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 
    
     # include 
     
      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;}