codeforces 660D D. Number of Parallelograms(计算几何)

题目链接：

D. Number of Parallelograms

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.

Input

The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.

Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.

Output

Print the only integer c — the number of parallelograms with the vertices at the given points.

Example

input

4
0 1
1 0
1 1
2 0

output

1


题意：

给了这么些点,问能形成多少个平行四边形;

思路：

把所有的线段找出来,按长度排序,平行四边形对边长度相等且互相平行,然后判断一下就好了,我的由于每个都算了4遍,所以最后/4;
比赛还没完我就写题解了,说不定还要被hack,好方,哈哈哈哈;

AC代码：

/*
2014300227    660D - 4    GNU C++11    Accepted    468 ms    80348 KB
*/
#include <bits/stdc++.h>
using namespace std;
const int N=1e6+20;
typedef long long ll;
const double PI=acos(-1.0);
int n,cnt=1,vis[4*N];
ll x[2005],y[2005];
struct Line
{
    int fi,se;
    ll le;
};
Line line[4*N];
int cmp(Line a,Line b)
{
    return a.le<b.le;
}
int findpos(ll num)
{
    int l=1,r=cnt-1,mid;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(line[mid].le<num)l=mid+1;
        else r=mid-1;
    }
    return l;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%I64d%I64d",&x[i],&y[i]);
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=i+1;j<=n;j++)
        {
            line[cnt].fi=i;
            line[cnt].se=j;
            line[cnt++].le=(x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]);
        }
    }

    sort(line+1,line+cnt,cmp);
    vis[1]=1;
    for(int i=2;i<cnt;i++)
    {
        if(line[i].le==line[i-1].le)
        {
            vis[i]=vis[i-1];
        }
        else vis[i]=i;
    }
    int ans=0;
    for(int i=1;i<cnt;i++)
    {
        int pos=vis[i];
        for(int j=pos;j<cnt;j++)
        {
            if(line[j].le>line[i].le)break;
            if(line[j].fi==line[i].se||line[j].se==line[i].fi||line[j].fi==line[i].fi||line[j].se==line[i].se)continue;
            int cx,cy,dx,dy;
            cx=line[i].fi;
            cy=line[i].se;
            dx=line[j].fi;
            dy=line[j].se;
            if((x[cx]-x[cy])*(y[dx]-y[dy])==(y[cx]-y[cy])*(x[dx]-x[dy]))
            ans++;
        }
    }
    cout<<ans/4<<"
";

    return 0;
}