hrbust 2024/hdu2899 Strange fuction【三分】

最新推荐文章于 2019-04-06 10:27:03 发布

原创最新推荐文章于 2019-04-06 10:27:03 发布 · 447 阅读

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本文介绍了一种求解特定形式复杂函数在指定区间内最小值的方法。通过使用三分法而非二分法来应对函数非单调的问题，给出了具体的C语言实现代码及样例输入输出。

Strange fuction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5275 Accepted Submission(s): 3767

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

因为有这样的一组数据，所以直接三分是不能实现的：

测试代码：

        for(int i=0;i<=100;i++)
        {
            double x=i;
            printf("%.3lf\n",f(x,y));
        }

因为不是单调的，所以直接二分是不行的。这里强调一下二分必须是单调的才可以直接使用~

所以这里上三分代码即可：

#include<stdio.h>
#include<string.h>
#include<math.h>
using namespace std;
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;
    scanf("%d",&t);
    while(t--)
    {
        double y;
        scanf("%lf",&y);
        double output=0x1f1f1f1f;
        double l=0;
        double r=100;
        double lm,rm;
        while(r-l>1e-6)
        {
            lm=(2*l+r)/3;
            rm=(l+2*r)/3;
            if(f(lm,y)>f(rm,y))l=lm;
            else r=rm;
        }
        printf("%.4lf\n",f(l,y));
    }
}