HDU 2899 Strange fuction（三分）

Strange fuction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 754 Accepted Submission(s): 612

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.

 

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

2100200

 

Sample Output

-74.4291-178.8534

 

Author

Redow

 

Recommend

lcy

这道题的数据很小，可以直接暴力过。

这道题的本意是二分查找。找导函数的零点。

这是暴力的代码：

#include<iostream>#include<string>#include<cstring>  #include<cmath>  #include<algorithm>  using namespace std;int main(){int t;cin >> t;while (t--){double y;cin >> y;double x;double min=0.0;for ( x = 0; x <10 ; x+=0.00001)//我先将Y=0时和Y=100时的情况枚举了一遍，发现最小值在0，10里面。{double temp = 6 * x*x*x*x*x*x*x + 8 * x*x*x*x*x*x + 7 * x*x*x + 5 * x*x - y* x;if (temp < min)min = temp;}printf("%.4lf\n", min);}return 0;}

三分的代码：

#include<iostream>#include<string>#include<cstring>  #include<cmath>  #include<algorithm>  using namespace std;double y;const double eps = 1e-6;double c(double x){return 6 * x*x*x*x*x*x*x + 8 * x*x*x*x*x*x + 7 * x*x*x + 5 * x*x - y* x;}int main(){int t;cin >> t;while (t--){cin >> y;double first = 0;double last =100;double mid1;double mid2;while (first+eps<last){mid1 = first + (last-first) / 3;mid2 = last - (last-first) / 3;if (c(mid1) > c(mid2))first = mid1;else last = mid2;}printf("%.4lf\n", c(mid2));}return 0;}