Problem Description

Now, here is a fuction:

F(x) = 6 * x^7+8*x6+7x^3+5*x2-yx (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

Author

Redow

---

思路

函数是：\(F(x) = 6x^7+8x^6+7x^3+5x^2-yx\)

导函数是：\(F'(x) = 42x^6 +48x^5+21x^2+10x-y\)

由题目条件可得，导函数单调递增，\(F'(100)\)最大，如果它还小于0，说明原函数单调递减，\(F(100)\)最小。其他情况是：二分查找导函数的零点，找到后代入原函数就可以得到答案

代码

#include<bits/stdc++.h>

using namespace std;

double y;

double f(double x)

{

	return 6*pow(x,7) + 8*pow(x,6) + 7*pow(x,3) + 5*pow(x,2) - y*x;

}//函数

double df(double x)

{

	return 42*pow(x,6) + 48*pow(x,5) + 21*pow(x,2) + 10*x - y;

}//导函数

int main()

{

	int n;

	cin >> n;

	while(n--)

	{

		cin >> y;

		if(df(100) <= 0)

		{

			printf("%.4lf\n",f(100));

			continue;

		}

		double l = 0, r = 100.0;

		double mid;

		while(r-l>=1e-8)

		{

			mid = (l+r)/2;

			if(df(mid)<0)

				l = mid;

			else

				r = mid;

		}

		printf("%.4lf\n",f(mid));

	}

	return 0;

}