（HDU - 2899）Strange fuction

最新推荐文章于 2020-02-14 16:14:55 发布

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本文介绍了一道名为Strangefuction的编程题，目标是在给定参数y的情况下，通过三分法查找区间[0,100]内多项式函数F(x)的最小值。该文提供了完整的C++实现代码。

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（HDU - 2899）Strange fuction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7975 Accepted Submission(s): 5470

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

题目大意：给定y，当 $x∈[0,100]$ 时，求函数 $f(x)= 6 * x^7+8*x^6+7*x^3+5*x^2-y*x$ 的最小值。

思路：三分求解函数最值模板题。

#include<cstdio>
using namespace std;

const double eps=1e-8;
double y;

double f(double x) 
{
    return 6*x*x*x*x*x*x*x+8*x*x*x*x*x*x+7*x*x*x+5*x*x-y*x;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf",&y);
        double lo=0,hi=100,mid,midmid;
        while(hi-lo>eps)
        {
            mid=(lo+hi)/2.0;
            midmid=(hi+mid)/2.0;
            if(f(mid)<f(midmid)) hi=midmid;
            else lo=mid;
        }
        printf("%.4f\n",f(lo));
    }
    return 0;
}