Daily AlpacaHack 2025-12-02 writeup【python - ord, chr, hex】

問題

chall.py

import os


flag = os.environ.get("FLAG", "not_a_flag")

# all prime numbers less than 300
primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293]
assert len(flag) <= len(primes)

ct = 1
for i, c in enumerate(flag):
    ct *= primes[i] ** (ord(c))

print(hex(ct))

output.txt

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解法

コードを見るに、flagをenumerate関数で要素のインデックスとその文字で取り出している。
i番目の素数をその文字のASCIIの値の数だけ累乗させている。
最終的にすべてを掛けて得られた整数値をhexで出力していると。

つまり

• 素因数分解する
• 累乗の数を控えておく
• chrで文字に戻す

この手順で復元すれば良さそう。

solve.py

for i in range(1, len(primes) + 1):
    ct = output
    flag = []
    for j in range(i):
        p = primes[j]
        exp = 0
        while ct % p == 0:
            ct //= p
            exp += 1
        flag.append(chr(exp))
    if ct == 1:
        print("".join(flag))
        break

解けた。

Python組み込み関数 - ord, hex, chr

ord(c) - 文字から数値へ

ord は "ordinal"（序数）の略。
文字がコンピュータ内で何番目の番号に割り当てられているかを調べます。

print(ord('A'))    # 65
print(ord('あ'))   # 12354
print(ord('😊'))   # 128522

chr(i) - 数値から文字へ

ord の逆の操作。指定した数値に対応する文字を返します。

print(chr(65))      # 'A'
print(chr(12354))   # 'あ'
print(chr(128522))  # '😊'

hex(x) - 10進数から16進数へ

print(hex(65))      # '0x41'
print(hex(12354))   # '0x3042'

これらは良く使われている気がする。
あとはpycryptodomeのlong_to_bytesとかbytes_to_longですね。