HDU 5105 Math Problem --数学，求导

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 ;

}

巴特西

HDU 5105 Math Problem --数学，求导

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