TCO 2B

Much easier than 2A, although I end up with -25.

p250, just keep the known ones in a set and do the processing.

you can use max in the known,
or use some prefix of unknown and the rest maxKnown,
or use several copies of all unknown and prefix + maxKnown.

My bug? I wrote
for(int k=1; k<=N && r>0; ++k, r-=k, ans++)
But in r-=k I need the k value before ++.
A hard lesson that cost my p250.

p500, the game is completely determined by move[0]. After deciding the partition of move[0] numbers, the rest is just a dp to pick best subset. The implementation is nontrivial and I need maybe 15min more to complete.