二分（解方程，求导）

最新推荐文章于 2020-10-15 15:51:38 发布

凌晨四点的洛杉矶fly

最新推荐文章于 2020-10-15 15:51:38 发布

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分类专栏：二分

本文链接：https://blog.youkuaiyun.com/qq_40859951/article/details/87469557

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本文介绍了一种通过二分查找法求解特定区间内多项式函数最小值的方法。该方法适用于形如F(x)=6*x^7+8*x^6+7*x^3+5*x^2-y*x(0<=x<=100)的多项式，其中y为输入参数。通过不断缩小搜索范围，算法能够精确到小数点后四位，快速找到函数在[0,100]区间内的最小值。

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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

#include<cstdio>
#include<cmath>
 #include<iostream>
 #include<algorithm>
double df(double x)
{
	return 42*pow(x,6)+48*pow(x,5)+21*x*x+10*x;
}
 
double f(double x,double y)
{
	return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*x*x-y*x;
}
 
int main()
{
	int t,i;
	double y,left,right,mid;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lf",&y);
		left=0,right=100;
		for(i=1;i<=100;i++)
		{
			mid=(left+right)/2;
			if(df(mid)>y)
				right=mid;
			else
				left=mid;
		}
		printf("%.4lf\n",f(mid,y));
	}
	return 0;
}