题目链接:http://codeforces.com/contest/749/problem/B
题意：给定平行四边形的3个点，输出所有可能的第四个点。
思路：枚举任意两个点形成的直线，然后利用这两个点计算偏移量用第三点求出第四个点。
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<stdio.h>
#include<queue>
#include<vector>
#include<stack>
#include<map>
#include<set>
#include<time.h>
#include<cmath>
using namespace std;
typedef long long int LL;
int main(){
//#ifdef kirito
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
//#endif
// int start = clock();
pair<int, int>point[];
while (scanf("%d %d", &point[].first,&point[].second) != EOF){
scanf("%d %d", &point[].first, &point[].second);
scanf("%d %d", &point[].first, &point[].second);
set<pair<int, int>>ans;
for (int i = ; i <= ; i++){
for (int j = i + ; j <= ; j++){
pair<int, int>p4;
int k = - i - j; //非i,j的点
int dx = point[i].first - point[j].first;
int dy = point[i].second - point[j].second;
p4.first = point[k].first + dx;
p4.second = point[k].second + dy;
ans.insert(p4); p4.first = point[k].first - dx;
p4.second = point[k].second - dy;
ans.insert(p4);
}
}
printf("%d\n", ans.size());
for (set<pair<int,int>>::iterator it=ans.begin(); it!= ans.end(); it++){
printf("%d %d\n", it->first, it->second);
}
}
//#ifdef LOCAL_TIME
// cout << "[Finished in " << clock() - start << " ms]" << endl;
//#endif
return ;
}