安洵杯 re + 其他部分题解

第11，比较小丑，差了一步
队伍wp应该会发吧，不知道，我先放点跟我有关系的

Re

mobilego

so的check看了一会比较南崩，但是看flag的密文形式很像简单位置替换
所以直接输编码表，jeb动调然后得到替换表
解密就行

flag密文的话有string.cmp，直接追踪资源就行

ori = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789ab'
key = 'V8L3ObJT4PK2MYZFQBSHUCXWI0AGa51D679NER'
v =   '49021}5f919038b440139g74b7Dc88330e5d{6'
for i in ori:print(v[key.index(i)],end='')
# D0g3{4c3b5903d11461f94478b7302980e958}

在这里插入图片描述

你见过蓝色的小鲸鱼

交给队友的一题：(
题目提示了blowfish加密，那么第一步要做的就是确认代码是否是一个正常的加密
然后密钥UzBtZTBuZV9EMGcz
密文dump出来：11A51F049550E2508F17E16CF1632B47
在线网站直接解密
拼接即可：D0g3{UzBtZTBuZV9EMGczQHRoZWJsdWVmMXNo}

感觉有点点简单

文件sys挺吓人，实际就一个改了的RC4和base64，都不需要运行
base改了爆破就行了，逆还得动脑子

#include <bits/stdc++.h>
using namespace std;int encode(char *a1, char *a2, int len)
{int v4; // [rsp+0h] [rbp-88h]int v5; // [rsp+4h] [rbp-84h]char table[80]; // [rsp+10h] [rbp-78h] BYREFstrcpy(table, "4KBbSzwWClkZ2gsr1qA+Qu0FtxOm6/iVcJHPY9GNp7EaRoDf8UvIjnL5MydTX3eh");v4 = 0;v5 = 0;while ( v4 < len ){a1[v5] = table[a2[v4] & 0x3F];a1[v5 + 1] = table[(4 * (a2[v4 + 1] & 0xF)) | ((a2[v4] & 0xC0) >> 6)];a1[v5 + 2] = table[(16 * (a2[v4 + 2] & 3)) | ((a2[v4 + 1] & 0xF0) >> 4)];a1[v5 + 3] = table[(a2[v4 + 2] & 0xFC) >> 2];v4 += 3;v5 += 4;}if ( len % 3 == 1 ){a1[v5 - 2] = '=';a1[v5 - 1] = '=';}else if ( len % 3 == 2 ){a1[v5 - 1] = '=';}string temp = "4Uw=";if (temp[0] == a1[0] && temp[1] == a1[1] && temp[2] == a1[2] && temp[3] == a1[3]){return 1;}return 0;
}
int main()
{string t = "4KBbSzwWClkZ2gsr1qA+Qu0FtxOm6/iVcJHPY9GNp7EaRoDf8UvIjnL5MydTX3eh";char a1[100];char a2[2] = {};for(int i = 0; i < 126; i++){for(int j = 0; j < 126; j++){for(int k = 0;k < 126; k++){a2[0] = i;a2[1] = j;if(encode(a1,a2,2) == 1){cout <<(int)a2[0] <<"," <<(int)a2[1]<<",";return 0;}}}}}
//92,33,123,51,81,51,56,40,58,43,48,64,22,44,51,37,54,4,56,70,81,60,37,74,19,51,57,59,105,39,77,41,51,20,51,70,48,49,50,64,108

RC4 64改写

#include <bits/stdc++.h>
typedef unsigned longULONG;
using namespace std;
/*初始化函数*/
void rc4_init(unsigned char*s, unsigned char*key, unsigned long Len)
{int i = 0, j = 0;char k[64] = { 0 };unsigned char tmp = 0;for (i = 0; i<64; i++){s[i] = i;k[i] = key[i%Len];}for (i = 0; i<64; i++){j = (j + s[i] + k[i]) % 64;tmp = s[i];s[i] = s[j];s[j] = tmp;}
}/*加解密*/
void rc4_crypt(unsigned char*s, unsigned char*Data, unsigned long Len)
{int i = 0, j = 0, t = 0;unsigned long k = 0;unsigned char tmp;for (k = 0; k<Len; k++){i = (i + 1) % 64;j = (j + s[i]) % 64;tmp = s[i];s[i] = s[j];s[j] = tmp;Data[k] ^= (i ^ j) & s[(((i ^ j) + s[i] + s[j]) % 64)];}
}int main()
{unsigned char s[64] = { 0 };char key[64] = {0x74, 0x68, 0x65, 0x5F, 0x6B, 0x65, 0x79, 0x5F};char pData[512] = {92,33,123,51,81,51,56,40,58,43,48,64,22,44,51,37,54,4,56,70,81,60,37,74,19,51,57,59,105,39,77,41,51,20,51,70,48,49,50,64,108};unsigned long len = strlen(pData);rc4_init(s, (unsigned char*)key, strlen(key));rc4_crypt(s, (unsigned char*)pData, 90);for(int i = 0; i < len; i++){printf("%c",pData[i]);}return 0;
}//D0g3{608292C4-15400BA4-B3299A5C-704C292D}

