Math Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 586    Accepted Submission(s): 142


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  
Source BestCoder Round #18




一道纯粹数学题，关于3次方函数求极致，运用求导以及简单的函数图像方面的知识。同数学求解的思路一样需要区分三种情况来做这道题目。




S #include #include #include int main() { double a, b, c, d, L, R, x, ans; while(scanf("%lf%lf%lf%lf%lf%lf", &a, &b, &c, &d, &L, &R) != EOF) { ans = -1; if (a != 0) { if ((4 * b * b - 12 * a * c) >= 0) { x = (-2 * b + sqrt(4 * b * b - 12 * a * c)) / (6 * a); if (x >= L && x <= R) { x = abs(a * x * x * x + b * x * x + c * x + d); if(x > ans) ans = x; } x = (-2 * b - sqrt(4 * b * b - 12 * a * c)) / (6 * a); if (x >= L && x <= R) { x = abs(a * x * x * x + b * x * x + c * x + d); if(x > ans) ans = x; } x = abs(a * L * L * L + b * L * L + c * L + d); if(x > ans) ans = x; x = abs(a * R * R * R + b * R * R + c * R + d); if(x > ans) ans = x; } else { x = abs(a * L * L * L + b * L * L + c * L + d); if(x > ans) ans = x; x = abs(a * R * R * R + b * R * R + c * R + d); if(x > ans) ans = x; } } else if (a == 0 && b != 0) { x = -c / (2 * b); x = abs(a * x * x * x + b * x * x + c * x + d); if(x > ans) ans = x; x = abs(a * L * L * L + b * L * L + c * L + d); if(x > ans) ans = x; x = abs(a * R * R * R + b * R * R + c * R + d); if(x > ans) ans = x; } else { x = abs(a * L * L * L + b * L * L + c * L + d); if(x > ans) ans = x; x = abs(a * R * R * R + b * R * R + c * R + d); if(x > ans) ans = x; } printf("%.2lf ", ans); } return 0; }