Strange fuction--hdu2899

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4538    Accepted Submission(s):
3261

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

 

分析：这个题是要求方程的最小值，首先我们来看一下他的导函数： F’(x) = 42 * x^6+48*x^5+21*x^2+10*x-y(0 <= x <=100)

很显然，导函数是递增的，那么只要求出其导函数的零点就行了，下面就是用二分法求零点！

 

 #include<stdio.h>
 #include<math.h>
 double hs(double x,double y)
 {
     return *pow(x,)+*pow(x,)+*pow(x,)+*x*x-y*x;//定义一个求函数值得函数
 }
 double ds(double x,double y)
 {
     return *pow(x,)+*pow(x,)+*pow(x,)+*x-y;//定义一个求导数的函数
 }
 int main()
 {
     int a;
     scanf("%d",&a);
     while(a--)
     {
         double b,x,y,z;
         scanf("%lf",&b);
         x=0.0;
         y=100.0;
         do
         {
             z=(x+y)/;
             if(ds(z,b)>)
             y=z;
             else
             x=z;
         }while(y-x>1e-);//求出一定精度内导数为0的大约值
         printf("%.4lf\n",hs(z,b));
     }
     return ;
  } 

 下面是我刚学的三分法：

 

 

 #include<stdio.h>
 #include<math.h>
 double hs(double x,double y)
 {
     return *pow(x,)+*pow(x,)+*pow(x,)+*x*x-y*x;
 }
 int main()
 {
     int n;
     scanf("%d",&n);
     while(n--)
     {
         double l,r,mid,midmid,a;
         scanf("%lf",&a);
         l=0.0;
         r=100.0;
         do
         {
             mid=(l+r)/;
             midmid=(mid+r)/;
             if(hs(mid,a)>hs(midmid,a))
             l=mid;
             else
             r=midmid;
         }while(r-l>1e-);
         printf("%.4lf\n",hs(mid,a));
     }
     return ;
 }