@Ronin_VB_King@Spidey_707 Not really u can use the property
That a^b=c then a=b^c
Now as we want A[i]^A[j] =k I>j
The we can rearrange it as
A[j]=A[i]^k
And keep last occ of each element
Then dp[i]=1+dp[lastOcc[A[j]]]
Answer is just max element in dp
@coolcoder56 First is just classic job scheduling problem u just have to do it twice on q and r
And in second one as it's functional graph u just need to calculate minimum value in each cycle
Made it to ICPC Kanpur & Amritapuri regionals, and somehow ended up in the Go for Gold camp. The past 2 days were the most pleasant days of the year…
@adxtya_jha @Priyansh_31Dec@aryanc403@i_pranavmehta We also registered for Amritapuri, but we're the second team from our college, while the first team has registered for Chennai and Amritapuri.