HDU 2899 Strange fuction（三分）

最新推荐文章于 2022-01-29 18:12:38 发布

tenglinxi

最新推荐文章于 2022-01-29 18:12:38 发布

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本文链接：https://blog.youkuaiyun.com/qq_32473657/article/details/51622105

本文探讨了一个特定的数学函数Strangefunction，并通过编程方式寻找该函数在x∈[0,100]区间内的最小值。输入为一个实数Y，输出为对应的最小值。

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Strange fuction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5755 Accepted Submission(s): 4081

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

Sample Output

  
   -74.4291
-178.8534

Author

Redow

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代码：

#include<stdio.h>
#include<string.h>
#include<cmath>
using namespace std;
double f(double x,double y)
{
    return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        double y;
        scanf("%lf",&y);
        double l=0,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));
    }
}