Day 42 A. Pashmak and Garden

最新推荐文章于 2021-11-06 22:58:45 发布

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该编程问题要求根据已知的两个正方形顶点坐标，判断能否形成一个正方形，并输出另外两个顶点的坐标。输入包含两个点的坐标，程序会检查它们是否能构成一个边长相等的正方形。如果可以，它将打印出剩余的两个顶点；否则，输出-1。示例展示了不同情况下的输出结果。

Problem
Pashmak has fallen in love with an attractive girl called Parmida since one year ago…

Today, Pashmak set up a meeting with his partner in a romantic garden. Unfortunately, Pashmak has forgotten where the garden is. But he remembers that the garden looks like a square with sides parallel to the coordinate axes. He also remembers that there is exactly one tree on each vertex of the square. Now, Pashmak knows the position of only two of the trees. Help him to find the position of two remaining ones.

Input
The first line contains four space-separated x1, y1, x2, y2 ( - 100 ≤ x1, y1, x2, y2 ≤ 100) integers, where x1 and y1 are coordinates of the first tree and x2 and y2 are coordinates of the second tree. It’s guaranteed that the given points are distinct.

Output
If there is no solution to the problem, print -1. Otherwise print four space-separated integers x3, y3, x4, y4 that correspond to the coordinates of the two other trees. If there are several solutions you can output any of them.

Note that x3, y3, x4, y4 must be in the range ( - 1000 ≤ x3, y3, x4, y4 ≤ 1000).

Examples
input
0 0 0 1
output
1 0 1 1
input
0 0 1 1
output
0 1 1 0
input
0 0 1 2
output
-1

题目大致意思为：给出两个点的坐标求是否能够成一个正方形如果能则输出另外两个点的坐标如果不能则输出-1

#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<stdio.h>
#include<limits.h>
#include<cmath>
#include<set>
#include<map>
#define ll long long
using namespace std;
int main()
{
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
	int dx = abs(x1 - x2);
	int dy = abs(y1 - y2);
	if (dx != 0 && dy != 0)
	{
		if (dx != dy)
		{
			cout << -1 << endl;
		}
		else 
		{
			cout << x1 << " " << y2 << " ";
			cout << x2 << " " << y1 << endl;
		}
	}
	else
	{
		cout << x1 + dy << " " << y1 + dx << " ";
		cout << x2 + dy << " " << y2 + dx << endl;
	}
    return 0;
}