HDUOJ--Strange fuction

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2098    Accepted Submission(s): 1577

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

http://acm.hdu.edu.cn/showproblem.php?pid=2899

代码：

 1 #include<iostream>

 2 #include<cmath>

 3 #include<cstdio>

 4 using namespace std;

 5 double y;

 6 double sum(double x)

 7 {

 8     return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;

 9 }

10 double func(double x)

11 {

12     return 42*pow(x,6)+48*pow(x,5)+21*x*x+10*x-y;

13 }

14 int main()

15 {

16     int t;

17     double mid,left,right;

18     cin>>t;

19     while(t--)

20     {

21         scanf("%lf",&y);

22         if(func(100)>0)

23         {

24         left=0.0,right=100.0;

25         while(right-left>1e-8)

26         {

27           mid=(right+left)/2.0;

28           if(func(mid)>0.0)

29                 right=mid;

30         else

31             if(func(mid)<0.0)

32                 left=mid;

33             else

34                 break;

35         }

36         printf("%.4lf\n",sum(mid));

37         }

38         else

39             printf("%.4lf\n",sum(100.0));

40     }

41     return 0;

42 }

View Code