Educational Codeforces Round 59 (Rated for Div. 2)

Problem  Educational Codeforces Round 59 (Rated for Div. 2) - D. Compression

Time Limit: 2500 mSec

Problem Description

Input

 

Output

 Print one number: maximum x such that an x-compression of the given matrix is possible.

Sample Input

8
E7
E7
E7
00
00
E7
E7
E7

Sample Output

1

题解：首先可以知道对于一个符合条件的x，原矩阵要能够划分成 x * x的小矩阵，每个小矩阵中的元素是相同的，这是充要条件。接下来就是从大到小遍历n的约数判断是否满足该条件即可，如果每次都暴力判断那最差的情况下是O(约数个数 * n^2)，约数的数量多一点就无法承受，主要到矩阵中的元素只可能是0和1，那么如果小矩阵中元素都相等，那么就都为0或者都为1，这时就可以通过小矩阵的元素和来判断是否元素都相同。想到这题目就结束了，二维前缀和预处理一下即可。

#include <bits/stdc++.h>

using namespace std;

#define REP(i, n) for (int i = 1; i <= (n); i++)
#define sqr(x) ((x) * (x))

const int maxn = 5200 + 5;
const int maxt = 600 + 10;
const int maxm = 200000 + 100;
const int maxs = 52;

typedef long long LL;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;

const LL unit = 1LL;
const int INF = 0x3f3f3f3f;
const LL Inf = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-14;
const double inf = 1e15;
const double pi = acos(-1.0);
const LL mod = 1000000007;

int n;
bool gra[maxn][maxn];
int dp[maxn][maxn];
char str[maxn];

inline int cal(char ch)
{
    if (isdigit(ch))
        return ch - '0';
    return ch - 'A' + 10;
}

void Fill(int x, int y, char ch)
{
    int num = cal(ch);
    for (int j = 3; j >= 0; j--)
    {
        if (num & (1 << j))
            gra[x][y + 3 - j] = true;
    }
}

void DP()
{
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + gra[i][j];
        }
    }
}

bool solve(int l)
{
    int tot = l * l;
    for (int i = l; i <= n; i += l)
    {
        for (int j = l; j <= n; j += l)
        {
            int sum = dp[i][j] - dp[i - l][j] - dp[i][j - l] + dp[i - l][j - l];
            if (sum && sum != tot)
            {
                return false;
            }
        }
    }
    return true;
}

void out()
{
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            cout << dp[i][j] << " ";
        }
        cout << endl;
    }
}

int main()
{
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    memset(gra, 0, sizeof(gra));
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
    {
        scanf("%s", str);
        for (int j = 0; j < n / 4; j++)
        {
            Fill(i, j * 4 + 1, str[j]);
        }
    }
    DP();
    //out();
    int ans = 0;
    for (int x = n; x >= 2; x--)
    {
        if (n % x)
            continue;
        if (solve(x))
        {
            ans = x;
            break;
        }
    }
    if (!ans)
    {
        cout << 1 << endl;
    }
    else
    {
        cout << ans << endl;
    }

    return 0;
}