Codeforces Round #263 （div2) A. Appleman and Easy Task-优快云博客

本文介绍了一个关于验证n×n大小的棋盘格中每个格子周围'o'字符数量是否为偶数的问题。通过遍历棋盘并计算每个格子相邻的'o'字符数量来判断整体布局的有效性。

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


判断每个格子是否有偶数个'o'




代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int a[110][110];
long long ans;
int main()
{
    char s[110];
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%s",s);
            for(int j=0;j<n;j++)//
            {
                if(s[j]=='o')
                a[i][j+1]=1;
                else
                a[i][j+1]=0;
            }
        }
        ans=0;
        int sign=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                ans=0;
                if(j>1&&a[i][j-1])
                {
                    ans++;
                }
                if(i>1&&a[i-1][j])
                {
                    ans++;
                }
                if(i<n&&a[i+1][j])
                {
                    ans++;
                }
                if(j<n&&a[i][j+1])
                {
                    ans++;
                }
                if(ans%2)
                {
                    sign=1;
                    break;
                }
            }
            if(sign)
            break;
        }
        if(sign)
        printf("NO\n");
        else
        printf("YES\n");
    }
    return 0;
}