1971.txt

北大ACM题目例程 详细的解答过程 程序实现 算法分析

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define debug 0

#define INF 30000

#if debug
	#define NMAX 10	
#else
	#define NMAX 1003
#endif
typedef struct
{
	int x,y;
}point;
point a[NMAX],m[NMAX*NMAX];
int cmp(const void*a,const void *b)
{
	if((*(point*)a).x==(*(point*)b).x)
		return (*(point*)a).y-(*(point*)b).y;
	else 
		return (*(point*)a).x-(*(point*)b).x;
}

int main()
{

#if debug 	
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
#endif
	int t,n,i,j,M,ans,c,k;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(i=0;i<n;i++)
		{
			scanf("%d%d",&a[i].x,&a[i].y);
		}
		k=0;
		for(i=0;i<n;i++) 
		{
			for(j=i+1;j<n;j++)
			{
				
				
					m[k].x=(a[i].x+a[j].x);
					m[k].y=(a[i].y+a[j].y);
					k++;
				 
			}
		}
		qsort(m,k,sizeof(point),cmp);
		ans=0;
		c=1;
		for(i=1;i<k;i++)
		{
			if(m[i].x==m[i-1].x&&m[i].y==m[i-1].y)
				c++;
			else
			{ 
				ans+=c*(c-1)/2;
				c=1;
			}
		}
//		ans+=c*(c-1)/2;
		printf("%d\n",ans);
	}
#if debug
	fclose(stdin);
	fclose(stdout);
#endif
	return 1;
}