2023-Moectf-wp

不包含web； misc；pwn；web misc太基础；pwn不会！！我是菜鸡

一：古典密码 1、ezrot

考点：ROT系列

题解：

（1）题目

@64E7LC@Ecf0:D0;FDE020D:>!=60=6EE6C0DF3DE:EFE:@?04:!96C0tsAJdEA6d;F}%0N

rot系列解密得到flag

oectf{rot47_is_just_a_simPle_letter_substitution_ciPher_EDpy5tpe5juNT_}

2、vigenere

考点：vigenere

题解：

（1）题目

scsfct{wOuSQNfF_IWdkNf_Jy_o_zLchmK_voumSs_zvoQ_loFyof_FRdiKf_4i4x4NLgDn}

开头是moectf；可以推测出密钥为goodjob

moectf{vIgENErE_CIphEr_Is_a_lIttlE_hardEr_thaN_caEsar_CIphEr_4u4u4EXfXz} 3、不是“皇帝的新密码”

考点：vigenere

题解：

（1)题目：

scsfct{wOuSQNfF_IWdkNf_Jy_o_zLchmK_voumSs_zvoQ_loFyof_FRdiKf_4i4x4NLgDn}

（2）还是vigenere

4、可可的新围墙

考点：栅栏密码

题解：

moectf{F3nc3_ciph3r_shiFTs_3ach_l3TT3r_By_a_Giv3n_nuMB3r_oF_plac3s_Ojpj} 5、猫言喵语

考点：

题解：

（1）题目

喵喵？ 喵喵喵喵喵喵喵喵喵喵喵喵 喵喵喵 喵喵喵喵喵喵喵喵？喵喵？喵喵喵喵喵？ 喵喵？喵喵喵喵喵？ 喵喵喵喵喵？ 喵喵喵喵喵？喵喵？ 喵喵喵喵喵？ 喵喵喵喵喵喵 喵喵喵喵喵喵 喵喵喵喵喵喵喵喵？喵喵？喵喵喵喵喵？ 喵喵？喵喵喵喵喵？喵喵喵 喵喵喵喵喵？ 喵喵？ 喵喵喵喵喵喵喵喵？喵喵？喵喵喵喵喵？ 喵喵？喵喵喵喵喵喵喵喵喵 喵喵喵喵喵喵喵喵？ 喵喵？ 喵喵喵喵喵喵喵喵？喵喵？喵喵喵喵喵？ 喵喵？喵喵喵喵喵喵喵喵喵 喵喵喵 喵喵喵喵喵喵喵喵？喵喵？喵喵喵喵喵？ 喵喵？喵喵喵喵喵？喵喵喵 喵喵喵喵喵？ 喵喵喵喵喵？喵喵喵喵喵喵 喵喵喵喵喵？喵喵喵喵喵喵 喵喵喵 喵喵？喵喵喵喵喵喵 喵喵喵喵喵喵喵喵？喵喵？喵喵喵喵喵？ 喵喵？喵喵？喵喵喵 喵喵？喵喵？喵喵？ 喵喵喵喵喵喵喵喵？ 喵喵？喵喵？喵喵喵喵喵喵 喵喵喵喵喵喵 喵喵喵喵喵喵喵喵？喵喵？喵喵喵喵喵？ 喵喵？喵喵喵喵喵喵喵喵喵 喵喵？喵喵喵喵喵？喵喵？ 喵喵喵喵喵喵喵喵？喵喵？喵喵喵喵喵？ 喵喵喵喵喵？喵喵喵 喵喵？喵喵喵喵喵喵喵喵？

（2）Morse code”，是摩斯密码；斯密码只有两种符号，所以得把文件内容根据特征划分成两块。内容中的空格相当于分隔符，根据观察，可以看出”喵喵？”是一组，”喵喵喵”是一组，根据这个规律进行转换，得到如下内容

./----/-/--..-./.-./-./-../-./--/--/--..-./.-.-/-././--..-./.---/--././--..-./.---/-/--..-./.-.-/-./-.--/-.--/-/.--/--..-./..-/.../--./..--/--/--..-./.---/.-../--..-./-.-/.--. moectf{THE_KAWAII_CAT_BUT_BE_CALLED_GOUZI_BY_RX} 二：Reverse 1、“天网”

考点：so文件提取分析

参考链接： blog.csdn.net/ULGANOY/article/details/136780906

2、ANDROID

考点：APK逆向

题解：

（1)签到题目

用户输入一个字符串；然后与key进行异或操作之后与enc进行比较

enc = [25, 7, 0, 14, 27, 3, 16, ord('/'), 24, 2, ord('\t'), ord(':'), 4, 1, ord(':'), ord('*'), 11, 29, 6, 7, ord('\f'), ord('\t'), ord('0'), ord('T'), 24, ord(':'), 28, 21, 27, 28, 16] key = ['t', 'h', 'e', 'm', 'o', 'e', 'k', 'e', 'y'] def decrypt(enc, key): result = [] key_length = len(key) for i in range(len(enc)): # 对每个字符进行 XOR 操作 decrypted_char = enc[i] ^ ord(key[i % key_length]) result.append(chr(decrypted_char)) return ''.join(result) decrypted_text = decrypt(enc, key) print("解密后的字符串:", decrypted_text) 3、Base64

