hdu5105Math Problem(分类讨论)

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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，直接比较端点即可。

2 a==0&&b!=0 那么函数就是二次函数，直接算出特征值，然后比较端点值即可。。

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，直接比较端点值即可。。

当大于0的时候，直接求出两个根，然后和端点值比较即可

ps：所有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)=|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

Source

BestCoder Round #18

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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 0x3f3f3f3fusing 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;}

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