hdu 2899 Strange fuction(导数+二分)

Strange fuction

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

Total Submission(s): 4000 Accepted Submission(s): 2888



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
题目分析:因为求最小值,所以求导数,然后二分导数结果为0的x值,代入原函数即可

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cmath>
#define eps 1e-8

using namespace std;

int t;
double n;

double f ( double x )
{
    return 42.0*pow( x , 6 ) + 48.0*pow( x , 5 ) + 21.0*pow( x , 2 ) + 10.0*x;
}

double g ( double x , double n )
{
    return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-n*x;
}

double search ( double n )
{
    double left = 0.0 , right = 100.0 , mid;
    while ( fabs(left-right) > eps )
    {
        mid = ( left + right ) / 2.0;
        if ( f(mid) < n ) left = mid;
        else right = mid;
    }
    return mid;
}

int main ( )
{
    scanf ( "%d" , &t );
    while ( t-- )
    {
        scanf ( "%lf" , &n );
        double ans = search ( n );
        printf ( "%.4lf\n" , g( ans , n ) );
    }
}