CodeForces462 A. Appleman and Easy Task

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A. Appleman and Easy Tasktime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

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

 

英語真的是差啊，沒辦法，沒理解，好好學英語，這學期過CET4，加油！

遍曆搞一下這道題就過了。

 1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5  6 using namespace std; 7 const int max_size = 105; 8  9 int d1[] = {-1, 0, 0, 1};10 int d2[] = {0, -1, 1, 0};11 12 bool check(int x, int y, int n)13 {14     if(x >= 0 && y >= 0 && x < n && y < n)15         return true;16     return false;17 }18 19 int main()20 {21     int n;22     int graph[max_size][max_size];23 24     while(scanf("%d%*c", &n) != EOF)25     {26         for(int i = 0; i < n; i++)27         {28             for(int j = 0; j < n; j++)29             {30                 char a = getchar();31                 if(a == ‘o‘)32                     graph[i][j] = 1;33                 else34                     graph[i][j] = 0;35             }36             getchar();37         }38 39         int tag = false;40         for(int i = 0; i < n; i++)41         {42             for(int j = 0; j < n; j++)43             {44                 int ans = 0;45                 int x, y;46                 for(int k = 0; k < 4; k ++)47                 {48                     x = i + d1[k];49                     y = j + d2[k];50                     if(check(x, y, n))51                     {52                         if(graph[x][y] == 1)53                             ans++;54                     }55                 }56                 if(ans % 2 != 0)57                 {58                     printf("NO\n");59                     tag = true;60                     break;61                 }62             }63             if(tag == true)64                 break;65         }66         if(tag == false)67             printf("YES\n");68     }69     return 0;70 }

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CodeForces462 A. Appleman and Easy Task