嘉新杯部分wp

于 2026-2-4 由晨星运营组发布

恒安嘉新杯

简单的逆向分析

打开IDA找到main函数，发现CIPHERTEXT，直接双击密文

这个地方一看就是加密信息

通过tab可以直接找到脚本，写一个脚本可以得出flag

EasyAES

通过脚本可以得出答案

from Crypto.Cipher import AES
from hashlib import sha256
import string
import itertools
def valid_padding(s):
    if len(s) == 0:
        return False
    n = s[-1]
    if n < 1 or n > 32:
        return False
    if len(s) < n:
        return False
    return all(s[-i] == n for i in range(1, n + 1))
def unpad(s):
    return s[:-s[-1]]
def decrypt_with_key(enc, key):
    iv = enc[:16]
    cipher = AES.new(key, AES.MODE_CBC, iv)
    dec = cipher.decrypt(enc[16:])
    return dec
def main():
    c3_hex = "62343dfc3e978a1d580b54f345e1ed719c85ab15781acfe8ba3bcef1560c9cf54f187bc204c302a5ed4ebb5b5454151ba9b8b73841e17dc391c30a637ef8cfa14a25d01765231ef93a6faede2d66bad5d124201a2d278522bfd416de294677046d47f2827580cdcb9c0d3b18e4c0c68c8948aaefe4e684c7386b426db7898b5c2090047ff433bb6a75b38beaf81b7ad9404d2f09c642179697e9d3721eefc0eb12ba8c780a8d07672f70b00b9cadef74"
    c3 = bytes.fromhex(c3_hex)
    chars = string.ascii_letters + string.digits
    candidates = []
    print("Brute forcing part2...")
    total = len(chars) ** 3
    count = 0
    for part2_tuple in itertools.product(chars, repeat=3):
        count += 1
        if count % 10000 == 0:
            print(f"Progress: {count}/{total}")
        part2 = ''.join(part2_tuple)
        key1 = sha256(part2.encode()).digest()
        try:
            dec3 = decrypt_with_key(c3, key1)
        except:
            continue
        if not valid_padding(dec3):
            continue
        c2 = unpad(dec3)
        if len(c2) < 16:
            continue
        try:
            dec2 = decrypt_with_key(c2, key1)
        except:
            continue
        if not valid_padding(dec2):
            continue
        c1 = unpad(dec2)
        if len(c1) < 16:
            continue
        try:
            dec1 = decrypt_with_key(c1, key1)
        except:
            continue
        if not valid_padding(dec1):
            continue
        c0 = unpad(dec1)
        if len(c0) < 16:
            continue
        candidates.append((part2, c0))
        print(f"Found candidate part2: {part2}")
    print(f"Number of candidates: {len(candidates)}")
    print("Brute forcing part1 for each candidate...")
    for part2, c0 in candidates:
        for part1_tuple in itertools.product(chars, repeat=3):
            part1 = ''.join(part1_tuple)
            key0 = sha256(part1.encode()).digest()
            try:
                dec0 = decrypt_with_key(c0, key0)
            except:
                continue
            if not valid_padding(dec0):
                continue
            flag_padded = unpad(dec0)
            try:
                flag_str = flag_padded.decode('utf-8')
            except:
                continue
            if len(flag_str) == 36 and flag_str.count('-') == 4:
                if flag_str[8] == '-' and flag_str[13] == '-' and flag_str[18] == '-' and flag_str[23] == '-':
                    print(f"Flag found: {flag_str}")
                    print(f"Part1: {part1}, Part2: {part2}")
                    return
    print("Done.")
if __name__ == '__main__':
main()