考点：bse64魔改 pyc在线反编译

题解：

(1)pyc文件；首先在线反编译拿到源码

import base64 from string import * str1 = 'yD9oB3Inv3YAB19YynIuJnUaAGB0um0=' string1 = 'ZYXWVUTSRQPONMLKJIHGFEDCBAzyxwvutsrqponmlkjihgfedcba0123456789+/' string2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/' flag = input('welcome to moectf\ninput your flag and I wiil check it:') enc_flag = base64.b64encode(flag.encode()).decode() enc_flag = enc_flag.translate(str.maketrans(string2, string1)) if enc_flag == str1: print('good job!!!!') else: print('something wrong???') exit(0)

（2）简单的一个换表base64；用户输入字符串之后base64加密与目标值进行对比；解密str1就行

import base64 str1 = 'yD9oB3Inv3YAB19YynIuJnUaAGB0um0=' string1 = 'ZYXWVUTSRQPONMLKJIHGFEDCBAzyxwvutsrqponmlkjihgfedcba0123456789+/' string2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/' translation_table = str.maketrans(string1, string2) replaced_str = str1.translate(translation_table) try: flag = base64.b64decode(replaced_str).decode() print("解密后的 flag:", flag) except Exception as e: print("解密失败:", e) 4、EQUATION

考点:Z3求解

5、GUI

考点：签到题目：只是增加了一个反调试而已

data= [0x0F, 0x3E, 0x30, 0x27, 0x13,0x01, 0x7D, 0x70, 0x70, 0x03,0x7D, 0x38, 0x0E, 0x7A, 0x23,0x7C, 0x0B, 0x1A, 0x3C, 0x7D,0x39, 0x7F, 0x3C, 0x4D, 0x4D,0x4D, 0x29] for i in data: print(chr((i ^ 0x51) + 5), end='') 6、RRRRRc4

考点：RC4

题解：

（1）题目

sub_7FF710DF3581("welcome to moectf!!!"); sub_7FF710DF3581("This is a very common algorithm "); sub_7FF710DF3581("show your flag:"); sub_7FF710DF27F8("%s", input); // 输入 if ( sumlen(input) == 37 ) // 验证输入长度 { RC4((int)v5, (int)v6, (int)input, 38, (__int64)v7, 10);// RC4加密 for ( j = 0; (unsigned __int64)j < 0x26; ++j ) { if ( enc[j] == (unsigned __int8)input[j] )// RC4加密之后与enc内容比较 ++v8; } } if ( v8 == 37 ) sub_7FF710DF3973("right!flag is your input!"); else sub_7FF710DF3973("try again~"); sub_7FF710DF4BCF(v3, &unk_7FF710EE2100); return 0i64;

用户就是输入字符串之后然后进行RC4加密；然后在与enc进行比较；现在就需要找到密钥就行，密文已经知道；简单调试以下就知道密钥是v7

然后直接RC4解密

def rc4_decrypt(ciphertext, key): S = list(range(256)) # 初始化 S-box j = 0 key_length = len(key) for i in range(256): j = (j + S[i] + key[i % key_length]) % 256 S[i], S[j] = S[j], S[i] # 交换 S[i] 和 S[j] i = j = 0 plaintext = [] for byte in ciphertext: i = (i + 1) % 256 j = (j + S[i]) % 256 S[i], S[j] = S[j], S[i] # 交换 S[i] 和 S[j] k = S[(S[i] + S[j]) % 256] # 生成伪随机字节 plaintext.append(byte ^ k) # 异或解密 return bytes(plaintext) enc = [ 27, 155, 251, 25, 6, 106, 181, 59, 124, 186, 3, 243, 145, 184, 182, 61, 138, 193, 72, 46, 80, 17, 231, 199, 79, 177, 39, 207, 243, 174, 3, 9, 178, 8, 251, 220, 34 ] key = b"moectf2023" plaintext = rc4_decrypt(enc, key) print("解密结果:", plaintext.decode('utf-8', errors='ignore')) 7、RUST

考点：XOR

题解：

（1）首先搜索定位main函数 （2）分析main函数；发现了数据和异或操作；输入的值和0x88进行异或操作

v25 = [ -27, -25, -19, -21, -4, -18, -13, -38, -3, -5, -4, -41, -6, -19, -2, -41, -1, -31, -28, -28, -41, -22, -19, -41, -23, -1, -18, -3, -71, -11 ] v25_unsigned = [x & 0xFF for x in v25] xor_key = 0x88 decrypted = [] for byte in v25_unsigned: decrypted.append(byte ^ xor_key) flag = ''.join([chr(b) for b in decrypted]) print("解密结果:", flag) 8、Reverse入门指北

考点：签到；flag直接放你脸上了

moectf{F1rst_St3p_1s_D0ne}

9、SMC

考点：动态调试 SMC

（1）main函数

sub_401087(aPlzInputYourFl, v4); sub_401023(aS, (char)v6); sub_4011E0();//SMC函数 if ( sub_401050(v6) ) sub_401087(aGood, v5); else sub_401087(aTryAgainPlease, v5); return 0; }

发现SMC函数对sub_401050函数进行了加密；无法查看sub_401050函数；需要先调试对sub_401050函数解密；然后再看解密逻辑；发现就一个异或的逻辑；解密就行

10、UPX!

