Strange fuction

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

题意：给你一个y值，x的范围是1~100；求所给方程的最小值；

解题思路：最小值就导数为零的点；

感悟：本来想用二分做，但是。。。。。。不会，后来想起来用导数做；

代码（G++）：

#include

#include

using namespace std;

double F(double x,double y)

{

    double
fx;

   
fx=6*x*x*x*x*x*x*x+8*x*x*x*x*x*x+7*x*x*x+5*x*x-x*y;

    return
fx;

}

double f(double x,double y)

{

    double
fx;

   
fx=42*x*x*x*x*x*x+48*x*x*x*x*x+21*x*x+10*x-y;

    return
fx;

}

int main()

{

   
//freopen("in.txt", "r", stdin);

    int n;

    double
y,x1,x2,x;

   
scanf("%d",&n);

    for(int
i=0;i

    {

       
scanf("%lf",&y);

       
x1=0;

       
x2=100;

       
while(true)

       
{

           
x=(x1+x2)/2;

           
if(x2-x1<1e-6)//这里精度最小就得是1e-6

           
{

               
printf("%.4lf\n",F(x,y));

               
break;

           
}

           
else

           
{

               
if(f(x,y)==0)

               
{

                   
printf("%.4lf\n",F(x,y));

                   
break;

               
}

               
else if(f(x,y)<0)

                   
x1=x;

               
else if(f(x,y)>0)

                   
x2=x;

           
}

       
}

    
}

    
return 0;

}