HDU 5105 Math Problem（BestCoder Round #18）

短命女 2021-12-20 08:23 189阅读 0赞

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

题意：求给出的三次方程绝对值（ f(x)=|a∗x3\+b∗x2\+c∗x\+d|(L≤x≤R)）的区间最大值。将这个三次方程求导得3a\*x^2+2\*b\*x+c，解这个二次方程得到两个解x1，x2，在x1，x2处就是f(x)的最大值，那么我们只需要判断这两个点是否在区间\[l，r\]中，并且判断区间端点的f\[l\]，f\[r\]，和f\[x1\]，f\[x2\]的大小即可。

#include<stdio.h>
    #include<string.h>
    #include<queue>
    #include<math.h>
    #include<stdlib.h>
    #include<algorithm>
    using namespace std;
    
    const int N=1e4+10;
    const int INF=0x3f3f3f3f;
    const int MOD=1e9+7;
    
    typedef long long LL;
    
    void Solve(double &x1, double &x2, int &flag, double a, double b, double c) ///求解二次方程
    {
        if (a == 0) ///要是二次项系数为0，只有一个根，且不能用求根公式
        {
            x1 = x2 = -c/b;
            flag = 1; ///flag标记该方程是否有根
            return ;
        }
        if (b*b-4*a*c >= 0)
        {
            x1 = (-b-sqrt(b*b-4*a*c))/(2*a);
            x2 = (-b+sqrt(b*b-4*a*c))/(2*a);
            flag = 1;
        }
    }
    
    void Result(double a, double b, double c, double d, double l, double r, double x, double &Max) 
    { ///代入求函数值
        if(x >= l && x <= r)
            Max = max(Max, fabs(a*x*x*x+b*x*x+c*x+d));
    }
    
    int main ()
    {
        double a, b, c, d, l, r, x1, x2, Max;
        int flag;
    
        while (scanf("%lf %lf %lf %lf %lf %lf", &a, &b, &c, &d, &l, &r) != EOF)
        {
            Max = -INF;
            flag = 0;
    
            Solve(x1, x2, flag, 3*a, 2*b, c);
    
            Result(a, b, c, d, l, r, l, Max);
            Result(a, b, c, d, l, r, r, Max);
    
            if (flag)
            {
                Result(a, b, c, d, l, r, x1, Max);
                Result(a, b, c, d, l, r, x2, Max);
            }
    
            printf("%.2f\n", Max);
        }
    
        return 0;
    }

转载于:https://www.cnblogs.com/syhandll/p/4939144.html