Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4795    Accepted Submission(s): 3420

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

简单二分+数学：求导后二分找与y最接近的值,此时原函数最小；
#include<stdio.h>
#include<math.h>
double y;
double hanshu(double x)//原函数值
{
	return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;
}
double daoshu(double x)//导数值
{
	return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y;
}
int main()
{
	int n;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%lf",&y);
		if(y>=daoshu(100))
		printf("%lf\n",hanshu(100));
		double right =100,left=0;
		while((right-left)>=1e-10)//二分
		{
			double mid=(right+left)/2;
			if(daoshu(mid)>0)
			right=mid;
			else
			left=mid;
		}
		printf("%.4lf\n",hanshu(right));
	}
	return 0;
}

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