【Link】:

【Description】



给你n个点整数点;

问你这n个点,能够组成多少个正多边形

【Solution】



整点只能构成正四边形.

则先把所有的边预处理出来;

枚举每某两条边为对角线的情况;

看看这两条对角线能否组成一个正方形;

可以的话,递增答案.



【NumberOf WA】



1



【Reviw】



(排序的时候边的数目和点的弄混了)



【Code】

#include <bits/stdc++.h>

#define int long long

using namespace std;

const int N = 500;

struct abc{

    int x[2],y[2],dis;

};

abc bian[N*N+100];

int x[N+10],y[N+10];

int n,tot;

int sqr(int x){

    return x*x;

}

int get_dis(int i,int j){

    return sqr(x[i]-x[j])+sqr(y[i]-y[j]);

}

bool cmp(abc a,abc b){

    return a.dis < b.dis;

}

main(){

    while (~scanf("%lld",&n)){

        int ans = 0;

        tot = 0;

        for (int i = 1;i <= n;i++)

            scanf("%lld%lld",&x[i],&y[i]);

        for (int i = 1;i <= n-1;i++)

            for (int j = i+1;j <= n;j++){

                tot++;

                bian[tot].x[0] = x[i],bian[tot].y[0] = y[i];

                bian[tot].x[1] = x[j],bian[tot].y[1] = y[j];

                bian[tot].dis = get_dis(i,j);

            }

        sort(bian+1,bian+1+tot,cmp);

        for (int i = 1;i <= tot;i++){

            int j = i;

            while (j+1 <= tot && bian[j+1].dis == bian[i].dis) j++;

            for (int k = i;k <= j-1;k++)

                for (int l = k+1;l <= j;l++){

                    int x1 = bian[k].x[0],y1 = bian[k].y[0];

                    int x2 = bian[k].x[1],y2 = bian[k].y[1];

                    int x3 = bian[l].x[0],y3 = bian[l].y[0];

                    int x4 = bian[l].x[1],y4 = bian[l].y[1];

                    if ( (x1+x2==x3+x4) && (y1+y2 == y3+y4) ){

                        if ( (x2-x1)*(x4-x3) + (y2-y1)*(y4-y3) == 0){

                            ans++;

                        }

                    }

                }

            i = j;

        }

        printf("%lld\n",ans);

    }

    return 0;

}