Pepero Tasting

For $N=2$ there is only one adjacent sum, so the range is $0$. For $N\geq3$, equal adjacent sums would imply $P_1+P_2=P_2+P_3$, hence $P_1=P_3$, a contradiction. Thus the optimum is at least $1$. Let $m=\lfloor(N+1)/2\rfloor$ and output the valid values in the order $m,m+1,m-1,m+2,m-2,\ldots$. Every value occurs once, and adjacent sums are either $2m$ or $2m+1$, attaining the lower bound. Time is $O(N)$ and auxiliary space is $O(1)$. Subtask 1 can enumerate all permutations in $O(N!N)$ time.

Solution written by GPT6.1