HDU-2899.三分初步-求最小值

最新推荐文章于 2021-05-08 12:30:04 发布

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本文介绍了一个特定的数学函数Strangefuction，并通过三分法算法找到该函数在0到100区间内的最小值。输入包括一系列测试案例，每个案例包含一个参数Y，输出则是对应情况下函数的最小值。

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Strange fuction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4362 Accepted Submission(s): 3145

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

一道很水的题目，用二分三分都可以，二分的话先求导，用二分法求导数为零的点，求最小值；三分法直接求最小值。

只贴三分的代码

#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;
#define eps 10e-7<span style="white-space:pre">		</span>//虽然保留四位小数，但是精度取10e-4的话是过不了的 -.-
double y;
double f(double x)
{
    return 6*pow(x,7) + 8*pow(x,6) + 7*pow(x,3) + 5*pow(x,2) -y*x;
}
double search()
{
    double left = 0, right = 100, mid1, mid2;
    while(right - left > eps)
    {
        mid1 = (2*left + right) / 3;
        mid2 = (left + right*2) / 3;
        if(f(mid1) < f(mid2))
        {
            right = mid2;
        }
        else if(f(mid1) > f(mid2))
        {
            left = mid1;
        }
    }
    return mid1;
}

int main()
{
    int T;
    cin >> T ;
    while(T--)
    {
        scanf("%lf", &y);
        printf("%.4f\n", f(search()));
    }
}