1 second
256 megabytes
standard input
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.
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.
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).
0 0 0 1
1 0 1 1
0 0 1 1
0 1 1 0
0 0 1 2
-1
大叫好我是紫名选手T T第二次akdiv2了
第一题神模拟……有一个边和xy轴平行的正方形，给两个点，要输出另两个点，不存在就输出-1
然后就是各种判断啦T T
x1==x2的时候分一类，y1==y2的时候分一类，否则是对角线的位置分一类，然后判一下-1就好了
直接把x1==x2和y1==y2的输出一大堆拷过去再改的，然后忘记交换x和y了QAQ
#include<iostream>
#include<cstdio>
#define LL long long
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
inline int abs(int a)
{return a<0?-a:a;}
int x1,x2,y1,y2;
int main()
{
x1=read();y1=read();x2=read();y2=read();
if (x1==x2)
{
printf("%d %d %d %d",abs(y2-y1)+x1,y1,abs(y2-y1)+x1,y2);
}else
if (y1==y2)
{
printf("%d %d %d %d",x1,abs(x2-x1)+y1,x2,abs(x2-x1)+y1);
}else
{
if (abs(x1-x2)!=abs(y1-y2)){printf("-1");return 0;}
printf("%d %d %d %d",x1,y2,x2,y1);
}
}