Codeforces Round #388(Div. 2)B. Parallelogram is Back【计算几何】

B. Parallelogram is Back

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Long time ago Alex created an interesting problem about parallelogram. The input data for this problem contained four integer points on the Cartesian plane, that defined the set of vertices of some non-degenerate (positive area) parallelogram. Points not necessary
were given in the order of clockwise or counterclockwise traversal.

Alex had very nice test for this problem, but is somehow happened that the last line of the input was lost and now he has only three out of four points of the original parallelogram. He remembers that test was so good that he asks you to restore it given only
these three points.

Input

The input consists of three lines, each containing a pair of integer coordinates xi and yi ( - 1000 ≤ xi, yi ≤ 1000).
It's guaranteed that these three points do not lie on the same line and no two of them coincide.

Output

First print integer k — the number of ways to add one new integer point such that the obtained set defines some parallelogram of positive
area. There is no requirement for the points to be arranged in any special order (like traversal), they just define the set of vertices.

Then print k lines, each containing a pair of integer — possible coordinates of the fourth point.

Example

input

0 0
1 0
0 1

output

3
1 -1
-1 1
1 1

Note

If you need clarification of what parallelogram is, please check Wikipedia page:
https://en.wikipedia.org/wiki/Parallelogram
题目大意：

给你三个不重复的点，并且这三个点不会共线，问你有几个第四个点，使其构成一个平行四边形。

思路：

1、小学生都知道一共会有三个点。

2、设定三个点编号为A,B,C；

A->B是怎么走的，那么C->D就怎么走；能够找到第一个点。

B->A是怎么走的，那么C->D就怎么走；能够找到第二个点。

B->C是怎么走的，那么D->A就怎么走；能够找到第三个点。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int x[5];
int y[5];
int main()
{
for(int i=1;i<=3;i++)
{
scanf("%d%d",&x[i],&y[i]);
}
int xx=x[1]-x[2];
int yy=y[1]-y[2];
int xxx=x[2]-x[3];
int yyy=y[2]-y[3];
printf("3\n");
printf("%d %d\n",x[3]+xx,y[3]+yy);
printf("%d %d\n",x[3]-xx,y[3]-yy);
printf("%d %d\n",x[1]+xxx,y[1]+yyy);
}