Educational Codeforces Round 26 Problem B

蔚落 2021-12-23 11:55 175阅读 0赞

> 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.

题目大意还不是很好表达。就是看能不能分成均等的三份；

只要遍历一遍就好了。

模拟吧，代码有点长

1 #include<iostream>
     2 #include<stdio.h>
     3 using namespace std;
     4 char a[105][105];
     5 int hang(int n,int m){           //验证行是否分成了三份
     6     char fir=a[1][1],sec=a[n/3+1][1],three=a[n][1];
     7     if(fir==sec||fir==three||sec==three){
     8         return false;
     9     }
    10     for(int i=1;i<=n;i++){
    11         for(int j=1;j<=m;j++){
    12             if(i<=n/3){
    13                 if(a[i][j]!=fir){
    14                     return false;
    15                 }
    16             }else if(i<=2*(n/3)){
    17                 if(a[i][j]!=sec){
    18                     return false;
    19                 }
    20             }else{
    21                 if(a[i][j]!=three){
    22                     return false;
    23                 }
    24             }
    25         }
    26     }
    27     return true;
    28 }
    29 int lie(int n,int m){       //验证列是否分成三份
    30     char fir=a[1][1],sec=a[1][m/3+1],three=a[1][m];
    31     if(fir==sec||fir==three||sec==three){    
    32         return false;
    33     }
    34     for(int i=1;i<=n;i++){
    35         for(int j=1;j<=m;j++){
    36             if(j<=m/3){
    37                 if(a[i][j]!=fir){
    38                     return false;
    39                 }
    40             }else if(j<=2*(m/3)){
    41                 if(a[i][j]!=sec){
    42                     return false;
    43                 }
    44             }else{
    45                 if(a[i][j]!=three){
    46                     return false;
    47                 }
    48             }
    49         }
    50     }
    51     return true;
    52 }
    53 int main(){
    54     int m,n;
    55     cin>>n>>m;
    56     for(int i=1;i<=n;i++){
    57         for(int j=1;j<=m;j++){
    58             cin>>a[i][j];
    59         }
    60     }
    61     if(n%3!=0&&m%3!=0){
    62         cout<<"NO"<<endl;
    63     }else{
    64         if(n%3==0&&m%3==0){
    65             if(hang(n,m)||lie(n,m)){
    66                 cout<<"YES"<<endl;
    67             }else{
    68             cout<<"NO"<<endl;
    69             }
    70         }else if(n%3==0){
    71             if(hang(n,m)){
    72                 cout<<"YES"<<endl;
    73             }else{
    74             cout<<"NO"<<endl;
    75             }
    76         }else{
    77             if(lie(n,m)){
    78                 cout<<"YES"<<endl;
    79             }else{
    80                 cout<<"NO"<<endl;
    81             }
    82         }
    83     }
    84     return 0;
    85 }

转载于:https://www.cnblogs.com/ISGuXing/p/7285051.html