【CUGBACM15级BC第18场 B】hdu 5105 Math Problem

Math Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3142    Accepted Submission(s): 762

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

 

别的没啥可说的，就是有唯一的大坑。。。前方高能= =：

http://blog.csdn.net/greatjames/article/details/77131177

#include <bits/stdc++.h>#define _ ios_base::sync_with_stdio(0);cin.tie(0);using namespace std;double a, b, c, d, L, R;double f(double x){    return abs(a * x * x * x + b * x * x + c * x + d);}bool judge(double x){    if (x >= L && x <= R)    {        return true;    }    else    {        return false;    }}int main(){    _    while (cin >> a >> b >> c >> d >> L >> R)    {        ///cout << 6.0 / 1.5 * 100 << endl;        ///cout << 6.0 / (1.5 * 100) << endl;        double ans = max(f(L), f(R));        if (a != 0)        {            double ter = 4 * b * b - 12 * a * c;            if (ter <= 0)            {                printf("%0.2f\n", ans);                continue;            }            double x1 = (-2 * b + (double)sqrt(ter)) / (6 * a);            double x2 = (-2 * b - (double)sqrt(ter)) / (6 * a);            double v;            if (judge(x1) == true && judge(x2) == true)            {                v = max(f(x1), f(x2));                ans = max(ans, v);            }            else if (judge(x1) == true && judge(x2) == false)            {                v = f(x1);                ans = max(ans, v);            }            else if (judge(x1) == false && judge(x2) == true)            {                v = f(x2);                ans = max(ans, v);            }        }        else if (a == 0 && b != 0)        {            double hh = -c / (2 * b);            if (judge(hh) == true)            {                ans = max(ans, f(hh));            }        }        printf("%0.2f\n", ans);    }    return 0;}