Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7230    Accepted Submission(s): 4996

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

二分先决条件是满足单调性，这里显然不满足不能直接二分，这道题目求导就可以解决，求出极值点，如果求导不方便的题目就得用三分搜索了。

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<math.h>
 4 #define exp 1e-8
 5
 6 using namespace std;
 7
 8 double y;
 9
10 double suan(double x)
11 {
12     return 42*pow(x,6)+48*pow(x,5)+21*x*x+10*x-y;
13 }
14 double binary(double low,double high)
15 {
16     while(high-low>exp)
17     {
18         double mid = (low+high)/2;
19         if(suan(mid)>0)
20             high=mid-exp;
21         else
22             low=mid+exp;
23     }
24     return low;
25 }
26 int main()
27 {
28     int t;
29     scanf("%d",&t);
30     while(t--)
31     {
32         scanf("%lf",&y);
33         double ans = binary(0,100);
34         printf("%.4f\n",6*pow(ans,7)+8*pow(ans,6)+7*pow(ans,3)+5*ans*ans-y*ans);
35     }
36     return 0;
37 }

三分查找

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<math.h>
 4 #define exp 1e-8
 5
 6 using namespace std;
 7
 8 double y;
 9
10 double suan(double x)
11 {
12     return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*x*x-y*x;
13 }
14 double binary(double low,double high)
15 {
16     double mid,mmid;
17     while(high-low>exp)
18     {
19         mid = (low+high)/2;
20         mmid = (high+mid)/2;
21         if(suan(mid)>suan(mmid))
22             low=mid+exp;
23         else
24             high=mmid-exp;
25     }
26     return (mid+mmid)/2;
27 }
28 int main()
29 {
30     int t;
31     scanf("%d",&t);
32     while(t--)
33     {
34         scanf("%lf",&y);
35         double ans = binary(0,100);
36         printf("%.4f\n",suan(ans));
37     }
38     return 0;
39 }