Codeforces Round #243 (Div. 2) B（思维模拟题）

http://codeforces.com/contest/426/problem/B

B. Sereja and Mirroring

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's assume that we are given a matrix b of size x × y, let's determine the operation of mirroring matrix b. The mirroring of matrix b is a 2x × y matrix c which has the following properties:

• the upper half of matrix c (rows with numbers from 1 to x) exactly matches b;
• the lower half of matrix c (rows with numbers from x + 1 to 2x) is symmetric to the upper one; the symmetry line is the line that separates two halves (the line that goes in the middle, between rows x and x + 1).

Sereja has an n × m matrix a. He wants to find such matrix b, that it can be transformed into matrix a, if we'll perform on itseveral (possibly zero) mirrorings. What minimum number of rows can such matrix contain?

Input

The first line contains two integers, n and m (1 ≤ n, m ≤ 100). Each of the next n lines contains m integers — the elements of matrix a. The i-th line contains integers ai1, ai2, ..., aim (0 ≤ aij ≤ 1) — the i-th row of the matrix a.

Output

In the single line, print the answer to the problem — the minimum number of rows of matrix b.

Sample test(s)

input

4 3
0 0 1
1 1 0
1 1 0
0 0 1

output

2

input

3 3
0 0 0
0 0 0
0 0 0

output

3

input

8 1
0
1
1
0
0
1
1
0

output

2

Note

In the first test sample the answer is a 2 × 3 matrix b:

001
110

If we perform a mirroring operation with this matrix, we get the matrix a that is given in the input:

001
110
110
001


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题意就是问你给定一个矩阵，求它的最小对称单元

#include<iostream>  
#include<cstdio>  
#include<cmath>  
#include<cstdlib>  
#include <algorithm>  
#include<cstring>  
#include<string>  
using namespace std;  
  
int a[100][100], b[100][100], c[100][100];  
int main()  
{  
    int n, m, i, j, line, temp = 0;  
    scanf("%d%d", &n, &m);  
    for (i = 0; i < n; i++)  
    {  
        for (j = 0; j < m; j++)  
        {  
            scanf("%d", &a[i][j]);  
        }  
    }  
    line = n;  
    while (!(line % 2))  
    {  
        for (i = 0; i < line / 2; i++)  
        {  
            for (j = 0; j < m; j++)  
            {  
                if (a[i][j] != a[line - 1 - i][j])  
                {  
                    temp++;  
                }  
            }  
        }  
        if (temp)  
        {  
            break;  
        }  
        else  
        {  
            line /= 2;  
        }  
    }  
    printf("%d
", line);  
    return 0;  
}