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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 //0MS    224K    590 B    G++
 /*

     题意：
         给出一关系式：
             F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
         和y的值，求x在范围(0,100)内时F(x)能取得的最小值

     二分法：
         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
     再代入原式即可求出最值   

 */
 #include<stdio.h>
 #define e 1e-7
 double fac(double a,double b)
 {
     return *a*a*a*a*a*a*a+*a*a*a*a*a*a+*a*a*a+*a*a-b*a;
 }
 double fac0(double a,double b)
 {
     return *a*a*a*a*a*a+*a*a*a*a*a+*a*a+*a-b;
 }
 int main(void)
 {
     int t;
     double n;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%lf",&n);
         double l=;
         double r=;
         while(r-l>e){
             double mid=(l+r)/;
             if(fac0(mid,n)>) r=mid;
             else l=mid;
         }
         printf("%.4lf\n",fac(r,n));
     }
     return ;
 }