HDU 2899 Strange fuction (二分)
Strange fuction
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3301 Accepted Submission(s): 2421
Problem Description
Now, here is a fuction:
F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
Can you find the minimum value when x is between 0 and 100.
F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
Can you find the minimum value when x is between 0 and 100.
Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has only one real numbers Y.(0 < Y <1e10)
Output
Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.
Sample Input
2
100
200
Sample Output
-74.4291
-178.8534
Author
Redow
Recommend
lcy
因为是求最小值,先求原函数的导数,
f '(x)=42x^6+48x^5+21x^2+10x-y
再次求导发现f(x)是一个单调递增函数。
当f '(0)>=0时,f(x)单调递增,最小值为f(0)。
当f '(100)<=0时,f(x)单调递减,最小值为f(100)。
当f '(0)<0 && f '(100)>0时,存在一个x,使得f '(x)=0。此时函数先递减再递增,所以最小值是f(x)。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #include<iostream> #include<cstring> #include<cmath> #include<stdlib.h> #include<algorithm> using namespace std; double num(double x,double y) { return 6*pow(x,7.0)+8*pow(x,6.0)+7*pow(x,3.0)+5*pow(x,2.0)-y*x; } double fun(double x,double y) { return 42*pow(x,6.0)+48*pow(x,5.0)+21*pow(x,2.0)+10*x-y; } int main() { int T; double Y; scanf("%d",&T); while(T--) { scanf("%lf",&Y); if(fun(0.0,Y)>=0) printf("%.4lf\n",num(0.0,Y)); else if(fun(100.0,Y)<=0) printf("%.4lf\n",num(100.0,Y)); else { double l=0.0,mid=50.0,r=100.0; double ans1,ans2,ans3; ans1=fun(l,Y); ans2=fun(mid,Y); ans3=fun(r,Y); while(fabs(ans1-ans2)>0.000001) { if(ans2>0) { ans3=ans2; r=mid; mid=(l+r)/2; ans2=fun(mid,Y); } else { ans1=ans2; l=mid; mid=(l+r)/2; ans2=fun(mid,Y); } } printf("%.4lf\n",num(mid,Y)); } } }