考点：upx脱壳

题解：

（1)首先脱壳处理；定位main函数 很简单；输入内容；然后与0x67异或；最后和enc比较；提取enc内容逆向

enc = [ 10, 8, 2, 4, 19, 1, 28, 87, 15, 56, 30, 87, 18, 56, 44, 9, 87, 16, 56, 47, 87, 16, 56, 19, 8, 56, 53, 2, 17, 84, 21, 20, 2, 56, 50, 55, 63, 70, 70, 70, 26 ] xor_key = 0x67 decrypted = [] for byte in enc: decrypted.append(byte ^ xor_key) flag = ''.join([chr(b) for b in decrypted]) print("解密结果:", flag) 11、Xor

考点：XOR

enc = [ 84, 86, 92, 90, 77, 95, 66, 96, 86, 76, 102, 82, 87, 9, 78, 102, 81, 9, 78, 102, 77, 9, 102, 97, 9, 107, 24, 68 ] xor_key = 0x39 decrypted = [] for byte in enc: decrypted.append(byte ^ xor_key) flag = ''.join([chr(b) for b in decrypted]) print("解密结果:", flag) 12、ezandroid

考点：APK逆向 迷宫问题 so文件

题解：

（1）定位代码：提示Try to reverse the native lib!

（2）将apk修改为zip找so文件；将so文件进行逆向

while ( 2 ) { v3 = 0; if ( *a1 ) v3 = *v4 != 42; if ( v3 ) { v1 = a1++; switch ( *v1 ) { case 'a': --v4; continue; case 'd': ++v4; continue; case 's': v4 += 15; continue; case 'w': v4 -= 15; continue; default: v6 = 0; break; } } else { v6 = *v4 == 35; } break; } return v6; }

找到了迷宫和迷宫函数；那么正确的路径就是flag (15个一行）ssaassssdddddwwddddssss

13、junk_code

考点：花指令

题解：

（1)sub_45A9A0((int)Str, 18)和sub_459EBF((int)v7, 18)加密函数被加了花指令；无法反汇编；首先nop掉花指令 （2）nop花指令 （3）分析sub_459EBF函数发现是简单的XOR操作

for ( i = 0; i < MaxCount; ++i ) Str2[i] ^= 0x66u;

还原得到一部分flag _th3_junk_c0d3!!!}

encrypted_data = [ 57, 18, 14, 85, 57, 12, 19, 8, 13, 57, 5, 86, 2, 85, 71, 71, 71, 27 ] xor_key = 0x66 decrypted_data = [] for byte in encrypted_data: decrypted_data.append(byte ^ xor_key) flag = ''.join([chr(b) for b in decrypted_data]) print("解密结果:", flag)

分析sub_45A9A0函数逆向还原得到第二部分flag

# 密文数组 encrypted_data = [ 0x68, 0x6A, 0x60, 0x5E, 0x6F, 0x61, 0x76, 0x74, 0x2B, 0x70, 0x5A, 0x6D, 0x60, 0x68, 0x2B, 0x71, 0x2E, 0x5F ] decrypted_data = [] for byte in encrypted_data: decrypted_data.append(byte + 5) flag = ''.join([chr(b) for b in decrypted_data]) print("解密结果:", flag) 14、unwind

考点：tea加密 反调试

题解：

（1）根据题目提示是tea加密和SEH；定位main函数 发现有混淆的操作和反调试；我先将MEMORY[0] = 0;的汇编代码nop掉就可以去除混淆操作了

nop掉这一段 然后nop掉mov；去除反调试函数 可以看出是TEA加密；提取一下密文

#include <stdint.h> #include<stdio.h> #include<iostream> using namespace std; void tea_decrypt(uint32_t* v, uint32_t* k) { uint32_t sum = 0xC6EF3720; uint32_t delta = 0x9e3779b9; uint32_t v0 = v[0], v1 = v[1]; uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3]; for (int i = 0; i < 32; i++) { v1 -= ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3); v0 -= ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1); sum -= delta; } v[0] = v0; v[1] = v1; } void fun(uint32_t *plaintext, uint32_t *key,int ok ) { tea_decrypt(plaintext, key); for (int i = 0 ; i < 2 ; i ++ ) { int tmp = plaintext[i]; while (tmp) { if(ok) printf("%c", tmp % 0x100); tmp /= 0x100; } } } int main() { //0x55, 0x1A, 0x27, 0x6D, 0xEA, 0x8E, 0x4C, 0x5A unsigned int byte_B8A000[16] = { 3832275802u, 1325565702u, 3251494723u, 3681261687u, 2571319439u, 3038942179u, 481361187u, 3471762405u, 732765487u, 4187493796u, 4196789208u, 1924604267u, 775095225u, 2079018201u, 4182288373u, 4231568784u, }; uint32_t key[18] = {959666244, 13872, 0, 0, 1952673636u, 3371631u, 0u, 0u, 827872582u, 1431916353u, 78u, 0u, 1702245202u, 1919248754u, 0u, 0u, }; int cd = 0; for (int i = 0 ; i < 8 ; i += 2) { fun(byte_B8A000 + i, key + cd , 1); cd += 4; } cd = 0; for (int i = 8 ; i < 16 ; i += 2) { fun(byte_B8A000 + i, key + cd , 0); cd += 4; } cd= 0; for (int i = 8 ; i < 16 ; i += 2 , 1) { fun(byte_B8A000 + i, key + cd,1); cd += 4; } }

