I am trying to solve [this](https://codeforces.me/contest/1797/problem/B) question. ↵
I am facing difficulty in understanding the diagram of test case 2.↵
As far as I have understood theis question, the question asks to check whether is it possible to create 180 degree rotated image by changing colour of any possible combination of blocks exactly k times. If we look at the second sample test case 2, it's image is↵
↵
![ ](https://i.imgur.com/ndMcvDS.png)↵
↵
So, it's 180 degree rotated image should look like this ↵
↵
![ ](https://i.imgur.com/NCDNL7k.png) ↵
↵
But, in the question the rotated image is give like this ↵
↵
![ ](https://i.imgur.com/c2j4caW.png) ↵
↵
I am unable to understand how is it possible?↵
Please explain the logic behind it.↵
↵
My next question is that suppose we have n=1. Then the answer should be "YES" if and only if k%2==0, because if we change the colour of the block odd times then the final colour of the only block will be changed (making it impossible to form 180 degree rotated image of the original block). But, I have made submission with this logic and the output was wrong answer.↵
Please correct me where am I thinking wrong?↵
↵
I will be thankful for any help!
I am facing difficulty in understanding the diagram of test case 2.↵
As far as I have understood th
↵
![ ](https://i.imgur.com/ndMcvDS.png)↵
↵
So, it's 180 degree rotated image should look like this ↵
↵
![ ](https://i.imgur.com/NCDNL7k.png)
↵
But, in the question the rotated image is give like this ↵
↵
![ ](https://i.imgur.com/c2j4caW.png) ↵
↵
I am unable to understand how is it possible?↵
Please explain the logic behind it.↵
↵
My next question is that suppose we have n=1. Then the answer should be "YES" if and only if k%2==0, because if we change the colour of the block odd times then the final colour of the only block will be changed (making it impossible to form 180 degree rotated image of the original block). But, I have made submission with this logic and the output was wrong answer.↵
Please correct me where am I thinking wrong?↵
↵
I will be thankful for any help!