Codeforces Round #263 (Div. 2)A. Appleman and Easy Task

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本文介绍了一个简单的编程挑战：验证一个n×n的棋盘上每个格子周围是否有偶数个'o'字符。通过遍历棋盘并计数相邻的'o'来判断条件是否满足，并给出了完整的C++实现代码。

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

Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two 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代码如下：#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;

int n;
char map[110][110];
const int dir[4][2] = {0,-1,0,1,-1,0,1,0};

int main()
{
    int ans;
    bool flag;
    int i,j;
    while(scanf("%d",&n)!=EOF)
    {
        getchar();
        memset(map,'x',sizeof(map));
        for(i = 1; i <= n; i++)
        {
            for(j = 1; j <= n; j++)
                map[i][j] = getchar();
            getchar();
        }
        flag = true;
        for(i = 1; i <= n; i++)
        {
            for(j = 1; j <= n; j++)
            {
                ans = 0;
                for(int k = 0; k < 4; k++)
                {
                    int x = i+dir[k][0];
                    int y = j+dir[k][1];
                    if(map[x][y] == 'o') ans++;
                }
                if(ans&1)
                {
                    flag = false;
                    break;
                }
            }
        }
        flag?puts("YES"):puts("NO");
    }
    return 0;
}