hdu Strange fuction

Strange fuction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 108 Accepted Submission(s): 96

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：

求函数的最小值，首先求导的导函数为：G(x) = 42 * x^6+48*x^5+21*x^2+10*x-y (0 <= x <=100)

分析导函数的，导函数为一个单调递增的函数。如果导函数的最大值小于0，那么原函数在区间内单调递减。

即F(100)最小；如果但函数的最小值大于0，那么原函数在区间内单调递增，即F(0)最小。如果导函数既有正又有负

又由于导函数是单增函数，所以必有先负后正，即原函数必有先减后增的性质。求出导函数的零点就是原函数的最小值点。

求导函数最小值方法是2分法.

#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;
double f(double x,double y)
{
    return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;
}
double func(double x, double y)
{
    return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x-y;
}
int main()
{
    int t;
    double y,left,right,mid,ans1,ans2,ans3;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf",&y);
        if(func(0.0,y)>=0)
        printf("%.4f\n",f(0.0,y));
        else if(func(100.0,y)<=0)
        printf("%.4f\n",f(100.0,y));
        else
        {
            left = 0.0;
            right = 100.0;
            mid = 50.0;
            ans1 = func(left,y);
            ans2 = func(mid,y);
            ans3 = func(right,y);
            while(fabs(ans1-ans2)>0.0001)
            {
                if(ans2>0)
                {
                    ans3 = ans2;
                    right = mid;
                    mid = (left+right)/2;
                }else
                {
                    ans1 = ans2;
                    left = mid;
                    mid = (left+right)/2;
                }
                ans2 = func(mid,y);
            }
            printf("%.4f\n",f(mid,y));
        }
    }
    return 0;
}