https://onlinejudge.u-aizu.ac.jp/beta/room.html#ACPC2018Day3/problems/B

I found an interesting solution. Most of the solutions use a linked list, but it is based on another property. The operation LxR->RxL is similar to LR->RL. LR->RL is well-known as rotation (C++ standard library contains it). Then, consider the following problem.

Given a permutation with N elements and Q queries. Each query is in the following form.

- Given k, rotate k elements. That is, change a[0],...,a[k-1],a[k],...,a[n-1] to a[k],...,a[n-1],a[0],...,a[k-1].

Please print the permutation after all queries are performed. This problem is really easy because just trace the beginning position. Actually, the original problem can also be solved by a similar technique.

That is why LxR->RxL can be treated as a rotation. Appending x to the end, it becomes LxRx -> RxLx.

#include <stdio.h> int a[100001]; int pos[100000]; int main() { int n, q; scanf("%d %d", &n, &q); for (int i = 0; i < n; i++) { scanf("%d", &a[i]); a[i]--; pos[a[i]] = i; } int k = n; while (q--) { int b; scanf("%d", &b); b--; a[k] = b; int tmp = pos[b]; pos[b] = k; k = tmp; } for (int i = 0; i < n; i++) { if (i) putchar(' '); printf("%d", a[(i + k + 1) % (n + 1)] + 1); } putchar('\n'); }