官方题解：
f(x)=|a∗x3+b∗x2+c∗x+d|, 求最大值。令g(x)=a∗x3+b∗x2+c∗x+d，f(x)的最大值即为g(x)的正最大值，或者是负最小值。a！=0时，
g′(x)=3∗a∗x2+2∗b∗x+c 求出g′(x)的根（若存在，x1,x2，由导数的性质知零点处有极值。ans=max(f(xi)|L≤xi≤R).然后考虑两个端点的特殊性有ans=max(ans,f(L),f(R)).
当时 x = -c/(2*b) 写成 x = -c/2*b 了，然后过pretest了。 然后。。你敢信？
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define eps 1e-8
using namespace std;
#define N 50017 int sgn(double x)
{
if(x > eps) return ;
if(x < -eps) return -;
return ;
} double a,b,c,d,L,R; double calc(double x) { return fabs(a*x*x*x + b*x*x + c*x + d); } int main()
{
while(scanf("%lf%lf%lf%lf%lf%lf",&a,&b,&c,&d,&L,&R)!=EOF)
{
if(sgn(a) == )
{
if(sgn(b) == )
{ if(sgn(fabs(calc(L))-fabs(calc(R))) >= )
printf("%.2f\n",calc(L));
else
printf("%.2f\n",calc(R));
}
else
{
double X = -c/(2.0*b);
double k1 = calc(L);
double k2 = calc(R);
double k3;
if(sgn(X-L) >= && sgn(X-R) <= )
k3 = calc(X);
else
k3 = 0.0;
printf("%.2f\n",max(max(k1,k2),k3));
}
continue;
}
double delta = 4.0*b*b - 12.0*a*c;
if(sgn(delta) <= )
{
if(sgn(fabs(calc(L))-fabs(calc(R))) >= )
printf("%.2f\n",calc(L));
else
printf("%.2f\n",calc(R));
}
else
{
double X1 = (-2.0*b + sqrt(delta))/(6.0*a);
double X2 = (-2.0*b - sqrt(delta))/(6.0*a);
double k1 = calc(L);
double k2 = calc(R);
double k3,k4;
if(sgn(X1-L) >= && sgn(X1-R) <= )
k3 = calc(X1);
else
k3 = 0.0;
if(sgn(X2-L) >= && sgn(X2-R) <= )
k4 = calc(X2);
else
k4 = 0.0;
printf("%.2f\n",max(max(max(k1,k2),k3),k4));
}
}
return ;
}