找出平行四边形的所有可能坐标
从给定的三个坐标中找出所有可能的坐标，形成一个非零面积的平行四边形。 我们称 A、B、C 为给定的三个点。我们只能有三种可能的情况:
(1) AB and AC are sides, and BC a diagonal
(2) AB and BC are sides, and AC a diagonal
(3) BC and AC are sides, and AB a diagonal
因此，我们可以说只有 3 个坐标是可能的，如果给出 3 个坐标，我们就可以从这 3 个坐标生成一个平行四边形。 为了证明这三点都不一样，我们假设它是错的。不失一般性地假设公元年和公元前年得到的分数相等。
考虑这两点相等的两个方程组:
Bx + Cx - Ax = Ax + Cx - Bx
By + Cy - Ay = Ay + Cy - By
It can be simplified as-
Ax = Bx
Ay = By
我们得到了一个矛盾，因为所有的点都是不同的。
示例:
Input : A = (0 0)
B = (1 0)
C = (0 1)
Output : 1 -1
-1 1
1 1
Input : A = (-1 -1)
B = (0 1)
C = (1 1)
Output : -2 -1
0 -1
2 3
由于对边相等，AD = BC，AB = CD，因此我们可以将缺失点(D)的坐标计算为:
AD = BC
(Dx - Ax, Dy - Ay) = (Cx - Bx, Cy - By)
Dx = Ax + Cx - Bx
Dy = Ay + Cy - By
对角线为 AD 和 BC、CD 和 AB 的情况以相同的方式处理。 参考:https://math . stackexchange . com/questions/1322535/能画出多少个不同的平行四边形如果给定三维三坐标
下面是上述方法的实现:
C++
// C++ program to all possible points
// of a parallelogram
#include <bits/stdc++.h>
using namespace std;
// main method
int main()
{
int ax = 5, ay = 0; //coordinates of A
int bx = 1, by = 1; //coordinates of B
int cx = 2, cy = 5; //coordinates of C
cout << ax + bx - cx << ", "
<< ay + by - cy <<endl;
cout << ax + cx - bx << ", "
<< ay + cy - by <<endl;
cout << cx + bx - ax << ", "
<< cy + by - ax <<endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to all possible
// points of a parallelogram
public class ParallelogramPoints{
// Driver code
public static void main(String[] s)
{
int ax = 5, ay = 0; //coordinates of A
int bx = 1, by = 1; //coordinates of B
int cx = 2, cy = 5; //coordinates of C
System.out.println(ax + bx - cx + ", "
+ (ay + by - cy));
System.out.println(ax + cx - bx + ", "
+ (ay + cy - by));
System.out.println(cx + bx - ax + ", "
+ (cy + by - ax));
}
}
// This code is contributed by Prerna Saini
Python 3
# Python3 program to find all possible points
# of a parallelogram
ax = 5
ay = 0 #coordinates of A
bx = 1
by = 1 #coordinates of B
cx = 2
cy = 5 #coordinates of C
print(ax + bx - cx, ", ", ay + by - cy)
print(ax + cx - bx, ", ", ay + cy - by)
print(cx + bx - ax, ", ", cy + by - ax)
C
// C# program to all possible
// points of a parallelogram
using System;
public class ParallelogramPoints
{
// Driver code
public static void Main()
{
//coordinates of A
int ax = 5, ay = 0;
//coordinates of B
int bx = 1, by = 1;
//coordinates of C
int cx = 2, cy = 5;
Console.WriteLine(ax + bx - cx + ", "
+ (ay + by - cy));
Console.WriteLine(ax + cx - bx + ", "
+ (ay + cy - by));
Console.WriteLine(cx + bx - ax + ", "
+ (cy + by - ax));
}
}
// This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to all
// possible points
// of a parallelogram
// Driver Code
//coordinates of A
$ax = 5; $ay = 0;
//coordinates of B
$bx = 1; $by = 1;
//coordinates of C
$cx = 2; $cy = 5;
echo $ax + $bx - $cx , ", "
, $ay + $by - $cy ,"\n";
echo $ax + $cx - $bx , ", "
, $ay + $cy - $by,"\n" ;
echo $cx + $bx - $ax , ", "
, $cy + $by - $ax ;
// This code is contributed by anuj_67.
?>
java 描述语言
<script>
// JavaScript program to all possible
// points of a parallelogram
// Driver Code
let ax = 5, ay = 0; // Coordinates of A
let bx = 1, by = 1; // Coordinates of B
let cx = 2, cy = 5; // Coordinates of C
document.write(ax + bx - cx + ", " +
(ay + by - cy) + "<br/>");
document.write(ax + cx - bx + ", " +
(ay + cy - by) + "<br/>");
document.write(cx + bx - ax + ", " +
(cy + by - ax) + "<br/>");
// This code is contributed by susmitakundugoaldanga
</script>
输出:
4, -4
6, 4
-2, 1
时间复杂度: O(1)
版权属于：月萌API www.moonapi.com，转载请注明出处
