AFA is a girl who likes running. Today,she downloaded an app about running .The app can record the trace of her running. AFA will start running in the park. There are many chairs in the park, and AFA will start his(her) running in a chair and end in this(another) chair. Between two chairs,she running(runs) in a line. she(She) wants the(/) the trace can (to) be a regular triangle or a square or a regular pentagon or a regular hexagon.
Please tell her how many ways can her(she) find.
Two ways are same if the set of chair that they contains are same.

There are multiply(multiple) cases.
In each case,there is an integer n(1 < = n < = 20)in a line.
In next n lines,there are two integers xi,yi(0 < = xi,yi < 9) in each line.

Output the number of ways.

4
0 0
0 1
1 0
1 1

1

BestCoder Round #50 (div.2)


题目大意：
    再给定的9*9方格线上有一些点，问能找到多少个正三角形，正方形，正五边形，正六边形。

解题：
    做的时候，已经猜到可能只有正方形才是可以的，只是来不及了，题解却说地球人都知道，无爱了....
    数据量这么小，怎么暴力怎么来咯！

代码：

#include <stdio.h>  
#include <iostream>
#include <cstring>
using namespace std;
struct point
{
	int x,y;
}store[25];
//该位置是否有点标记
bool status[10][10];
//判断是否出界
bool inside(int x,int y)
{
	if(x>=0&&x<9&&y>=0&&y<9)
		return true;
	else
		return false;
}
int main()
{
    int t,tx1,ty1,tx2,ty2,tx3,ty3,ans,xd,yd;
	while(~scanf("%d",&t))
	{
		memset(status,0,sizeof(status));
		ans=0;
		for(int i=0;i<t;i++)
		{
			scanf("%d %d",&store[i].x,&store[i].y);
			//标记
			status[store[i].x][store[i].y]=1;
		}
		for(int i=0;i<t-1;i++)
		{
			for(int j=i+1;j<t;j++)
			{
               tx1=store[i].x;
			   tx2=store[j].x;
			   ty1=store[i].y;
			   ty2=store[j].y;
			   xd=tx1-tx2;
			   yd=ty1-ty2;
			   //所有点枚举
               for(int k=0;k<=8;k++)
			   {
				   for(int m=0;m<=8;m++)
				   {
					   //该位置有点
					   if(status[k][m])
					   {
						   //且不是枚举的两个点
						  if((k==tx1&&m==ty1)||(k==tx2&&m==ty2))
							  continue;
						  //垂直
                          if((xd*(k-tx2)+yd*(m-ty2))==0)
						  {
							  //距离相等
							  if(((k-tx2)*(k-tx2)+(m-ty2)*(m-ty2))==(xd*xd+yd*yd))
							  {
							    tx3=k+xd;
							    ty3=m+yd;
								//第四个点在界内，且存在
							    if(inside(tx3,ty3)&&status[tx3][ty3])
								   ans++;
							  }
						  }
					   }
				   }
			   }
			}
		}
		//一个正方形四条边都会计算一次
		ans/=4;
		printf("%d\n",ans);
	}
	return 0;
}