HDU2899Strange fuction （二分）

 

Strange fuction

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

Total Submission(s): 4004    Accepted Submission(s): 2891

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

2100 200

 

Sample Output

-74.4291 -178.8534

 
Author
Redow
 
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题意：求F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)当x的值为多少时F(X)有最小值

思路：F(x)为单调递增函数。所以当F(x)的导数=0时有极值也为最小值
Can you find：
 

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
double node(double x)
{
    return  6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2);   //令f(x)=0,则有等式<1>   6 * x^7+8*x^6+7*x^3+5*x^2=y*x
}
double nodes(double x)
{
    return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x;    //对等式<1>求导42*X^6+ 48* x^5+ 21* x^2+10*x=y
}
int main()
{
    double y;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf",&y);
        double left=0;
        double right=100;
        double mid;
        while(right-left>1e-10)
        {
            mid=(left+right)/2;
            if(nodes(mid)<y)
            left=mid+1e-10;
            else
            right=mid-1e-10;
        }
        printf("%.4lf\n",node(mid)-y*mid);
    }
    return 0;
}