Original question link: C1. Heidi and the Turing Test (Easy)
I'll give you a plane rectangular coordinate system, and the coordinates of some points (there are \ (4 \cdot n + 1 \). You know that there must be a square, and there is an extra point beside the square. Now please find out that point (Note: the points on each side are not less than \ (n \))
Problem solving: this problem is really a pit. When I first saw this problem, the first reaction was to find the number of points on the row and column. If it was greater than or equal to \ (n \), it was determined as a side. However, various judgments made me defenseless, and I gave up the idea.
This method can't work. Then violence is OK. The data range is so small. We can enumerate the deleted points, and then judge by violence.

#include <cstdio>
#include <algorithm>
using namespace std;
#define Maxn 41
struct Node{
    int x,y;
    friend bool operator <(Node p,Node q){
        if(p.x==q.x){
            return p.y<q.y;
        }
        return p.x<q.x;
    }
}a[Maxn+5];
int mn(int a,int b){
    return a<b?a:b;
}
int mx(int a,int b){
    return a>b?a:b;
}
void swp(int &a,int &b){
    int t=a;
    a=b;
    b=t;
}
int n,m;
bool cmp(Node p,Node q){
    if(p.y==q.y){
        return p.x<q.x;
    }
    return p.y<q.y;
}
int x[Maxn+5],y[Maxn+5];
bool check(){
    int mn_x=55,mn_y=55,mx_x=-1,mx_y=-1;
    for(int i=1;i<m;i++){
        mn_x=mn(mn_x,x[i]);
        mn_y=mn(mn_y,y[i]);
        mx_x=mx(mx_x,x[i]);
        mx_y=mx(mx_y,y[i]);
    }
    if(mx_x-mn_x!=mx_y-mn_y){
        return 0;
    }
    for(int i=1;i<m;i++){
        if(x[i]!=mn_x&&x[i]!=mx_x&&y[i]!=mn_y&&y[i]!=mx_y){
            return 0;
        }
    }
    return 1;
}
int main(){
    scanf("%d",&n);
    m=(n<<2)+1;
    for(int i=1;i<=(n<<2)+1;i++){
        scanf("%d%d",&a[i].x,&a[i].y);
    }
    sort(a+1,a+1+(n<<2)+1);
    for(int i=1;i<=m;i++){
        for(int j=1;j<=m;j++){
            if(i==j){
                continue;
            }
            if(j>i){
                x[j-1]=a[j].x;
                y[j-1]=a[j].y;
            }
            else{
                x[j]=a[j].x;
                y[j]=a[j].y;
            }
        }
        if(check()){
            printf("%d %d\n",a[i].x,a[i].y);
            break;
        }
    }
    return 0;
}