Codeforces Round #261 (Div. 2) A. Pashmak and Garden

Pashmak需要找到一个浪漫花园的确切位置来与他的爱人Parmida会面。他只知道两棵树的位置,而这两棵树位于一个正方形的两个顶点上。本篇将介绍如何利用这些信息找出其余两个顶点的位置。

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A. Pashmak and Garden
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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).

Sample test(s)
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
 
题意: 已知正方形的2个顶点的位置,求另外两个顶点的位置。(多解随意输出其中一种)
 
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
	int x1,y1,x2,y2;
	scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
	int deltax=abs(x1-x2),deltay=abs(y1-y2);
	if(deltax==0)
	{
		int y3=y1,y4=y2;
		int x3=x1+deltay,x4=x2+deltay;
		printf("%d %d %d %d\n",x3,y3,x4,y4);
	}
	else if(deltay==0)
	{
		int x3=x1,x4=x2;
		int y3=y1+deltax,y4=y2+deltax;
		printf("%d %d %d %d\n",x3,y3,x4,y4);
	}
	else if(deltax!=deltay)
	{
		printf("-1\n");
	}
	else
	{
		if(x1-x2==y1-y2)
		{
			printf("%d %d %d %d\n",min(x1,x2),max(y1,y2),max(x1,x2),min(y1,y2));
		}
		else
		{
			printf("%d %d %d %d\n",min(x1,x2),min(y1,y2),max(x1,x2),max(y1,y2));
		}
	}
	return 0;
}

 
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