HDU 2899 Strange fuction (二分)

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1840    Accepted Submission(s): 1364

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

 

 

Recommend

lcy

 

 

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<cstdio>

#include<cmath>



using namespace std;



#define eps 1e-10



double y;



double g(double x){

    return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x;

}



double f(double x){

    return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-y*x;;

}



int main(){



    //freopen("input.txt","r",stdin);



    int t;

    double left,right,mid;

    scanf("%d",&t);

    while(t--){

        scanf("%lf",&y);

        if(g(100.0)-y<=0){

            printf("%.4lf\n",f(100.0));

            continue;

        }

        left=0;

        right=100;

        while(right-left>=eps){

            mid=(right+left)/2;

            if(g(mid)-y<eps)

                left=mid;

            else

                right=mid;

        }

        printf("%.4lf\n",f(mid));

    }

    return 0;

}