Codeforces 1333 A. Little Artem

[1][1]格子是W，其余都是B即可

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• 题目：
  nxm的网格，只能是黑白两种颜色，有至少一个相邻格子为白色的黑色数目为B，至少一个相邻格子是黑色的数目是W；
  输入n,m。输出一种B = W + 1的涂色方法。

• 题解：
  分类讨论n,m是奇偶的情况即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<queue>
#include<vector>
#include<string>
#include<fstream>
 
using namespace std;
typedef long long ll;
const int N = 3e6 + 105;
const int mod = 998244353;
const double Pi = acos(- 1.0);
const int INF = 0x3f3f3f3f;
const int G = 3, Gi = 332748118;
 
//
 
//main()
ll T, n, m, x, y;
ll a[N];
ll res[150][150];
 
int main()
{
    scanf("%lld",&T);
    while(T --){
        scanf("%lld%lld",&n,&m);
            for(int i = 1; i <= n; ++ i){
                for(int j = 1; j <= m; j += 2){
                    res[i][j] = 1;
                    res[i][j + 1] = 0;
                }
            }
            if(m % 2 == 0) {
                res[n - 1][m] = 1;
                res[n - 1][m - 1] = 0;
                res[n][m] = 1;
            }
            else{
                for(int i = 1; i <= n; i += 2){
                    res[i][m] = 1;
                    res[i + 1][m] = 0;
                }
                if(n % 2 == 0) res[n][2] = 1;
            }

            for(int i = 1; i <= n; ++ i){
                for(int j = 1; j <= m; ++ j){
                    if(res[i][j]) printf("B");
                    else printf("W");
                }
                printf("\n");
            }
    }
    return 0;
}