A. Pashmak and Garden

简单题：给正方形的两个点，求剩下两个点。因为边和坐标轴是平行的，所以分四种情况。

一开始还在两个点是输出顺序上纠结了半天。囧。。。。

//A#include<iostream>#include<cmath>using namespace std;int main (){    int x1,y1,x2,y2;    while(cin>>x1>>y1>>x2>>y2)    {        if(x1==x2)            cout<<x2+abs(y2-y1)<<" "<<y2<<" "<<x1+abs(y2-y1)<<" "<<y1<<endl;        else if(y1==y2)            cout<<x1<<" "<<y1+abs(x2-x1)<<" "<<x2<<" "<<y2+abs(x2-x1)<<endl;        else if(abs(x2-x1)==abs(y2-y1))            cout<<x1<<" "<<y2<<" "<<x2<<" "<<y1<<endl;        else            cout<<-1<<endl;    }return 0;}