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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））

2024-08-02 08:12:08   251人阅读

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;}

 

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））


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