题目本质：如果答案是i，那么从行和列两维都会满足：以i的倍数分块，矩阵值相同。

一种解决方法：

1.首先题目里说了要在n的约数里找orzorz……

2.块中需要一整排都相同。用“与前一排相同否？”来判定，而每块的第一排允许与上一排不同。复杂度还是n^2。

1 rep(i, 2, n) {
2         ec[i] = true;
3         rep(j, 1, n) {
4             if (Matrix[i][j] != Matrix[i - 1][j]) {
5                 ec[i] = false;
6                 break;
7             }
8         }
9     }

行和列都弄一遍以上代码。

 

  1 #pragma comment(linker, "/STACK:1024000000,1024000000")
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cstdlib>
  5 #include <cmath>
  6 #include <ctime>
  7 #include <cctype>
  8 #include <climits>
  9 #include <iostream>
 10 #include <iomanip>
 11 #include <algorithm>
 12 #include <string>
 13 #include <sstream>
 14 #include <stack>
 15 #include <queue>
 16 #include <set>
 17 #include <map>
 18 #include <vector>
 19 #include <list>
 20 #include <fstream>
 21 #define ri readint()
 22 #define gc getchar()
 23 #define R(x) scanf("%d", &x)
 24 #define W(x) printf("%d\n", x)
 25 #define init(a, b) memset(a, b, sizeof(a))
 26 #define rep(i, a, b) for (int i = a; i <= b; i++)
 27 #define irep(i, a, b) for (int i = a; i >= b; i--)
 28 #define ls p << 1
 29 #define rs p << 1 | 1
 30 using namespace std;
 31 
 32 typedef double db;
 33 typedef long long ll;
 34 typedef unsigned long long ull;
 35 typedef pair<int, int> P;
 36 const int inf = 0x3f3f3f3f;
 37 const ll INF = 1e18;
 38 
 39 inline int readint() {
 40     int x = 0, s = 1, c = gc;
 41     while (c <= 32)    c = gc;
 42     if (c == '-')    s = -1, c = gc;
 43     for (; isdigit(c); c = gc)
 44         x = x * 10 + c - 48;
 45     return x * s;
 46 }
 47 
 48 const int maxn = 5205;
 49 int n;
 50 string s;
 51 int Matrix[maxn][maxn];
 52 bool ec[maxn], er[maxn];
 53 
 54 void Deal(int i, string s) {
 55     rep(ss, 0, s.length() - 1) {
 56         int k = isdigit(s[ss]) ? s[ss] - '0' : 10 + s[ss] - 'A';
 57         rep(t, 0, 3) {
 58             Matrix[i][(ss << 2) + t + 1] = k >> (3 - t) & 1;
 59         }
 60     }
 61 }
 62 
 63 int solve() {
 64     rep(i, 2, n) {
 65         ec[i] = true;
 66         rep(j, 1, n) {
 67             if (Matrix[i][j] != Matrix[i - 1][j]) {
 68                 ec[i] = false;
 69                 break;
 70             }
 71         }
 72     }
 73     rep(j, 2, n) {
 74         er[j] = true;
 75         rep(i, 1, n) {
 76             if (Matrix[i][j] != Matrix[i][j - 1]) {
 77                 er[j] = false;
 78                 break;
 79             }    
 80         }
 81     }
 82 
 83     irep(i, n, 1) {
 84         if (n % i)    continue;
 85         bool flag = true;
 86         rep(j, 1, n) {
 87             if ((j - 1) % i != 0 && (not ec[j] || not er[j])) {
 88                 flag = false;
 89                 break;
 90             }
 91         }
 92         if (flag)    return i;
 93     }
 94 }
 95 
 96 int main() {
 97     ios_base::sync_with_stdio(false);
 98     cin.tie(0);
 99 
100     cin >> n;
101     rep(i, 1, n) {
102         cin >> s;
103         Deal(i, s);
104     }
105 
106     cout << solve() << endl;
107     return 0;
108 }