Scratch brush HDOJ (3) [HDOJ2899 - Strange fuction]

Scratch brush HDOJ (3) [HDOJ2899 - Strange function]

Face questions

Strange fuction

Time limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7321 Accepted Submission(s): 5057

Problem Description

Now, here is a function:

$$F(x) = 6x^7+8x^6+7x^3+5x^2-yx (0 ≤ x ≤100)$$

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

Submit Portal: the Submit

translation

When it is to seek$y$When you enter a value equal to$F(x)$[0, 100] on zero.

Thinking

In fact, this function is unimodal in [0, 100] on.

Proof:

​ $\because (kx^\mu)' = k\mu (x)^{(\mu - 1)}$

​ $\therefore F'(x) = 42x^6 + 48x^5 + 21x^2 + 10x – y$

​ $\because y=ax^6在[0, 100]上单调增，y=ax^5在[0, 100]上单调增，y=ax^2在[0, 100]上单调增，y=ax在[0, 100]上单调增$

​ $\therefore F'(x)在[0, 100]上单调增$

Since we can prove$F(x)$The derivative function$F'(x)$Monotonically increasing in the domain, then we can define the binary zero field (refer to the specific method for a high school mathematics compulsory), find the value of this point is the point we are asking for. Final output$F(ans)$Enough.

Code

#include <cmath>
#include <iostream>

// F(x) = 6x7 + 8x6 + 7x3 + 5x2 – xy 

double y;

inline double f(const double x)
{
    return 6 * std::pow(x, 7) + 8 * std::pow(x, 6) + 7 * std::pow(x, 3) + 5 * std::pow(x, 2) - y * x;
}

inline double F(const double x)
{
    return 42 * std::pow(x, 6) + 48 * std::pow(x, 5) + 21 * std::pow(x, 2) + 10 * x;
}

int main(int argc, char ** argv)
{
    std::ios_base::sync_with_stdio(false);
    std::cout.setf(std::ios_base::fixed);
    std::cout.precision(4);
    double ans;
    double l = 0.0, r = 100.0;
    int times = 1000;
    int T;
    std::cin >> T;
    while (T--)
    {
        std::cin >> y;
        l = 0.0;
        r = 100.0;
        times = 10000;
        double eps = 1e-10;
        while (times--)
        {
            double mid = (l + r) / 2.0;
            if (F(mid)>y)
                r = mid;
            else
                l = mid;
        }
        std::cout << f(l) << std::endl;
    }
    std::cin.get();
    std::cin.get();
    return 0;
}