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;
}