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. InputThe 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) OutputJust 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

看了大神的題解（沒錯我就是每道題都要看題解的菜菜）才知道要用求導算，得自己解，不能讓程式解。虧我之前還傻乎乎的想讓程式給我求出來……hehe

：

#include<stdio.h>  
#include<math.h>  
  
double df(double x) //定義double型別函式，宣告變數x 
{  
    return 42*pow(x,6)+48*pow(x,5)+21*x*x+10*x;  //返回這個函式的一階導數 -（y*x）沒了 
}  
  
double f(double x,double y)//再定義一個函式， 
{  
    return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*x*x-y*x;  //返回原函式
}  
  
int main()  
{  
    int T, i;  
    double y, left, right, mid;  //二分 
    scanf("%d", &T);  
    while(T--)  
    {  
        scanf("%lf",&y);  
        left = 0, right = 100;  //x的範圍 
        for(i = 1; i <= 100; i++)  //三分法 
        {  
            mid = (left + right) / 2;  
            if(df(mid) > y)  
                right = mid;  
            else  
                left = mid;  
        }  
        printf("%.4lf\n", f(mid, y));  
    }  
    return 0;  
}