CodeForces462 A. Appleman and Easy Task

A. Appleman and easy tasktime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard ou Tput

Toastman came up and a very easy task. He gives it to Appleman, but Appleman doesn ' t know what to solve it. Can you help him?

Given a n x n Checkerboard. Each cell of the board has either character 'X ', or character 'o '. Is it true this each cell of the board have even number of adjacent cells with 'o '? Both cells of the board are adjacent if they share a side.

Input

The first line contains an integer n (1≤ n ≤100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x ' or 'o ') without spaces.

Output

Print "YES" or "NO" (without the quotes) depending on the answer to the problem.

Sample Test (s) input

3
Xxo
XOX
Oxx

Output

YES

Input

4
Xxxo
xoxo
Oxox
Xxxx

Output

NO

English is really poor ah, no way, did not understand, learn English well, this semester over CET4, refueling!

It's too soon to go through this problem.

1#include <cstdio>2#include <cstring>3#include <iostream>4#include <algorithm>5 6 using namespacestd;7 Const intMax_size = the;8 9 intD1[] = {-1,0,0,1};Ten intD2[] = {0, -1,1,0}; One  A BOOLCheckintXintYintN) - { -     if(x >=0&& y >=0&& x < n && y <N) the         return true; -     return false; - } -  + intMain () - { +     intN; A     intGraph[max_size][max_size]; at  -      while(SCANF ("%d%*c", &n)! =EOF) -     { -          for(inti =0; I < n; i++) -         { -              for(intj =0; J < N; J + +) in             { -                 CharA =GetChar (); to                 if(A = ='o') +GRAPH[I][J] =1; -                 Else theGRAPH[I][J] =0; *             } $ GetChar ();Panax Notoginseng         } -  the         intTag =false; +          for(inti =0; I < n; i++) A         { the              for(intj =0; J < N; J + +) +             { -                 intAns =0; $                 intx, y; $                  for(intK =0; K <4; K + +) -                 { -x = i +D1[k]; they = j +D2[k]; -                     if(check (x, y, n))Wuyi                     { the                         if(Graph[x][y] = =1) -ans++; Wu                     } -                 } About                 if(Ans%2!=0) $                 { -printf"no\n"); -Tag =true; -                      Break; A                 } +             } the             if(Tag = =true) -                  Break; $         } the         if(Tag = =false) theprintf"yes\n"); the     } the     return 0; -}

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CodeForces462 A. Appleman and Easy Task