CF 459A(Pashmak and Garden-正方形给出2点求2点)

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

分为同行。同列，对角线，无解4种情况

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int main()
{
//	freopen("a.in","r",stdin);
//	freopen("a.out","w",stdout);
	int x1_,x2_,y1_,y2_;
	cin>>x1_>>y1_>>x2_>>y2_;
	if (x1_==x2_)
	{
		int d=abs(y1_-y2_);
		printf("%d %d %d %d\n",x1_+d,y1_,x2_+d,y2_);
	}
	else
	if (y1_==y2_)
	{
		int d=abs(x1_-x2_);
		printf("%d %d %d %d\n",x1_,y1_+d,x2_,y2_+d);
	}
	else
	{
		if (abs(x1_-x2_)==abs(y1_-y2_))
		{
			printf("%d %d %d %d\n",x2_,y1_,x1_,y2_);
		}
		else cout<<"-1\n";
	}
	return 0;
}