Hdu5105Math Problem (classification discussion)
Question link:
Huangjing
Question:
Ideas:
The equation is given. First, the highest coefficient is discussed.
1: If a = 0 & B = 0, then the function is linear and you can directly compare the endpoint.
2 a = 0 & B! If the value is 0, the function is a quadratic function. Calculate the feature value directly, and then compare the endpoint value ..
3! = 0 and there are several cases, then when the feature root B * B-4 * a * c <0 indicates that the function is monotonous, directly compare the endpoint value ..
When the value is greater than 0, you can directly find two roots and then compare them with the endpoint value.
Ps: All feature roots are valid, that is, they must be between [L, R ..
Question:
Math Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 943 Accepted Submission (s): 250
Problem DescriptionHere has an function:
F (x) = |? X3 + B? X2 + c? X + d | (L ≤ x ≤ R)
Please figure out the maximum result of f (x ).
InputMultiple 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)
OutputFor 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
SourceBestCoder Round #18
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Code:
#include
#include
#include
#include#include
#include
#include
#include
#include
#define eps 1e-9#define ll long long#define INF 0x3f3f3f3fusing namespace std;priority_queue
,greater
>Q;double a,b,c,d,l,r;double f(double x){ return fabs(a*x*x*x+b*x*x+c*x+d);}int main(){ double ans; while(~scanf("%lf%lf%lf%lf%lf%lf",&a,&b,&c,&d,&l,&r)) { if(a==0&&b!=0) { double x=-c/(2*b); ans=max(f(l),f(r)); if(x>=l&&x<=r) ans=max(ans,f(x)); } else if(a==0&&b==0) ans=max(f(l),f(r)); else if(a!=0) { double xx=4*b*b-12*a*c; if(xx<0) ans=max(f(l),f(r)); else { double x1=(-2*b+sqrt(xx))/(6*a); double x2=(-2*b-sqrt(xx))/(6*a); ans=max(f(l),f(r)); if(x1>=l&&x1<=r) ans=max(ans,f(x1)); if(x2>=l&&x2<=r) ans=max(ans,f(x2)); } } printf("%.2lf\n",ans); } return 0;}