strange function

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本文介绍了一种通过求导及二分法逼近的方法来寻找特定区间内多项式函数的最小值。利用C++实现该算法，并通过样例验证了其准确性。

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Strange fuction

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 45 Accepted Submission(s) : 36

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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：找出那个函数在x属于0~100内的最小值。
思路：首先想到的应该是求导求驻点，然后用二分法逼近。
#include "cstdio"
#include "cstring"
#include "cmath"
#define mi 0.00001
using namespace std;
double x,y;
double f(double x)
{
	return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y;
}
double fun(double x){
	return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-x*y;
}
int main()
{
	int T;
	
	scanf("%d",&T);
	while(T--)
	{
		scanf("%lf",&y);
		double l=0,r=100;
		double min;
		double mid=(l+r)/2;
		while(fabs(f(mid))>mi)
		{
			mid=(l+r)/2;
			double a=f(mid);
			if(a>=0)
			{
				r=mid;
			}
			else if(a<0)
			{
				l=mid;
			}
		}
		printf("%.4lf\n",fun(mid));
	}
	return 0;
}