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;
}