1e-6)
  {
    mid=(L+R)/2;
    midmid=(mid+R)/2;
    if(fun(mid)<fun(midmid))
    R=midmid;
    else
    L=mid; 
  }
  return L;
}
int main()
{
  int T;
  scanf("%d",&T);
  while(T--)
  {
    scanf("%d",&y);
    printf("%.4lf\n",fun(search()));
  } 
}