moectf{WoOo00Oow_S0_interesting_y0U_C4n_C41l_M3tW1c3_BY_Unw1Nd~}

三：Crypto 1、ABC

考点：二进制转二维码

题解：

（1)题目给出了a.npy b.npy c.npy Code in My Matrix A∗B∗CA∗B∗C；题目要求计算矩阵的乘积 A×B×CA×B×C，其中 AA、BB、CC 是从 .npy 文件中加载的矩阵

import numpy as np # 从 .npy 文件中加载矩阵 A = np.load('a.npy') B = np.load('b.npy') C = np.load('c.npy') # 计算矩阵乘积 A * B * C result = np.dot(np.dot(A, B), C) # 输出结果 print("Result of A * B * C:") print(result)

得到了一个矩阵只包含 -1 1；猜测是二进制转二维码

import numpy as np # 给定的二进制矩阵 binary_matrix = np.array([ [0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0], [0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0], [0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0], [0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0], [0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0], [0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0], [1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1], [1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0], [1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1], [0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0], [1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1], [0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0], [0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1], [1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0], [0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1], [0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0], [0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1], [1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0], [0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1], [1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1], [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0], [0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0], [0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1], [0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0], [0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1], [0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1], [0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1] ]) # 将二进制矩阵转换为二进制字符串 binary_string = ''.join(str(bit) for row in binary_matrix for bit in row) print("Binary String:", binary_string)

然后二进制转二维码（在线）

扫码得到flag

moectf{U_C4n_D0_uR_AipH4_B_See_2023}

2、Crypto 入门指北

考点：rsa

题解：

（1）题目给了一个python脚本；运行就是flag

moectf{weLCome_To_moeCTf_CRypTo_And_enjoy_THis_gAme!_THis_is_yoUR_fLAg!} 3、baby_e

考点：RSA—小e攻击（低解密指数攻击）

题解：

（1）属于比较弱智的题目：直接上脚本

from gmpy2 import iroot import libnum e= n= k = 0 while 1: res=iroot(c+k*n,7) if(res[1]==True): print(libnum.n2s(int(res[0]))) break k=k+1 3、bad_E

考点：RSA-e phi有限域不互素

from gmpy2 import * from Crypto.Util.number import * e = 65537 p= q= c= n = p * q phi_n = (p - 1) * (q - 1) print(gcd(e, p - 1)) d = invert(e, (q - 1)) m = pow(c, d, q) print(long_to_bytes(m)) 4、bad_random

考点：MD5爆破

from itertools import product import string import hashlib target_hash = "83990600a2ddb3a57dd150050df128f3" suffix = "ZhZe" # 生成所有4位字母+数字组合（共36^4=1,679,616种） charset = string.digits + string.ascii_letters # 0-9 + a-z + A-Z for chars in product(charset, repeat=4): candidate = ''.join(chars) + suffix if hashlib.md5(candidate.encode()).hexdigest() == target_hash: print(f"Found: {''.join(chars)}") exit() 5、crypto指北

考点：RSA

from Crypto.Util.number import * # 一个非常好用的crypto库 p = 0xe82a76eeb5ac63e054128e040171630b993feb33e0d3d38fbb7c0b54df3a2fb9b5589d1205e0e4240b8fcb4363acaa4c3c44dd6e186225ebf3ce881c7070afa7 q = 0xae5c2e450dbce36c8d6d1a5c989598fc01438f009f9b4c29352d43fd998d10984d402637d7657d772fb9f5e4f4feee63b267b401b67704979d519ad7f0a044eb c = 0x4016bf1fe655c863dd6c08cbe70e3bb4e6d4feefacaaebf1cfa2a8d94051d21e51919ea754c1aa7bd1674c5330020a99e2401cb1f232331a2da61cb4329446a17e3b9d6b59e831211b231454e81cc8352986e05d44ae9fcd30d68d0ce288c65e0d22ce0e6e83122621d2b96543cec4828f590af9486aa57727c5fcd8e74bd296 e = 65537 n = p*q phi = (p-1) * (q-1) # 你知道什么是 欧拉函数吗 [1] d = pow(e, -1, phi) # 什么是乘法逆元？ [2] m = pow(c,d,n) print(long_to_bytes(m)) 6、ez_chain

考点：AES加密

题解：不会！

7、factor_signin

考点:RSA分解n

8、factorize_me!

