CF ECR59div2 D

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。

 rep(i, , n) {
         ec[i] = true;
         rep(j, , n) {
             if (Matrix[i][j] != Matrix[i - ][j]) {
                 ec[i] = false;
                 break;
             }
         }
     }

行和列都弄一遍以上代码。

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <ctime>
 #include <cctype>
 #include <climits>
 #include <iostream>
 #include <iomanip>
 #include <algorithm>
 #include <string>
 #include <sstream>
 #include <stack>
 #include <queue>
 #include <set>
 #include <map>
 #include <vector>
 #include <list>
 #include <fstream>
 #define ri readint()
 #define gc getchar()
 #define R(x) scanf("%d", &x)
 #define W(x) printf("%d\n", x)
 #define init(a, b) memset(a, b, sizeof(a))
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define irep(i, a, b) for (int i = a; i >= b; i--)
 #define ls p << 1
 #define rs p << 1 | 1
 using namespace std;

 typedef double db;
 typedef long long ll;
 typedef unsigned long long ull;
 typedef pair<int, int> P;
 const int inf = 0x3f3f3f3f;
 const ll INF = 1e18;

 inline int readint() {
     int x = , s = , c = gc;
     while (c <= )    c = gc;
     if (c == '-')    s = -, c = gc;
     for (; isdigit(c); c = gc)
         x = x *  + c - ;
     return x * s;
 }

 const int maxn = ;
 int n;
 string s;
 int Matrix[maxn][maxn];
 bool ec[maxn], er[maxn];

 void Deal(int i, string s) {
     rep(ss, , s.length() - ) {
         int k = isdigit(s[ss]) ? s[ss] - '' :  + s[ss] - 'A';
         rep(t, , ) {
             Matrix[i][(ss << ) + t + ] = k >> ( - t) & ;
         }
     }
 }

 int solve() {
     rep(i, , n) {
         ec[i] = true;
         rep(j, , n) {
             if (Matrix[i][j] != Matrix[i - ][j]) {
                 ec[i] = false;
                 break;
             }
         }
     }
     rep(j, , n) {
         er[j] = true;
         rep(i, , n) {
             if (Matrix[i][j] != Matrix[i][j - ]) {
                 er[j] = false;
                 break;
             }
         }
     }

     irep(i, n, ) {
         if (n % i)    continue;
         bool flag = true;
         rep(j, , n) {
             if ((j - ) % i !=  && (not ec[j] || not er[j])) {
                 flag = false;
                 break;
             }
         }
         if (flag)    return i;
     }
 }

 int main() {
     ios_base::sync_with_stdio(false);
     cin.tie();

     cin >> n;
     rep(i, , n) {
         cin >> s;
         Deal(i, s);
     }

     cout << solve() << endl;
     return ;
 }