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来自洛谷，原作者为

	阔耐滴小云呀 眼里有光，心中有爱

搬运于2025-08-24 22:17:14，当前版本为作者最后更新于2021-08-19 08:33:43，作者可能在搬运后再次修改，您可在原文处查看最新版

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以下是正文

本题怎么这么少人提交

我们先考虑构造出原正方形经过 $4$ 种轴对称变换以及 $2$ 种旋转变换之后的正方形都构造出来，然后对所得的 $7$ 个正方形都跑一遍二维哈希。

这样我们就可以通过哈希，在 $O(n^2)$ 时间内判断原正方形中是否存在某一类型的某一大小的子正方形。

但是如果我们枚举边长，复杂度就会达到 $O(n^3)$ 级别，显然过不了。

考虑优化：我们发现对于任意一种类型的正方形，它把最外面一圈去掉之后还是满足原来的性质，所以我们可以二分来求。需要注意的是我们不好同时计算奇数边长和偶数边长的正方形，所以要二分两次。

参考代码：

#include <cstdio>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
template < class T > inline T max(T a, T b) { return a > b ? a : b; }
template < class T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}
 
typedef unsigned long long ull;
const int _ = 502;
const ull base1 = 19491001, base2 = 19260817;
int n; ull pw1[_], pw2[_], h[7][_][_];
 
inline ull gethash(int k, int x1, int y1, int x2, int y2) {
    int w = x2 - x1 + 1;
    return h[k][x2][y2] - h[k][x1 - 1][y2] * pw1[w] - h[k][x2][y1 - 1] * pw2[w] + h[k][x1 - 1][y1 - 1] * pw1[w] * pw2[w];
}
 
inline bool check_90(int x1, int y1, int x2, int y2) {
    return gethash(0, x1, y1, x2, y2) == gethash(5, y1, n - x2 + 1, y2, n - x1 + 1);
}
 
inline bool check_180(int x1, int y1, int x2, int y2) {
    return gethash(0, x1, y1, x2, y2) == gethash(6, n - x2 + 1, n - y2 + 1, n - x1 + 1, n - y1 + 1);
}
 
inline bool check_diag(int x1, int y1, int x2, int y2) {
    return gethash(0, x1, y1, x2, y2) == gethash(1, x1, n - y2 + 1, x2, n - y1 + 1)
        || gethash(0, x1, y1, x2, y2) == gethash(2, n - x2 + 1, y1, n - x1 + 1, y2)
        || gethash(0, x1, y1, x2, y2) == gethash(3, y1, x1, y2, x2)
        || gethash(0, x1, y1, x2, y2) == gethash(4, n - y2 + 1, n - x2 + 1, n - y1 + 1, n - x1 + 1);
}
 
inline bool check_4(int x1, int y1, int x2, int y2) {
    return check_180(x1, y1, x2, y2) && check_diag(x1, y1, x2, y2);
}
 
inline bool check_8(int x1, int y1, int x2, int y2) {
    return check_90(x1, y1, x2, y2) && check_diag(x1, y1, x2, y2);
}
 
template < class T > inline bool check(T f, int x) {
    for (rg int i = x; i <= n; ++i)
        for (rg int j = x; j <= n; ++j)
            if  (f(i - x + 1, j - x + 1, i, j)) return 1;
    return 0;
}
 
template < class T > inline int solve(T f) {
    int res = 0, l, r, mid;
    l = 0, r = (n - 1) >> 1;
    while (l < r) {
        mid = (l + r + 1) >> 1;
        if (check(f, mid << 1 | 1)) l = mid; else r = mid - 1;
    }
    res = max(res, l << 1 | 1);
    l = 1, r = n >> 1;
    while (l < r) {
        mid = (l + r + 1) >> 1;
        if (check(f, mid << 1)) l = mid; else r = mid - 1;
    }
    res = max(res, l << 1);
    return res;
}
 
int main() {
#ifndef ONLINE_JUDGE
    file("cpp");
#endif
    read(n);
    for (rg int i = 1; i <= n; ++i)
        for (rg int j = 1; j <= n; ++j) {
            scanf("%1d", &h[0][i][j]);
            h[1][i][n - j + 1] = h[0][i][j];
            h[2][n - i + 1][j] = h[0][i][j];
            h[3][j][i] = h[0][i][j];
            h[4][n - j + 1][n - i + 1] = h[0][i][j];
            h[5][j][n - i + 1] = h[0][i][j];
            h[6][n - i + 1][n - j + 1] = h[0][i][j];
        }
    pw1[0] = 1; for (rg int i = 1; i <= n; ++i) pw1[i] = pw1[i - 1] * base1;
    pw2[0] = 1; for (rg int i = 1; i <= n; ++i) pw2[i] = pw2[i - 1] * base2;
    for (rg int k = 0; k < 7; ++k) {
        for (rg int i = 1; i <= n; ++i) for (rg int j = 1; j <= n; ++j) h[k][i][j] += h[k][i - 1][j] * base1;
        for (rg int i = 1; i <= n; ++i) for (rg int j = 1; j <= n; ++j) h[k][i][j] += h[k][i][j - 1] * base2;
    }
    printf("%d %d %d %d %d\n", solve(check_8), solve(check_90), solve(check_4), solve(check_180), solve(check_diag));
    return 0;
}