题解：RSA

(1)题目

from Crypto.Util.number import getPrime from math import prod from sympy import nextprime from random import choices with open('flag.txt', 'rb') as fs: flag = fs.read().strip() primes = [getPrime(512) for _ in range(9)]#生成9个512素数 print(f"{prod(primes) = }")#输出素数的乘积 print(f"{prod(p - 1 for p in primes) = }")#每个素数减1的成绩 primes2 = [nextprime(p) for p in choices(primes, k=3)]#选取三个素数 n = prod(primes2)#三个素数的成绩 e = 65537 c = pow(int.from_bytes(flag, 'big'), e, n) print(f'n = {n}') print(f'e = {e}') print(f'c = {c}')

（2）解密：很简单的一个逻辑

from Crypto.Util.number import * e = 65537 c = 841335863342518623856757469220437045493934999201203757845757404101093751603513457430254875658199946020695655428637035628085973393246970440054477600379027466651143466332405520374224855994531411584946074861018245519106776529260649700756908093025092104292223745612991818151040610497258923925952531383407297026038305824754456660932812929344928080812670596607694776017112795053283695891798940700646874515366341575417161087304105309794441077774052357656529143940010140 p = 6991223361118904775931217829045348785013077549030883418924453538830605687999480005714979700653172534877541317997174968789510984315425270755055110913347349 q = 9987009117206906203158749743824168660291275882852229158070368815160479543708376165641735042845357978292384303332559592302507789120810447986634662721490849 r = 12876877424944854147075816504195994138450356002779004886384584287813869165469217718717854027672044903401715370348223932937626725119320180795716270261309141 n = p * q * r phi = (p-1)*(q-1)*(r-1) d = pow(e,-1,phi) m = pow(c,d,n) print(long_to_bytes(m)) #moectf{you_KNow_how_to_faCtorize_N_right?_9?WPIBung6?WPIBung6?WPIBund6?} 9、feistel

考点：DES Feistel算法

题解：（1）有点类似于==梅森旋转算法==

题目：

from Crypto.Util.number import * round = 2 flag = open("./secret", "rb").read().strip() def f(m, key): m = m ^ (m >> 4) m = m ^ (m << 5) m = m ^ (m >> 8) m ^= key m = (m * 1145 + 14) % 2**64 m = (m * 1919 + 810) % 2**64 m = (m * key) % 2**64 return m def enc(m, key, round): key = bytes_to_long(key) left = bytes_to_long(m[:8]) right = bytes_to_long(m[8:]) for i in range(round): left, right = right, f(right, key) ^ left left, right = right, left return long_to_bytes(left).rjust(8, b"\x00") + long_to_bytes(right).rjust(8, b"\x00") def padding(m): mlen = len(m) pad = 16 - mlen % 16 return m + pad * bytes([pad]) def ecb_enc(m, key): m = padding(m) mlen = len(m) c = b"" for i in range(mlen // 16): c += enc(m[i * 16 : i * 16 + 16], key, round) return c print(ecb_enc(flag, b"wulidego")) # b'\x0b\xa7\xc6J\xf6\x80T\xc6\xfbq\xaa\xd8\xcc\x95\xad[\x1e\'W5\xce\x92Y\xd3\xa0\x1fL\xe8\xe1"^\xad'

feistel加密结构，加解密过程一样，就是轮密钥相反；而这里每轮密钥都是wulidego，所以说这个加解密是完全一样的

moectf{M@g1cA1_Encr1tion!!!} 10、feistel_promax

考点：DES Feistel算法

题解：

（1）相比于上一题，这题key不知，仅根据此条件根本无从下手，不过观察仔细的话会发现最后一行，加密的主体是padding(flag)，而加密过程中又进行了一次padding，所以说进行了两次padding

from Crypto.Util.number import * round = 2 def f(m, key): m = m ^ (m >> 4) m = m ^ (m << 5) m = m ^ (m >> 8) m ^= key m = (m * 1145 + 14) % 2**64 m = (m * 1919 + 810) % 2**64 m = (m * key) % 2**64 return m def dec(m, key, round): key = bytes_to_long(key) left = bytes_to_long(m[:8]) right = bytes_to_long(m[8:]) for i in range(round): left, right = right, f(right, key) ^ left left, right = right, left return long_to_bytes(left).rjust(8, b"\x00") + long_to_bytes(right).rjust(8, b"\x00") def ecb_dec(c, key): clen = len(c) m = b"" for i in range(clen // 16): m += dec(c[i * 16 : i * 16 + 16], key, 2) return m m = b'\x10\x10\x10\x10\x10\x10\x10\x10' ct = b'B\xf5\xd8gy\x0f\xaf\xc7\xdf\xabn9\xbb\xd0\xe3\x1e0\x9eR\xa9\x1c\xb7\xad\xe5H\x8cC\x07\xd5w9Ms\x03\x06\xec\xb4\x8d\x80\xcb}\xa9\x8a\xcc\xd1W\x82[\xd3\xdc\xb4\x83P\xda5\xac\x9e\xb0)\x98R\x1c\xb3h' c = bytes_to_long(ct[-8:]) ^ bytes_to_long(m) bin_c = bin(c)[2:].rjust(64,'0') # = f(M[8:],key)，其中M = m * 2 k1 = [b''] for i in range(1,5): k2 = [] for KEY in k1: for j in range(2**16): key1 = long_to_bytes(j) + KEY cc = f(bytes_to_long(m),bytes_to_long(key1)) # = f(M[8:],key) bin_cc = bin(cc)[2:].rjust(64,'0') if bin_cc[-16*i:] == bin_c[-16*i:]: k2.append(key1) k1 = k2 # print(k1) # k1= [b'4t*zFD\xac\xb4', b'\xb4t*zFD\xac\xb4', b'\\q\x0f\x00w\xeb\xc1"', b'\xdcq\x0f\x00w\xeb\xc1"'] for KEY in k1: flag = ecb_dec(ct,KEY) print(flag) # moectf{F_func_1s_n1t_Ve5y_$EcU%e} 11、giant_e

