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

 

Author

Redow

 

很水的题 但是WA了几次

虽说最后要求1e-4的精度 

但是因为求最小值 所以注意所求的x值精度 至少要1e-6 所以请注意

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <cstring>

#include <ctime>

#include <algorithm>

#include <iostream>

#include <sstream>

#include <string>

#define oo 0x13131313

using namespace std;

double y;

void init()

{

	freopen("a.in","r",stdin);

	freopen("a.out","w",stdout);

}

double cal1(double x)

{

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

}

double  cal2(double x)

{

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

}

void solve()

{

	double L=0,R=100,ans=1e10;

	double temp1=cal1(L),temp2=cal1(R);

	if(temp1<ans) ans=temp1;

	if(temp2<ans) ans=temp2;

	double ANS;

	while(R-L>1e-6)

	{

		double m=(R+L)/2;

		if(cal2(m)>0) R=m;

		else if(cal2(m)<0) L=m;

		else { ANS=m;break; }

	}

	ANS=R;

	if(cal1(ANS)<ans) ans=cal1(ANS);

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

}

int main()

{

//	init();

	int T;

	cin>>T;

	while(T--)

	{

		cin>>y;

		solve();

	}

	return 0;

}