正常没有正方形的限制下，值为i的点个数4i
那么从0开始遍历，第一个不为4i的值就是min(x, y)
由于对称性我们姑且令x为这个值

我们先列举n*m=t的各种情况
对于一对n, m。我们已经知道n,m,x
再由于对称性，我们假设距离(x,y)最远的点在(n, m)。(当然也可能在(1,m))
现在知道了(n,m)到(x,y)为max(a[I])
列方程就能求出y了
然后再暴力验证就好了

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <map>
using namespace std;
const int N = 1e6 + 5;
const int INF = 0x3f3f3f3f;
typedef long long ll;
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1


int A[N];
int mp[N];
int maxNum;

struct Line{
    int k, b;
    Line(int a=0, int _b=0):k(a), b(_b){}
    int Count(int x1, int x2, int y1, int y2) {
        int L = k*x1 + b;
        int R = k*x2 + b;
        if(L > R) swap(L, R);
        
        x1 = L; x2 = R;
        if(y1 < x1) {
            if(y2 < x1) return 0;
            else if(y2 <= x2) 
                return y2 - x1 + 1;
            else return x2 - x1 + 1;
        } else if(y1 <= x2) {
            if(y2 <= x2) 
                return y2 - y1 + 1;
            else return x2 - y1 + 1;
        } else 
            return 0;
    }
};
bool Test(int n, int m, int x) {
    
    if(n < 2*x-1 || m < 2*x-1)
        return false;
    int y = n + m - maxNum - x;
    if(y < x || m - y < x - 1) 
        return false;

//    printf("%d %d\n", n, m);
    for(int i = x; i <= maxNum; ++i) {
        int all = 0;
        int l1 = max(1, x - i);
        int l2 = min(n, x + i);
        int r1 = max(1, y - i);
        int r2 = min(m, y + i);

        Line t1(-1, i+x+y);
        all += t1.Count(l1, l2, r1, r2);

        Line t2(1, i-x+y);
        all += t2.Count(l1, l2, r1, r2);

        Line t3(-1, -i+x+y);
        all += t3.Count(l1, l2, r1, r2);

        Line t4(1, -i-x+y);
        all += t4.Count(l1, l2, r1, r2);

    //    printf("%d ", all);
        if(n - x >= i) all --;
        if(y > i) all --;
        if(m - y >= i) all --;
    //    printf("%d %d\n", i, all);
        if(all != mp[i])
            return false;
    }
    std::printf("%d %d\n%d %d\n", n, m, x, y);
    return true;
}
int main() {
    int t;
    while(~scanf("%d", &t)) {
        memset(mp, 0, sizeof(mp));
        maxNum = -1;
        for(int i = 0; i < t; ++i) {
            scanf("%d", &A[i]);
            mp[A[i]] ++;
            maxNum = max(maxNum, A[i]);
        }
        
        bool flag = true; int X;
        for(int i = 0; ; ++i) {
            int id = i; int num = mp[i];
            if(id == 0) {
                flag &= num == 1;
            } else {
                flag &= num == 4*i;
            }

            if(flag == false) {
                X = i;
                break;
            }
        }
        // for(int i = 0; i < t; ++i) printf("%d ", mp[i]); printf("\n");
        // printf("%d\n", X);

        if(X == 0) {
            printf("-1\n");
            continue;
        }
        flag = false;
        for(int i = 1; i <= sqrt(t) && !flag; ++i) {
            if(t % i == 0) {
                int n = i; int m = t / i;

                flag = Test(n, m, X);
                if(!flag) {
                    flag = Test(m, n, X);
                }
            }
        }
        if(!flag) {
            printf("-1\n");
        }   
    }
    return 0;
}