PE

有点可惜，但也正常，太久不看的人是这样的
动调call恢复function可以进入主逻辑，加密过程也很简单
就剩一张当时截的图，v5 < 0，8个字节一组进行处理
巨佬的脚本

def decrypt(value, key):assert key & 1key |= 1 << 64for i in range(64):if value & 1:value = (value ^ key) >> 1else:value = value >> 1return valueout = bytes.fromhex('4db87629f5a99e595556b1c42f212c30b3797817a8edf7dbe153f0dbe903515e09c100dff096fcc1b5e6629501000000')
flag = b''
for i in range(0, len(out), 8):v = int.from_bytes(out[i: i + 8], 'little')flag += decrypt(v, 0x54aa4a9).to_bytes(8, 'little')print(flag)

牢大想你了

混淆的挺好，下次别混淆了
manager/data定位核心dll，dotpeek反编译
找到Tea的密文和密钥，直接解密【花里胡哨的没用代码一大坨】
继续交给队友:(

void decrypt(unsigned int v[], unsigned int k[]) {unsigned int v0 = v[0];unsigned int v1 = v[1];unsigned int delta = 2654435769;unsigned int sum1 = delta * 32;int i;for (i = 0; i < 32; i++) {v1 -= ((v0 << 4) + k[2]) ^ (v0 + sum1) ^ ((v0 >> 5) + k[3]);v0 -= ((v1 << 4) + k[0]) ^ (v1 + sum1) ^ ((v1 >> 5) + k[1]);sum1 -= delta;}for (i = 0; i < 4; i++) {printf("%c", (v0 >> (8 * i)) & 0xff);}for (i = 0; i < 4; i++) {printf("%c", (v1 >> (8 * i)) & 0xff);}
}int main() {unsigned int data[] = {3363017039, 1247970816, 549943836, 445086378, 3606751618, 1624361316, 3112717362, 705210466,3343515702, 2402214294, 4010321577, 2743404694};unsigned int key[] = {286331153, 286331153, 286331153, 286331153};int i;for (i = 0; i < sizeof(data) / sizeof(unsigned int); i += 2) {decrypt(&data[i], key);}return 0;
}

ok，re结束

Crypto

010101

nc得到三个参数npc
然后看逻辑就是p替换了两个bit输出来了，爆破就行了【这个地方比赛环境当时有点问题，我卡了有半个小时，脚本没动，数据不对】

from Crypto.Util.number import long_to_bytes
from gmpy2 import invertn = 
s = '1101100101111101001110011011001110011101111100000010100010100010110110100000001001110011111...'
c = 
e = 0x10001
s1 = s[:1024]
s2 = s[1024:]
print(s1)
print(s2)
for i in range(len(s1)):if s1[i] == '0':for j in range(len(s2)):if s2[j] == '1':temp = s1[:i] + '1' + s1[i+1:] + s2[:j] + '0' + s2[j+1:]p = int(temp, 2)if n % p == 0:q = n //pphi = (p-1) * (q-1)d = invert(e,phi)m = pow(c,d,n)print(long_to_bytes(m))

Misc

疯狂的麦克斯

docx变zip，后面文字是rot22，加上提示base64
直接把列表所有元素rot22然后base64，再放进去zip爆破

s = 
import base64
result = []def sort_by_numbers(item):return int(item[0])def rot22_encrypt(text):encrypted_text = ""for char in text:if char.isalpha():ascii_offset = 65 if char.isupper() else 97encrypted_char = chr((ord(char) - ascii_offset + 22) % 26 + ascii_offset)encrypted_text += encrypted_charelse:encrypted_text += charreturn encrypted_text# 打印结果
for i in s:m = str(base64.b64encode(rot22_encrypt(i).encode()))print(m[2:-1])

在这里插入图片描述

Nahida

逆转搞出图片

# 打开输入文件和输出文件
with open('Nahida!', 'rb') as input_file, open('output.jpg', 'wb') as output_file:# 读取输入文件的所有字节data = input_file.read()for i in data[::-1]:output_file.write(int.to_bytes(((i >> 4) | (i << 4))&0xff) )