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, 2, n) {
        ec[i] = true;
        rep(j, 1, n) {
            if (Matrix[i][j] != Matrix[i - 1][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
", 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 = 0, s = 1, c = gc;
    while (c <= 32)    c = gc;
    if (c == '-')    s = -1, c = gc;
    for (; isdigit(c); c = gc)
        x = x * 10 + c - 48;
    return x * s;
}

const int maxn = 5205;
int n;
string s;
int Matrix[maxn][maxn];
bool ec[maxn], er[maxn];

void Deal(int i, string s) {
    rep(ss, 0, s.length() - 1) {
        int k = isdigit(s[ss]) ? s[ss] - '0' : 10 + s[ss] - 'A';
        rep(t, 0, 3) {
            Matrix[i][(ss << 2) + t + 1] = k >> (3 - t) & 1;
        }
    }
}

int solve() {
    rep(i, 2, n) {
        ec[i] = true;
        rep(j, 1, n) {
            if (Matrix[i][j] != Matrix[i - 1][j]) {
                ec[i] = false;
                break;
            }
        }
    }
    rep(j, 2, n) {
        er[j] = true;
        rep(i, 1, n) {
            if (Matrix[i][j] != Matrix[i][j - 1]) {
                er[j] = false;
                break;
            }    
        }
    }

    irep(i, n, 1) {
        if (n % i)    continue;
        bool flag = true;
        rep(j, 1, n) {
            if ((j - 1) % i != 0 && (not ec[j] || not er[j])) {
                flag = false;
                break;
            }
        }
        if (flag)    return i;
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    cin >> n;
    rep(i, 1, n) {
        cin >> s;
        Deal(i, s);
    }

    cout << solve() << endl;
    return 0;
}