Educational Codeforces Round 26 Problem B
B. Flag of Berlandtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe 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).
InputThe 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.
OutputPrint "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes).
Examplesinput6 5
RRRRR
RRRRR
BBBBB
BBBBB
GGGGG
GGGGGoutputYESinput4 3
BRG
BRG
BRG
BRGoutputYESinput6 7
RRRGGGG
RRRGGGG
RRRGGGG
RRRBBBB
RRRBBBB
RRRBBBBoutputNOinput4 4
RRRR
RRRR
BBBB
GGGGoutputNONoteThe 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 }