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

생성일: 2026년 4월 12일

답변에 사용된 모델 GPT-5.4 Thinking by Chat01

질문

image.png

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']))

사고

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

답변

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

이 Q&A 공유하기