Traveling Across All Realms: Transcending Transcendence

I still have to go to school, so I don't have much time to write. I'll write again when I come back

Luo Qinghan

Support Meow

P1(p2(p3(p4(p5(p5(p6(p7(p8(p9))))))))[2] We operate on this expression. The operation process is as follows: The number in square brackets is recorded as k, find the rightmost pn, and check it: If n=1: If it is in the outermost layer, that is, there is nothing wrapping it, then remove it directly and add k by one Otherwise, find the nearest pm wrapping it, and get pm ($+p1), transform it into pm($)+pm($)+..., There are k pms in total, and then add one to k If n≠1: Find the nearest pm wrapping pn, get pm($+pn), transform it into pm($+pm($+pm($+...))), K layers. Then add k by one Loop and operate until pn does not exist in the expression, end the operation, and output k [Note: $ can be an empty expression, and pn is equivalent to pn (empty), pm (pn) is equivalent to pm (empty + pn)] Example: p1(p1(p1+p1))[2] =p1(p1(p1)+p1(p1))[3] =p1(p1(p1)+p1+p1+p1)[4] =p1(p1(p1)+p1+p1)+p1(p1(p1)+p1+p1)+p1(p1 (p1)+p1+p1)+p1(p1(p1)+p1+p1)[5] =…… Second example: p1(p2(p2+p2+p1))[ 2] =p1(p2(p2+p2)+p2(p2+p2))[3] =p1(p2(p2+p2)+p2(p2+p1(p2(p2+p2 )+p2(p2+p1(p2(p2+p2)+p2(p2))))))[4] =p1(p2(p2+p2)+p2(p2+p1(p2(p2+p 2)+p2(p2+p1(p2(p2+p2)+p2(p1(p2(p2+p2)+p2(p1(p2(p2+p2)+p2(p1(p2(p2+p2) )+p2)))))))))))[5] =…… Then, p1(p2(p3(p4(p5(p6(p7(p8(p9))))))))[2]

Weimo is a man