hdu 2899 Strange fuction (二分法)

Strange fuction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2278 Accepted Submission(s): 1697

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

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 1 //0MS    224K    590 B    G++
 2 /*
 3     
 4     题意：
 5         给出一关系式： 
 6             F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100) 
 7         和y的值，求x在范围(0,100)内时F(x)能取得的最小值
 8     
 9     二分法：
10         很容易可以看出函数F(x)在(0,100)内是先减后增的，故用二分法求其倒数的零值点，然后用零值
11     再代入原式即可求出最值   
12          
13 */
14 #include<stdio.h>
15 #define e 1e-7
16 double fac(double a,double b)
17 {
18     return 6*a*a*a*a*a*a*a+8*a*a*a*a*a*a+7*a*a*a+5*a*a-b*a;    
19 } 
20 double fac0(double a,double b)
21 {
22     return 42*a*a*a*a*a*a+48*a*a*a*a*a+21*a*a+10*a-b;       
23 }
24 int main(void)
25 {
26     int t;
27     double n;
28     scanf("%d",&t);
29     while(t--)
30     {
31         scanf("%lf",&n);
32         double l=0;
33         double r=100;
34         while(r-l>e){
35             double mid=(l+r)/2;
36             if(fac0(mid,n)>0) r=mid;
37             else l=mid;
38         }
39         printf("%.4lf
",fac(r,n));
40     }
41     return 0;
42 }