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AtCoder Beginner Contest 167 [ D - Teleporter ] https://atcoder.jp/contests/abc167/tasks/abc167_d
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''' | |
[問題] | |
https://atcoder.jp/contests/abc167/tasks/abc167_d | |
[解説] | |
https://blog.hamayanhamayan.com/entry/2020/05/10/232914 | |
dpでダブリングを用いるのが想定解答のようだが、周期性を用いたコードで何とか解けた。 | |
''' | |
import sys | |
sys.setrecursionlimit(10 ** 6) # 再帰上限の引き上げ | |
input = sys.stdin.readline | |
INF = 2 ** 63 - 1 | |
N, K = map(int, input().split()) | |
A = list(map(int, input().split())) | |
dict = {} # ループになる数字のインデックスを保存 | |
loop_mae = 0 # ループになる数字の前の数 | |
loop = 0 # ループの長さ | |
i = 0 # 次にいく場所のインデックス | |
R = [1] # 町番号の出現順に保存 | |
dict[1] = 0 # 1番目の町はインデックス0 | |
j = 1 # 何個目か。 | |
while True: | |
a = A[i] | |
if (a in dict) == False: | |
dict[a] = j | |
i = A[i] - 1 | |
R.append(a) | |
else: | |
loop = j - dict[a] | |
loop_mae = dict[a] | |
break | |
j += 1 | |
if j >= N: | |
break | |
if loop_mae >= K: | |
# ループは発見されているがKよりループ前のほうが長い | |
print(R[K]) | |
else: | |
# 素直なループ計算 | |
num = (K - loop_mae) % loop | |
print(R[loop_mae + num]) |
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