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| Time Limit: 1000MS | Memory Limit: 262144KB | 64bit IO Format: %I64d & %I64u |
Description
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 a2x × y matrix c which has the following properties:
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 it several(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 Input
4 3
0 0 1
1 1 0
1 1 0
0 0 1
2
3 3
0 0 0
0 0 0
0 0 0
3
8 1
0
1
1
0
0
1
1
0
2
Hint
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
Source
#include <iostream> #include <algorithm> #include <string.h> #include <stdlib.h> #include <math.h> #include <stdio.h> using namespace std; int a[105][105]; int main() { int i,j,ii,jj,m,n,tmp,ans; scanf("%d%d",&n,&m); ans=n; for(i=0;i<n;i++) for(j=0;j<m;j++) scanf("%d",&a[i][j]); int flag=1; while(n%2==0) { for(i=0;i<n/2;i++) { for(j=0;j<m;j++) { if(a[i][j]!=a[n-1-i][j]) { flag=0; break; } } if(flag==0) break; } if(flag) n/=2; else break; } printf("%d\n",n); return 0; }
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原文地址：http://www.cnblogs.com/Ritchie/p/5425257.html
