【hdu2899】Strange fuction——三分

最新推荐文章于 2024-08-22 20:10:34 发布

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最新推荐文章于 2024-08-22 20:10:34 发布

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分类专栏：题解心得三分文章标签： hdu 三分

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本文介绍了一种使用三分枚举算法解决给定高次函数在特定区间内的最小值问题的方法，详细解释了算法原理，并通过实例展示了如何应用此算法求解实际问题。

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题目：

Strange fuction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5360 Accepted Submission(s): 3832

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

描述：给定含参数的高次函数，求最小值

题解：单峰函数的最值问题一般用三分枚举就好了

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const double eps = 1e-8;
double f(double x, double y)
{
	return 6.0 * pow(x, 7) + 8 * pow(x, 6) + 7 * pow(x, 3) + 5 * pow(x, 2) - y * x;
	// F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
}
int main()
{
	//freopen("input.txt", "r", stdin);
	int T;
	scanf("%d", &T);
	while (T--)
	{
		double y;
		scanf("%lf", &y);
		double l = 0.0;
		double r = 100.0;
		while (r - l > eps)
		{
			double mid1 = (r + l) / 2;
			double mid2 = (r + mid1) / 2;
			if (f(mid1, y) > f(mid2, y))
				l = mid1;
			else
				r = mid2;
		}
		printf("%.4lf\n", f(r,y));
	}
	return 0;
}