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假设AES算法的原始密钥K和明文P分别如下,计算AES的第一轮输出，并进行解密 给出解密文档

Créé le : 12 avril 2026

Répondu en utilisant GPT-5.4 Thinking par Chat01

Question

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假设AES算法的原始密钥K和明文P分别如下,计算AES的第一轮输出，并进行解密
给出解密文档

Python

python
# Let's implement AES round states for AES-128
Sbox = [
0x63,0x7c,0x77,0x7b,0xf2,0x6b,0x6f,0xc5,0x30,0x01,0x67,0x2b,0xfe,0xd7,0xab,0x76,
0xca,0x82,0xc9,0x7d,0xfa,0x59,0x47,0xf0,0xad,0xd4,0xa2,0xaf,0x9c,0xa4,0x72,0xc0,
0xb7,0xfd,0x93,0x26,0x36,0x3f,0xf7,0xcc,0x34,0xa5,0xe5,0xf1,0x71,0xd8,0x31,0x15,
0x04,0xc7,0x23,0xc3,0x18,0x96,0x05,0x9a,0x07,0x12,0x80,0xe2,0xeb,0x27,0xb2,0x75,
0x09,0x83,0x2c,0x1a,0x1b,0x6e,0x5a,0xa0,0x52,0x3b,0xd6,0xb3,0x29,0xe3,0x2f,0x84,
0x53,0xd1,0x00,0xed,0x20,0xfc,0xb1,0x5b,0x6a,0xcb,0xbe,0x39,0x4a,0x4c,0x58,0xcf,
0xd0,0xef,0xaa,0xfb,0x43,0x4d,0x33,0x85,0x45,0xf9,0x02,0x7f,0x50,0x3c,

Python

python
# complete AES encryption/decryption
state=add_round_key(bytes2state(pt),rks[0])
round_states=[]
for rnd in range(1,10):
    state=sub_bytes(state)
    state=shift_rows(state)
    state=mix_columns(state)
    state=add_round_key(state,rks[rnd])
    round_states.append((rnd,state_hex_words(state)))
state=sub_bytes(state); state=shift_rows(state); state=add_round_key(state,rks[10])
ct=state2bytes(state)
ct.hex().upper(), round_states[:2], round_states[-1]


('526C1ACCC320C5226C25617C107D07B3',
 [(1, ['C43311D3', 'B826F4B4', '94B93A17', '4BB7366B']),
  (2, ['37CFE45F', '3B3F2368', 'F24E9836', '1F5DE836'])],
 (9, ['11E9B30E', 'A3C19EC6', 'B4EC4D2F', 'C543A08C']))

Réflexion

Checking page 2

Page 1 looks good, so now I need to check out page 2. I want to make sure everything flows as expected, especially with how the sections and table are placed. Just need to verify how the content looks and whether any further adjustments are necessary for smooth transitions.

Thought for 6m 4s

Réponse

已完成计算并整理成文档。

第 1 轮输出为： {C43311D3, B826F4B4, 94B93A17, 4BB7366B}

逆向解密恢复的明文为： {01234567, 89ABCDEF, FEDCBA98, 76543210}

文档下载： 下载 AES 解密文档

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