考点：rsa-维纳攻击

题解：

（1）题目给出的e很大；猜测可能是维纳攻击

12、minipack

考点：

13、n&n

考点：RSA—共模攻击

import gmpy2 from Crypto.Util.number import getPrime,long_to_bytes e1 = e2 = n = c1 = c2 = _,s1, s2 = gmpy2.gcdext(e1, e2) m = pow(c1, s1, n) * pow(c2, s2, n) % n print(long_to_bytes(m)) 14、rsa_signin

考点：RSA—广播攻击

题解：

（1）题目给出了多组nc ；可能是广播攻击或者中国剩余定理；最后发现是广播攻击

from Crypto.Util.number import * from gmpy2 import * e = 65537 n1 = 17524722204224696445172535263975543817720644608816706978363749891469511686943372362091928951563219068859089058278944528021615923888948698587206920445508493551162845371086030869059282352535451058203615402089133135136481314666971507135484450966505425514285114192275051972496161810571035753943880190780759479521486741046704043699838021850105638224212696697865987677760179564370167062037563913329993433080123575434871852732981112883423565015771421868680113407260917902892944119552200927337996135278491046562185003012971570532979090484837684759828977460570826320870379601193678304983534424368152743368343335213808684523217 c1 = 6870605439714128574950893771863182370595667973241984289208050776870220326525943524507319708560433091378319367164606150977103661770065561661544375425887970907060665421562712515902428061727268441585629591525591001533188276465911918724808701356962871139957343861919730086334623932624184172272488406793955068827527130338853980609365042071290967556159598511667974987218999253443575482949258292953639729393456515185185102248985930422080581185292420347510600574229080211050520146551505605537486989306457793451086767402197128573781597156939709237045132856159368959981648969874765462190363842275826077556314448408825308218451 n2 = 24974121071274650888046048586598797033399902532613815354986756278905133499432183463847175542164798764762683121930786715931063152122056911933710481566265603626437742951648885379847799327315791800670175616973945640322985175516271373004547752061826574576722667907302681961850865961386200909397231865804894418194711076667760169256682834206788730947602211228930301853348503098156592000286467190760378847541148772869356389938999094673945092387627113807899212568399028514283219850734634544982646070106811651490010946670117927664594365986238107951837041859682547029079035013475238052160645871718246031144694712586073789250183 c2 = 10324627733161143472233272675096997859064721978612320424254305978486200326061730105384511258706433940176741256952824288120499229240005823611541292676234913505775165761543820764046537413943393325463602612485849366939102550336256797820440347815027443410399157963547486098366749815425187247171697678576246606105486928212486117878157055321965270364583625270716186820068538749425299073309429589410882809098930213978117176627031795312102177342499674234163614021182116065492884880492891668658240362567156235958605768725892407536211503981819707919444725863397622629226309480836486427388484176463279384813974310500625102568341 n3 = 14215826065753265334521416948225868542990756976323308408298887797364519400310818641526401662106853573185085731682502059761982246604277475488691297554851873224516934619888327644352138127883043558424300092247604877819821625587944308487310522092440517150600171819145803937177931473336108429889165189521078678397694303305705260759351843006130968234071638035667854938070597400634242396852782331461576526836227336952718230741560369621645218729592233657856104560425642219241082727756696967324334634822771842625681505869025740662258929200756109704988223034840699133778958569054445520305361142302393767439478256174414187983763 c3 = 415916446053083522663299405080903121619846594209033663622616979372099135281363175464579440520262612010099820951944229484417996994283898028928384268216113118778734726335389504987546718739928112684600918108591759061734340607527889972020273454098314620790710425294297542021830654957828983606433731988998097351888879368160881316237557097381718444193741788664735559392675419489952796677690968481917700683813252460912749931286739585465657312416977086336732056497161860235343155953578618273940135486362350057858779130960380833359506761436212727289297656191243565734621757889931250689354508999144817518599291078968866323093 n4 = 12221355905532691305226996552124162033756814028292708728711809229588190407700199452617060657420166395065565154239801465361510672853972152857415394695376825120759202857555325904640144375262531345320714166285999668052224661520834318497234299585219832943519644095197479639328120838919035625832361810964127485907587199925564724081163804724975965691571850962714258888527902920462746795712011579424322515292865504642938090200503979483095345893697972170153990274670257331483858538617460680462369680572833191232126527727222302641204529110948993583190295067970240051042000918629138767209918572311469915774910003970381965123241 c4 = 2248834602646305164283014556051672824689884721514190813323189875541899566338153534858709617544459297836048770439230174669883719627734394673012731609952869246171300132019334542245094425654362711870373095782083791160029789553806741967408922001051006100049326921742208757147339981269528740944842177729701945606827918253016001436218891580980192743564642120923356793292885805519110411357830040053435569937296612987581482128241218218550319154933831743819546558930918761162723110000328532730751591375727881221199739397698390594797621758011191224528339478784930214820615602510460640307707682865125229937141010351138099874025 n5 = 18152103454920389919231636321286527841833809319334215885641536161086810144890443857211776387914779781628740172079478910188540146498426564211851629962338413488555121865779016981727229209606498886170396500155102635962395243364899026418106378234307821492609778555173516000309435730752571818439328803899462791834490025768785383592935046996428331508608555503567191807692523852530836008436655164751054189301721070209363416058642811329040202582026786024825518381761299547703962502636888833428457116986351812252188468878701301184044948733274488264320930936362549028124581962244201377136969591119942276742760215403738913067567 c5 = 2797812094994121597295362327809389195134238119144547570610194659000554967367804835006774413888965325870488368112707535584687083342412367127561646136089638402907513075405746055834487062923240856950047936297155455745928810738711368950139327254040579266046642851362228893522740216519732851152162928545416236075387903789535000820423985522550638100049857678600662008021574841083416323980817348573062083159710189689337626277009675683473560325178417766400002763719953723259300977655801234386662217462862844994462505601804422871991694828697337752697234180117437785537788728412520613916334045368736691714704501962513954509705 n6 = 22877887459293720334652698748191453972019668578065068224653972884599636421200068659750242304040301306798039254241668648594556654589309801728248683586229288074709849246660525799452637187132633064172425677552176203292787732404537215347782229753837476655088638984496409603054524994383358547132112778403912563916886533181616856401929346567686400616307916690806467019665390260267596320840786982457521423178851498130935577260638269429250197050326097193841333205073650802709022947551398142692735680419453533128176592587955634333425401930362881423044363132586170013458300714163531162544301477356808388416864173949089028317961 