HDU2899：Strange fuction（二分 + 数学）

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6874    Accepted Submission(s): 4807

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

 

题意：给定y值，求上面函数在[0, 100]内的最小值。

思路：先对函数求导，发现导函数是递增的，若f'(100) > 0，二分求出导函数的零点，代入原函数即可。

# include <stdio.h>
# include <math.h>
int y;
double fun1(double x)
{
    return 6*pow(x, 7)+8*pow(x, 6)+7*pow(x, 3)+5*pow(x, 2)-y*x;
}

double fun2(double x)
{
    return 42*pow(x, 6)+48*pow(x, 5)+21*pow(x, 2)+10*x-y;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&y);
        if(fun2(100) <= 0)
            printf("%.4f\n",fun1(100));
        else
        {
            double l = 0, r = 100, mid;
            while(r-l > 1e-7)
            {
                mid = (l+r)/2;
                if(fun2(mid) > 0)
                    r = mid;
                else
                    l = mid;
            }
            printf("%.4f\n",fun1(mid));
        }
    }
    return 0;
}