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July 12, 2021 05:14
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AtCoder Beginner Contest 171 [ C - One Quadrillion and One Dalmatians ] https://atcoder.jp/contests/abc171/tasks/abc171_c
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''' | |
https://atcoder.jp/contests/abc171/tasks/abc171_c | |
問題の例に「0をどう変換するか?」書いてないところがポイントなのかなー? | |
そこから一歩進んだ考察するべき? | |
[参考] | |
https://blog.hamayanhamayan.com/entry/2020/06/23/193042 | |
基本はまやんさんのコードをPythonにコンバート | |
https://drken1215.hatenablog.com/entry/2020/06/21/225500 | |
27=aa と入力例ではなってる。 | |
26進数では | |
26=10 だからab?0をどう表現するかが入力例からは謎なところも考察ポイントか | |
https://youtu.be/TUdZT1wIbe8?t=585 | |
''' | |
import sys | |
sys.setrecursionlimit(10 ** 6) # 再帰上限の引き上げ | |
input = sys.stdin.readline | |
INF = 2 ** 63 - 1 | |
# a - zまでのテーブルを作成 | |
S_TABLE = [] | |
for i in range(26): | |
S_TABLE.append(chr(ord('a') + i)) | |
N = int(input()) | |
ans = [] | |
while 0 < N: | |
# Nから1引いて26で割った余りが文字となる | |
N -= 1 | |
m = int(N%26) | |
ans.append(S_TABLE[m]) | |
N //= 26 | |
ans.reverse() | |
print("".join(ans)) |
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