A. Pashmak and Garden
time limit per test
memory limit per test
input
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
x1, y1, x2, y2 ( - 100 ≤ x1, y1, x2, y2 integers, where x1 and y1 are coordinates of the first tree and x2 and y2
Output
-1. Otherwise print four space-separated integers x3, y3, x4, y4
x3, y3, x4, y4 must be in the range ( - 1000 ≤ x3, y3, x4, y4.
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;
}
