zoukankan      html  css  js  c++  java
  • CodeForces

    Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u

     Status

    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:

    • 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 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

    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

    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);
        return 0;
    }
  • 相关阅读:
    170929-关于md5加密
    170911-关于maven的知识点
    opencv-python 学习初探1
    使用PDFminer3k解析pdf为文字遇到:WARING:root:GBK-EUC-H
    Python time strptime()与time strftime()
    chromedriver下载安装
    计数
    高效的几个小技巧
    phantomjs在win10下的安装
    win10下安装lxml
  • 原文地址:https://www.cnblogs.com/Ritchie/p/5425257.html
Copyright © 2011-2022 走看看