hdu5105Math Problem(分类讨论)

题目链接:

huangjing

题目:

思路:

给出的是一个方程,首先讨论最高项系数。
1:a==0&& b==0  那么函数就是线性的。直接比較端点就可以。

2   a==0&&b!=0  那么函数就是二次函数。直接算出特征值,然后比較端点值就可以。。

3  a!=0  又有几种情况,那么当特征根  b*b-4*a*c<0 时  说明愿函数是单调,直接比較端点值就可以。

当大于0的时候,直接求出两个根,然后和端点值比較就可以

ps:全部的特征根都要是有效的,即都要在[L,R]之间。



题目:

Math Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 943    Accepted Submission(s): 250


Problem Description
Here has an function:
  f(x)=|ax3+bx2+cx+d|(LxR)
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. (10a,b,c,d10,100LR100)
 

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
 

Source
 

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代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
priority_queue<int,vector<int>,greater<int> >Q;

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

double f(double x)
{
    return fabs(a*x*x*x+b*x*x+c*x+d);
}

int main()
{
    double ans;
    while(~scanf("%lf%lf%lf%lf%lf%lf",&a,&b,&c,&d,&l,&r))
    {
        if(a==0&&b!=0)
        {
            double x=-c/(2*b);
            ans=max(f(l),f(r));
            if(x>=l&&x<=r)
               ans=max(ans,f(x));
        }
        else if(a==0&&b==0)
            ans=max(f(l),f(r));
        else if(a!=0)
        {
            double xx=4*b*b-12*a*c;
            if(xx<0)
                ans=max(f(l),f(r));
            else
            {
                double x1=(-2*b+sqrt(xx))/(6*a);
                double x2=(-2*b-sqrt(xx))/(6*a);
                ans=max(f(l),f(r));
                if(x1>=l&&x1<=r)
                     ans=max(ans,f(x1));
                if(x2>=l&&x2<=r)
                     ans=max(ans,f(x2));
            }
        }
        printf("%.2lf\n",ans);
    }
    return 0;
}


posted on 2017-07-19 11:59  yjbjingcha  阅读(129)  评论(0编辑  收藏  举报

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