Educational Round 26 B. Flag of Berland

B. Flag of Berland
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

The flag of Berland is such rectangular field n?×?m that satisfies following conditions:

  • Flag consists of three colors which correspond to letters ‘R‘, ‘G‘ and ‘B‘.
  • Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color.
  • Each color should be used in exactly one stripe.

You are given a field n?×?m, consisting of characters ‘R‘, ‘G‘ and ‘B‘. Output "YES" (without quotes) if this field corresponds to correct flag of Berland. Otherwise, print "NO" (without quotes).

Input

The first line contains two integer numbers n and m (1?≤?n,?m?≤?100) — the sizes of the field.

Each of the following n lines consisting of m characters ‘R‘, ‘G‘ and ‘B‘ — the description of the field.

Output

Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes).

Examples
Input
6 5
RRRRR
RRRRR
BBBBB
BBBBB
GGGGG
GGGGG
Output
YES
Input
4 3
BRG
BRG
BRG
BRG
Output
YES
Input
6 7
RRRGGGG
RRRGGGG
RRRGGGG
RRRBBBB
RRRBBBB
RRRBBBB
Output
NO
Input
4 4
RRRR
RRRR
BBBB
GGGG
Output
NO
Note

The field in the third example doesn‘t have three parralel stripes.

Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1.

 

 

B R G 数量相同，并且只有一个区域，计下数量，然后记下B G R 出现的位置的最大(x,y)和最小(x,y) 看看这个区域的面积是否和记下的B G R的数量相同就行。。

 

 1 #include <iostream>
 2 #include <stdio.h>
 3 using namespace std;
 4 int n, m;
 5 char s[120][120];
 6 char s2[120][120];
 7 int check(int n, int m) {
 8     if (n % 3 != 0)
 9         return 0;
10     int d = n / 3;
11     if (s[0][0] == s[d][0] || s[0][0] == s[2 * d][0] || s[d][0] == s[2 * d][0])
12         return 0;
13     for (int i = 0; i < n; ++i)
14         for (int j = 0; j < m; ++j)
15             if (s[i][j] != s[(i / d) * d][0])
16                 return 0;
17     return 1;
18 }
19 
20 int main() {
21     cin >> n >> m;
22     for (int i = 0; i < n; ++i)
23         scanf(" %s", s[i]);   //scanf(" %s",s[i])
24     if (check(n, m)) {
25         cout << "YES\n";
26     }
27     else {
28         for (int i = 0; i < n; ++i)
29             for (int j = 0; j < m; ++j)
30                 s2[j][i] = s[i][j];
31         swap(n, m);
32         for (int i = 0; i < n; ++i)
33             for (int j = 0; j < m; ++j)
34                 s[i][j] = s2[i][j];
35         if (check(n, m)) {
36             cout << "YES\n";
37         }
38         else {
39             cout << "NO\n";
40         }
41     }
42     return 0;
43 
44 }

 

posted @ 2017-08-08 16:46  浅忆~  阅读(129)  评论(0)  编辑  收藏  举报