Strange fuction

最新推荐文章于 2024-04-19 20:40:32 发布

原创最新推荐文章于 2024-04-19 20:40:32 发布 · 935 阅读

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本文介绍了一种通过求导和二分法寻找特定区间内多项式函数最小值的方法，并提供了完整的C++代码实现。

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

100

200

Sample Output

-74.4291

-178.8534

分析：这个题是要求方程的最小值，首先我们来看一下他的导函数： F’(x) = 42 * x^6+48*x^5+21*x^2+10*x-y(0 <= x <=100)

很显然，导函数是递增的，那么只要求出其导函数的零点就行了，下面就是用二分法求零点！

#include <iostream>
#include<cstdio>
#include<cmath>
const double eps = 1e-6;
using namespace std;
double hs(double x,double y)
{
	return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;
}
double ds(double x,double y)
{
	return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*pow(x,1)-y;	
}
int main(){
	int a;
	scanf("%d",&a);
	while(a--)
	{
		double b,x,y,z;
		scanf("%lf",&b);
		x=0.0;
		y=100.0;
		while(y-x>eps){
			z=(x+y)/2;
			if(ds(z,b)>0){
				y=z;
			}
			else{
				x=z;
			}
		}
		printf("%.4lf\n",hs(z,b));
	}
	return 0;
}