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You are given a binary matrix A of size n×n. Let’s denote an x-compression of the given matrix as a matrix B of size nx×nx such that for every i∈[1,n],j∈[1,n] the condition A[i][j]=B[⌈ix⌉][⌈jx⌉] is met.
Obviously, x-compression is possible only if x divides n, but this condition is not enough. For example, the following matrix of size 2×2 does not have any 2-compression:
01
10
For the given matrix A, find maximum x such that an x-compression of this matrix is possible.
Note that the input is given in compressed form. But even though it is compressed, you’d better use fast input.
Input
The first line contains one number n (4≤n≤5200) — the number of rows and columns in the matrix A. It is guaranteed that n is divisible by 4.
Then the representation of matrix follows. Each of n next lines contains n4 one-digit hexadecimal numbers (that is, these numbers can be represented either as digits from 0 to 9 or as uppercase Latin letters from A to F). Binary representation of each of these numbers denotes next 4 elements of the matrix in the corresponding row. For example, if the number B is given, then the corresponding elements are 1011, and if the number is 5, then the corresponding elements are 0101.
Elements are not separated by whitespaces.
Output
Print one number: maximum x such that an x-compression of the given matrix is possible.
Examples
inputCopy
8
E7
E7
E7
00
00
E7
E7
E7
outputCopy
1
inputCopy
4
7
F
F
F
outputCopy
1
题意：
给你n*n的矩阵，但是是用16进制来给，每一位都是0或者1，问你它最大的压缩比是多少，压缩就是 i∈[1,n],j∈[1,n] the condition A[i][j]=B[⌈i/x⌉][⌈j/x⌉] 向上取整。
题解：
那么我们可以得到，压缩之后，就是将A矩阵分成了一块一块的小矩阵，并且小矩阵里所有元素是一样的，那我们枚举n的所有因子，再用前缀和检查每一个矩形是否全1或者全0
#include<bits/stdc++.h>
using namespace std;
char mp[5205][5205];
int sum[5205][5205];
int main()
{int n;scanf("%d",&n);char a;for(int i=1;i<=n;i++){getchar();for(int j=1;j<=n;j+=4){scanf("%c",&a);int x=a>'9'?a-'A'+10:a-'0';for(int k=0;k<4;k++)sum[i][j+k]=(x>>(3-k))&1;}}for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)sum[i][j]+=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1];for(int i=n;i>=1;i--){if(n%i)continue;int flag=1;for(int j=1;j*i<=n;j++){for(int k=1;k*i<=n;k++){int s=sum[i*j][i*k]-sum[i*j][i*(k-1)]-sum[i*(j-1)][i*k]+sum[i*(j-1)][i*(k-1)];flag&=(s==0||s==i*i);}}if(flag)return 0*printf("%d\n",i);}
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