limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Young boy Artem tries to paint a picture, and he asks his mother Medina to help him. Medina is very busy, that’s why she asked for your help.Artem wants to paint an n×mn×m board. Each cell of the board should be colored in white or black.Lets BB be the number of black cells that have at least one white neighbor adjacent by the side. Let WW be the number of white cells that have at least one black neighbor adjacent by the side. A coloring is called good if B=W+1B=W+1.The first coloring shown below has B=5B=5 and W=4W=4 (all cells have at least one neighbor with the opposite color). However, the second coloring is not good as it has B=4B=4, W=4W=4 (only the bottom right cell doesn’t have a neighbor with the opposite color).
Please, help Medina to find any good coloring. It’s guaranteed that under given constraints the solution always exists. If there are several solutions, output any of them.
Input
Each test contains multiple test cases.The first line contains the number of test cases tt (1≤t≤201≤t≤20). Each of the next tt lines contains two integers n,mn,m (2≤n,m≤1002≤n,m≤100) — the number of rows and the number of columns in the grid.
Output
For each test case print nn lines, each of length mm, where ii-th line is the ii-th row of your colored matrix (cell labeled with ‘B’ means that the cell is black, and ‘W’ means white). Do not use quotes.It’s guaranteed that under given constraints the solution always exists.
Example
Input
2
3 2
3 3
Output
BW
WB
BB
BWB
BWW
BWB
NoteIn the first testcase, B=3B=3, W=2W=2.In the second testcase, B=5B=5, W=4W=4. You can see the coloring in the statement.
大家好,我是弱弱,为什么要深夜放毒呢,自然是因为我想纪念一下自己不仔细审题(没长脑子),当然了这跟我的差劲的英文也是有关的(要是我的英文很好,我估计会更喜欢打CF,不至于像现在这么艰难/(ㄒoㄒ)/~~),所以说好好学英语吧,加油(ง •_•)ง!!
思路
就是找规律,主要是题意,懂了题,自然会做
给你两个样例你就明白了
3 2
BB
BW
BW
3 3
BBB
BWW
BWW
好了,样例在这,就不用我多说了吧,人家没规定B的数量就得是最多的(我就是犯的这错误)
最后还得感谢ybl大佬为弱弱我指点迷津(叩谢)、
哦,不能忘了代码:
#include<iostream>
using namespace std;
int main()
{
ios::sync_with_stdio(false);
int t,m,n;
cin>>t;
while(t--)
{
cin>>m>>n;
for(int i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
if(i==1) cout<<'B';
else if(j==1)
{
cout<<'B';
}
else cout<<'W';
}
cout<<endl;
}
}
return 0;
}