Bestcoder round 18----B题（一元三次方程确定区间的最大值（包含极值比较））

Math Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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

 

 

代码：

#include <math.h>

#include <string.h>

#include <stdio.h>

#include <iostream>

#include <string>

#include <algorithm>



using namespace std;



// f(x)=|a∗x3+b∗x2+c∗x+d|(L≤x≤R)



int main()

{

    double a, b, c, d, ll, r;

    double mm, dd, ff;

    double gg, hh;



    while(cin>>a)

    {

        cin>>b>>c>>d>>ll>>r;



        dd=(a*pow(ll, 3.0)+b*pow(ll, 2.0)+c*ll+d);

        if(dd<0)

          dd=-dd;

        ff=(a*pow(r, 3.0)+b*pow(r, 2.0)+c*r+d);

        if(ff<0)

          ff=-ff;

        mm=max(dd, ff);



        if((4*b*b - 12*a*c)<0)

        {

            //无解

            printf("%.2lf\n", mm);

            continue;

        }

        else if( (4*b*b - 12*a*c)==0 )

        {

            //有一个解

            if( (-(b)/(3*a))>=ll && (-(b)/(3*a))<=r )

            {

                 gg=-1*(b/3*a);

                 hh=a*pow(gg,3)+b*pow(gg, 2)+c*gg+d;

                 if(hh<0)

                   hh=-hh;

                 if(hh>mm)

                 mm=hh;

                 printf("%.2lf\n", mm);

                 continue;

            }

        }

        else

        {

            //有2个解

            gg=(-b+sqrt(b*b-4*a*c))/2*a;

            if( gg>=ll && gg<=r )

            {

                double q;

            q=a*pow(gg,3)+b*pow(gg, 2)+c*gg+d;

            if(q<0)

              q=-q;

            if(q>mm)

              mm=q;

            }



            hh=(-b*sqrt(b*b-4*a*c))/2*a;

            if(hh>=ll && hh<=r)

            {

               double w;

            w=a*pow(hh,3)+b*pow(hh, 2)+c*hh+d;

            if(w<0)

              w=-w;

            if(w>mm)

              mm=w;

            }



            printf("%.2lf\n", mm);

        }

    }



    return 0;

}