2005 TCO Online Round 1 - RectangleError

最新推荐文章于 2021-11-29 16:26:13 发布

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最新推荐文章于 2021-11-29 16:26:13 发布

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原文链接：https://yq.aliyun.com/articles/441227

本文介绍了一个计算矩形绘制中底边长度误差的问题。通过给定矩形顶边、左边及右边长度的范围，求解底边斜长最大值与最小值之差。采用简单高效的算法实现，时间复杂度为O(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\n",ans);
    }
    return 0;
}
/*

*/