Codefroces Educational Round 26 837 B. Flag of Berland

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).

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.

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

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.

大模拟，n m被三整除

#include <iostream>

#include <algorithm>

#include <cstring>

#include <cstdio>

#include <vector>

#include <queue>

#include <cstdlib>

#include <iomanip>

#include <cmath>

#include <ctime>

#include <map>

#include <set>

using namespace std;

#define lowbit(x) (x&(-x))

#define max(x,y) (x>y?x:y)

#define min(x,y) (x<y?x:y)

#define MAX 100000000000000000

#define MOD 1000000007

#define pi acos(-1.0)

#define ei exp(1)

#define PI 3.141592653589793238462

#define ios() ios::sync_with_stdio(false)

#define INF 0x3f3f3f3f

#define mem(a) (memset(a,0,sizeof(a)))

typedef long long ll;

int n,m;

char s[][];

char name[]={'B','G','R'};

bool check(int x,int n,int y,int m,char c)

{

    for(int i=x;i<=n;i++)

    {

        for(int j=y;j<=m;j++)

        {

            if(s[i][j]!=c) return false;

        }

    }

    return true;

}

int main()

{

    scanf("%d%d",&n,&m);

    for(int i=;i<=n;i++)

        scanf("%s",s[i]+);

    bool flag=false;

    do

    {

        if(n%==)

        {

            bool falg=true;

            for(int i=;i<;i++)

            {

                if(!check(i*n/+,(i+)*n/,,m,name[i])) falg=false;

            }

            if(falg)flag=true;

        }

        if(m%==)

        {

            bool falg=true;

            for(int i=;i<;i++)

            {

                if(!check(,n,i*m/+,(+i)*m/,name[i])) falg=false;

            }

            if(falg)flag=true;

        }

    }while(next_permutation(name,name+));

    puts(flag?"YES":"NO");

    return ;

}