HDU 2899

Strange fuction

Problem Description Now, here is a fuction:
F(x) = 6 * x^7+8*x6+7x^3+5*x2-yx (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 Input2
100
200

Sample Output
-74.4291
-178.8534

F(x) = 6 * x^7+8*x6+7x^3+5*x2-yx (0 <= x <=100)
有两种方法
第一种 二分查找 用函数的导数来写
第二种 三分查找

第一种

#include <stdio.h>
#include <math.h> 
double y;
double F(double x)
{
    return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;
}
double f(double x)
{
    return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y;
}
int main(int argc, char *argv[])
{
    int t;
    double m,l,r;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf",&y);
        if(f(100)<=0)
        {
            printf("%.4lf\n",F(100.0));
        }
        else 
        {
            l=0.0;r=100.0;
            while(r-l>1e-8)
            {
                m=(l+r)/2.0;
                if(f(m)<0)
                    l=m;
                else 
                    r=m;
            }
            printf("%.4lf\n",F(m));
        }            
    }
    return 0;
}

第二种
三分查找 （数据需要是凸型的）

#include<stdio.h>
#include<math.h>
int y;
double sum;
double cal(double x)
{
 sum=6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;
    return sum;
}
double search()
{
 double left=0,right=100,mid1,mid2;
 while(right-left>1e-6)
 {
  mid1=(left*2+right)/3;
     mid2=(left+right*2)/3;
  if(cal(mid1)>cal(mid2))
   left=mid1;
  else right=mid2;
 }
 return  cal(right);
  
}
int main()
{
    int num,i;
    scanf("%d",&num);
    for(i=0;i<num;i++)
 {
 scanf("%d",&y);
 search();
 printf("%.4lf\n",sum);
 }
}