c6 = 12271947322974809255127222556723394446467844330408506340843897575503534175121932185624776713618037572593449207329510171212097269297133492090526270770286000839978630002819714376964416081198925899119135271459404333829811516667576167576916805217016117373027245648473458331936273975110163065432285322832123169216976420362833557809289561705091817949915218278430834098156335989014645979633658818904753942786129126233956314517292746008579152368541316795082120147520597254020266752859205131887527661767589367756335766220841483940854397440079467053684289006956034944336788288196391829411432383541473132962783883758561108297747 n7 = 19844333358004073542783728196775487079202832688982038135532362073659058674903791697765527614270399097276261983744620537925712167578187109058145015032736796457938148615396547198728652435169126585595701228287449135664667959433491335769206692390262797325133960778920452511673878233190120432257482339068405290918739453464061987163074129048150451046315248186376609350095502130018696275764450248681787926130463463923862832714969425813770847493135627599129546112143050369344208092649256659330284904392961574494907186727388685504929586018639846040474616307662546605623294842316524163106100888851228858194942825157286544846177 c7 = 9531264751315473345056673937611382755236533664089452852716992791452558274873158812669513178040971923528201631609089069182049526587423864397527252061341857426422965190913745048414029690931254119437249218321954899956104589066479231204536856131403590472063496956452030342299863907499976917750846369802185896519725837163530049157920978007252920334447236842959033879772444475877613295594785710745889554296655932909212643500877218304116451889820444820534937901427158918411546484157737612926382420354101675658160847653151539420222526999426483473829341628599881460824765758346670633385844187252696874025582747177333702736465 n8 = 16956880944655068255446705024149899655327230949463546092744762226005904114738078692036960935391303255804754787864713189658290361949509917704853428701870609882427423574672772606814823959758208695540116440342488334213300943604780971422918744381486937517952553797134323570131582724393100092308466968491068503301604506186521656059375518680612292667310641047190088814753025794048591445267711939066523165042651430468971452726568222388482323097260496415484997546126185688914792795834046855221759289007609518312601640548469651358391745947588643697900883634533872314566389446271647587564348026861264979727062157272541149018781 c8 = 16110326928338602237561005337578085623028116490564329920738844771341250444164294693848130674347672763073995755532723894042946521372321947507527854966013459795492930736187058535665041545095683801386814190612817128504426590828954205050425979880047802547011117626354405687170961272200066258220699329112978151044633994329352673342582175349200008181837211288847301836681860817044391028992501763375849046751094019224570802498414368189170656992427042010362385494565216988561215657424755648213390551881450141899860811844684546992754530755092358644968088017107313907435586729574798046187046145596726569637758312033849476689378 n9 = 16472195897077185060734002588086375750797253422014472876266294484788862733424113898147596402056889527985731623940969291811284437034420929030659419753779530635563455664549165618528767491631867637613948406196511848103083967995689432928779805192695209899686072900265108597626632371718430059561807147486376536203800038054012500244392964187780217667805308512187849789773573138494622201856638931435423778275004491853486855300574479177472267767506041000072575623287557610576406578525902565241580838652860552046216587141709709405062150243990097835181557208274750462554811004137033087430556692966525170882625891516050207318491 c9 = 11867731823522211833301190385669833752050387304375114576570892885641949969365352586215693183003550684262313893105989683214739695968039039944442567581277252581988489020834299896625977474857889570528169919064941042132119301236852358823696947330423679033138054012027878783478922023431469564210485180679933264749281963405243082505688901662659030897104957499953192201440290084373968716271056483463909282407034181891901928790601973222643210525000717355062752079302291729448234374709852429885984987094307177760741403086538949190424454337896501402430653783597070178968921411867485584517214777073301007918941216316241784521708 n10 = 13890749889361612188368868998653029697326614782260719535555306236512452110708495623964530174188871342332417484996749651846510646453983388637377706674890018646246874688969342600780781646175634455109757266442675502522791531161284420286435654971819525519296719668701529481662071464145515727217108362496784024871976015116522898184301395037566514980846499856316532479656908169681719288258287756566886281183699239684997698487409138330229321935477734921670373632304542254938831218652340699024011371979519574576890581492623709896310465567043899767342676912434857372520308852745792360420376574037705943820090308501053778144141 c10 = 6250115196713939477947942995075509357173312813431601073354390451609559579925704891503987992181988654989477525811826607070378476102616752398280691012244301950194800995432882828020405062344160270290542566163969692748126314259624623341922057435728127596172871894887055305291345372720594481096374310285437492746765510292863238933163142677773310305789984897974266961231555124787205980411992251387207335655129551950825339766848166539671565212408741432649813058363660321480995187545006718837863674527475323414266732366507905974800565463011676462244368010182725161416783875646259625352308599198614681446394427674340328493047 n11 = 21457499145521259498911107987303777576783467581104197687610588208126845121702391694574491025398113729462454256070437978257494064504146718372095872819969887408622112906108590961892923178192792218161103488204912792358327748493857104191029765218471874759376809136402361582721860433355338373725980783308091544879562698835405262108188595630215081260699112737457564998798692048522706388318528370551365364702529068656665853097899157141017378975007689790000067275142731212069030175682911154288533716549782283859340452266837760560153014200605378914071410125895494331253564598702942990036163269043699029806343766286247742865671 c11 = 6269656777204332618433779865483197625538144405832409880710764183039800286008967127279281167109250083159801218370191973055663058165456565194979210256278526713608759141588082614531352489547674696723140599892318118960648862531538435596775798128845789504910467783731144808685373807716609662688064728614003904579841055786083326311313295311152563668422289435606771091246147867715987583149743032723028324394173498623642539175178996531881058274717907066845565199058931743481410454382746158558886667761300257488769795092777021292335562818583719708133179974425584610403335487082478848975656282384575767178925517257692365828720 for i in range(1, 12): for j in range(i + 1, 12): ni = eval("n" + str(i)) nj = eval("n" + str(j)) p = gcd(ni, nj) if p > 1: c = eval("c" + str(i)) q = ni // p d = invert(e, (p - 1) * (q - 1)) flag = long_to_bytes(pow(c, d, ni)) print(flag) 14、|p-q|

