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  • 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
    ";
    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
    ";
    37         }
    38         else {
    39             cout << "NO
    ";
    40         }
    41     }
    42     return 0;
    43 
    44 }
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  • 原文地址:https://www.cnblogs.com/z-712/p/7307650.html
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