hdu2899Strange fuction(解方程+二分)

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10462    Accepted Submission(s): 6996

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

给出y，求出F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x的最小值，x在0~100之间。

需要先求一下他的导函数42x^6+48x^5+21x^2+10x-y。最小值就是这个导函数为0

这样就可以用二分求一下了，具体看代码

#include<bits/stdc++.h>
using namespace std;
const int inf=0x3f3f3f3f;
double y;
double f(double x)
{
    return 6*x*x*x*x*x*x*x+8*x*x*x*x*x*x+7*x*x*x+5*x*x-y*x;
}
double f2(double x)
{
    return 42*x*x*x*x*x*x+48*x*x*x*x*x+21*x*x+10*x;
}
int main()
{
    int t;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            
            scanf("%lf",&y);
            double l=0.0,r=100.0,mid=50.0;
            while(fabs(f2(mid)-y)>1e-5)
            {
                if(f2(mid)<y)
                {
                    l=mid;
                    mid=(l+r)/2.0;
                }
                if(f2(mid)>y)
                {
                    r=mid;
                    mid=(l+r)/2.0;
                }
                
            }
            printf("%.4lf
",f(mid));
        }
    }
    return 0;
}