考点：rsa p q很接近

题解：

（1）核心代码

p = getPrime(2048) q = next_prime(p)

发现p 和q 很接近；直接对n进行开方得到p；然后计算下一个素数可以算出p

from Crypto.Util.number import * from gmpy2 import * flag = 307746143297103281117512771170735061509547958991947416701685589829711285274762039205145422734327595082350457374530975854337055433998982493020603245187129916580627539476324521854057990929173492940833073106540441902619425074887573232779899379436737429823569006431370954961865581168635086246592539153824456681688944066925973182272443586463636373955966146029489121226571408532284480270826510961605206483011204059402338926815599691009406841471142048842308786000059979977645988396524814553253493672729395573658564825709547262230219183672493306100392069182994445509803952976016630731417479238769736432223194249245020320183199001774879893442186017555682902409661647546547835345461056900610391514595370600575845979413984555709077635397717741521573798309855584473259503981955303774208127361309229536010653615696850725905168242705387575720694946072789441481191449772933265705810128547553027708513478130258801233619669699177901566688737559102165508239876805822898509541232565766265491283807922473440397456701500524925191214292669986798631732639221198138026031561329502985577205314190565609214349344303324429408234237832110076900414483795318189628198913032900272406887003325858236057373096880675754802725017537119549989304878960436575670784578550 n = 329960318345010350458589325571454799968957932130539403944044204698872359769449414256378111233592533561892402020955736786563103586897940757198920737583107357264433730515123570697570757034221232010688796344257587359198400915567115397034901247038275403825404094129637119512164953012131445747740645183682571690806238508035172474685818036517880994658466362305677430221344381425792427288500814551334928982040579744048907401043058567486871621293983772331951723963911377839286050368715384227640638031857101612517441295926821712605955984000617738833973829140899288164786111118033301974794123637285172303688427806450817155786233788027512244397952849209700013205803489334055814513866650854230478124920442832221946442593769555237909177172933634236392800414176981780444770542047378630756636857018730168151824307814244094763132088236333995807013617801783919113541391133267230410179444855465611792191833319172887852945902960736744468250550722314565805440432977225703650102517531531476188269635151281661081058374242768608270563131619806585194608795817118466680430500830137335634289617464844004904410907221482919453859885955054140320857757297655475489972268282336250384384926216818756762307686391740965586168590784252524275489515352125321398406426217 q = next_prime(isqrt(n)) p = n // q assert p*q == n e = 0x10001 phi = (p - 1)*(q - 1) d = inverse(e, phi) m = pow(flag, d, n) print(long_to_bytes(m))

不包含web； misc；pwn；web misc太基础；pwn不会！！我是菜鸡