hdu 5105 Math Problem（数学）

Math Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2518    Accepted Submission(s): 605

Problem Description

Here has an function:
  f(x)=|a∗x3+b∗x2+c∗x+d|(L≤x≤R)
Please figure out the maximum result of f(x).

 

Input

Multiple test cases(less than 100). For each test case, there will be only 1 line contains 6 numbers a, b, c, d, L and R. (−10≤a,b,c,d≤10,−100≤L≤R≤100)

 

Output

For each test case, print the answer that was rounded to 2 digits after decimal point in 1 line.

 

Sample Input

1.00 2.00 3.00 4.00 5.00 6.00

 

Sample Output

310.00

题意：求题里给的f(x)

思路：直接求导判断是否有极值即可

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define eps 1e-8
int main()
{
    double a,b,c,d,l,r;
    while(~scanf("%lf %lf %lf %lf %lf %lf",&a,&b,&c,&d,&l,&r))
    {
        double ans;
        ans=max(fabs(a*l*l*l+b*l*l+c*l+d),fabs(a*r*r*r+b*r*r+c*r+d));
        if(fabs(a)<=eps&&fabs(b)>eps)
        {
                double k=-c/(2*b);
                if(k>l&&k<r)
                    ans=max(ans,fabs(b*k*k+c*k+d));
        }
        else
        {
            if(4*b*b-12*a*c>=0)
            {
                double q=sqrt(4*b*b-12*a*c);
                double x1=(-2*b-q)/(6*a),x2=(-2*b+q)/(6*a);
                if(x1>l&&x1<r) ans=max(ans,fabs(a*x1*x1*x1+b*x1*x1+c*x1+d));
                if(x2>l&&x2<r) ans=max(ans,fabs(a*x2*x2*x2+b*x2*x2+c*x2+d));
            }
        }
        printf("%.2lf\n",ans);
    }
    return 0;
}