4.1.2Strange fuction详细解答

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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

 

Recommend

lcy

 

此题大意：F(x) = 6*x^7+8*x^6+7*x^3+5*x^2-y*x, x ∈ [0, 100], 对于y ∈ (0, 10^10)，求F(x)的最小值

首先，对该式子，求一阶导，得到F'(x) = 42*x^6 + 48*x^5+21*x^2+10*x-y，可以发现，抛开y不管，那么42*x^6 + 48*x^5+21*x^2+10*x是递增的，范围在[0, 10^13]，所以我们可以发现无论y取什么值，总可以找到x1∈[0, 100]，使F'(x1)=0，又因为此时x∈[0, x1]时，函数递减；x∈[x1, 100]时，函数递增。所以F(x1)便是极小值，也是最小值。

综上，先对F'(x)使用二分法即可，再输出F(x)的值，便是最小值。

代码如下

//author: mission
#include <stdio.h>
#include <math.h>
double y;
double df(double x) // f'(x)
{
    return 42 * pow(x, 6) + 48 * pow(x, 5) + 21 * pow(x, 2) + 10 * x - y;
}
double f(double x)  //f(x)
{
    return 6 * pow(x, 7) + 8 * pow(x, 6) + 7 * pow(x, 3) + 5 * pow(x, 2) - y * x;
}
int main()
{
    int T;
    double x1, x2, x3, y1, y2, y3;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%lf", &y);
        x1 = 0;
        x2 = 100;
        y1 = df(x1);
        y2 = df(x2);
        //dichotomy
        while (fabs(x1 - x2) > 1e-6)
        {
            x3 = (x1 + x2) / 2;
            y3 = df(x3);
            if (y3 > 0)
                x2 = x3;
            else
                x1 = x3;
        }
        printf("%.4f\n", f(x3));
    }
    return 0;
}

 

    1e-6