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 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*x*x-y*x;//定义一个求函数值得函数
}
double ds(double x,double y)
{
    return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*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)/2;
            if(ds(z,b)>0)
            y=z;
            else
            x=z;
        }while(y-x>1e-6);//求出一定精度内导数为0的大约值
        printf("%.4lf\n",hs(z,b));
    }
    return 0;
 } 

 

 下面是我刚学的三分法：

 

 

#include<stdio.h>
#include<math.h>
double hs(double x,double y)
{
    return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*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)/2;
            midmid=(mid+r)/2;
            if(hs(mid,a)>hs(midmid,a))
            l=mid;
            else
            r=midmid;
        }while(r-l>1e-6);
        printf("%.4lf\n",hs(mid,a));
    }
    return 0;
}