HDU 5105 Math Problem

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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求函数f(x)的最大值。大体思路是先求f(L)和f(R)两个边界值，取较大的，然后判断a是否为0。如果a不为0，即为一元三次方程，求f'(x)=0时的两个点，若点在[L,R]这个区间,则分别与边界值中较大的继续比较。如果a为0，即为一元二次方程，求出最值与边界值中较大的比较。这题有人枚举0.01精度居然过了，还叉不掉，也是醉了- -#include<iostream>#include<stdio.h>#include<cstring>#include<cstdlib>#include<cmath>using namespace std;int max(const int a,const int b){    return a>b?a:b;}int main(){    double a,b,c,d,l,r;    double a1,a2,x1,x2,mid;    double A,B,C;    double ans;    while(scanf("%lf%lf%lf%lf%lf%lf",&a,&b,&c,&d,&l,&r)!=EOF)    {        a1=fabs(a*l*l*l+b*l*l+c*l+d);        a2=fabs(a*r*r*r+b*r*r+c*r+d);        ans=max(a1,a2);        if(a)        {            A=3*a;            B=2*b;            C=c;            x1=(-B+sqrt(B*B-4*A*C))/(2*A);            x2=(-B-sqrt(B*B-4*A*C))/(2*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));        }        else        {            A=b;            B=c;            C=d;            mid=B/(-2*A);            ans=max(ans,fabs(A*mid*mid+B*mid+C));                }        printf("%.2f\n",ans);    }    return 0;}

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