2005 TCO Online Round 1

 

RectangleError 

Problem's Link

Problem Statement

     You want to draw a rectangle on a piece of paper. Unfortunately, you are not a perfect draftsman. The lines you make, although straight, do not always have the correct lengths. The top edge has length in the inclusive range [topMin,topMax], the left edge in the inclusive range [leftMin,leftMax], and the right edge in the inclusive range [rightMin,rightMax]. Fortunately, the left, top and right edges are at right angles to each other and meet (where applicable) at their ends. The bottom edge is made by connecting the bottom end of the left edge to the bottom end of the right edge. Return the maximum length the bottom edge could be minus the minimum length the bottom edge could be.

Definition

-Class: RectangleError

-Method: bottomRange

-Parameters: double, double, double, double, double, double

-Returns: double

-Method signature: double bottomRange(double topMin, double topMax, double leftMin, double leftMax, double rightMin, double rightMax)

-(be sure your method is public)

Notes

- Your return value must have an absolute or relative error less than 1e-9.

Constraints

- Each input will be between 5 and 100 inclusive.

- topMin will not be greater than topMax.

- leftMin will not be greater than leftMax.

- rightMin will not be greater than rightMax.

----------------------------------------------------------------------------

Mean: 

给定一个矩形的顶边、左边、右边的长度范围，求连接左边和右边下顶点的斜边长的最大可能长度与最小可能长度的差.

analyse:

Time complexity: O(1)

 

view code

#include <bits/stdc++.h>
using namespace std;

class RectangleError
{
public:
   double bottomRange(double topMin, double topMax, double leftMin, double leftMax, double rightMin, double rightMax)
   {

       double Max = max(hypot(topMax, leftMin-rightMax) , hypot(topMax, leftMax-rightMin));

       double y;
       if(rightMin >= leftMax)
           y = rightMin - leftMax;
       else if(leftMin >= rightMax)
           y = rightMax - leftMin;
       else
           y = 0;

       double Min = hypot(topMin, y);
       return Max-Min;
   }
};

int main()
{
    double topMin,topMax,leftMin, leftMax, rightMin,rightMax;
    while(cin>>topMin>>topMax>>leftMin>>leftMax>>rightMin>>rightMax)
    {
        RectangleError rectangleError;
        double ans=rectangleError.bottomRange(topMin,topMax,leftMin,leftMax,rightMin,rightMax);
        printf("%f
",ans);
    }
    return 0;
}
/*

*/