一，给阿姨倒一杯卡布奇诺

是一道魔改TEA加密

给出了一些初始化，然后输入的flag拆分，两两一组，通过for循环放入encrypt加密函数

#include <stdio.h>
#define uint32_t unsigned intvoid decrypt(uint32_t *v, uint32_t *key)
{static uint32_t data1 = 0x5F797274;static uint32_t data2 = 0x64726168;int i;   // [rsp+20h] [rbp-10h]uint32_t sum; // [rsp+24h] [rbp-Ch]uint32_t v1;  // [rsp+28h] [rbp-8h]uint32_t v0;  // [rsp+2Ch] [rbp-4h]sum = 0x6E75316C * 32;uint32_t data1_tmp = v[0];uint32_t data2_tmp = v[1];v0 = v[0];v1 = v[1];for (i = 31; i >= 0; i--){v1 -= ((v0 >> 5) + key[3]) ^ (v0 + sum) ^ (key[2] + 16 * v0) ^ (sum + i);v0 -= ((v1 >> 5) + key[1]) ^ (v1 + sum) ^ (key[0] + 16 * v1) ^ (sum + i);sum -= 0x6E75316C;}v[0] = v0 ^ data1;v[1] = v1 ^ data2;data1 = data1_tmp;data2 = data2_tmp;
}int main()
{uint32_t key[4];   // [rsp+60h] [rbp-40h] BYREFuint32_t array[8]; // [rsp+70h] [rbp-30h]array[0] = 0x9B28ED45;array[1] = 0x145EC6E9;array[2] = 0x5B27A6C3;array[3] = 0xE59E75D5;array[4] = 0xE82C2500;array[5] = 0xA4211D92;array[6] = 0xCD8A4B62;array[7] = 0xA668F440;key[0] = 0x65766967;key[1] = 0x756F795F;key[2] = 0x7075635F;key[3] = 0x6165745F;for (int i = 0; i <= 7; i += 2){decrypt(array + i, key);}for(int i=0; i<32; i++){printf("%c", ((char*)array)[i]);}return 0;
}// 133bffe401d223a02385d90c5f1ca377

二，ez_rand

int __cdecl main(int argc, const char **argv, const char **envp)
{unsigned __int64 v3; // rbx ，无符号64位整数型unsigned __int16 v4; // ax  ， 无符号16位整数型int v5; // edi__int64 v6; // rsiint v7; // eaxint v9[7]; // [rsp+20h] [rbp-50h]char v10; // [rsp+3Ch] [rbp-34h]__int16 v11; // [rsp+3Dh] [rbp-33h]__int128 v12; // [rsp+40h] [rbp-30h]__int64 v13; // [rsp+50h] [rbp-20h]int v14; // [rsp+58h] [rbp-18h]__int16 v15; // [rsp+5Ch] [rbp-14h]char v16; // [rsp+5Eh] [rbp-12h]v13 = 0i64;v12 = 0i64;v14 = 0;v15 = 0;v16 = 0;print((char *)&Format);scanf("%s");v9[0] = -362017699;v11 = 0;v3 = -1i64;v9[1] = 888936774;v9[2] = 119759538;v9[3] = -76668318;v9[4] = -1443698508;v9[5] = -2044652911;v9[6] = 1139379931;v10 = 77;do++v3;while ( *((_BYTE *)&v12 + v3) );v4 = time64(0i64);srand(v4);v5 = 0;if ( v3 ){v6 = 0i64;do{v7 = rand();if ( (*((_BYTE *)&v12 + v6) ^ (unsigned __int8)(v7+ ((((unsigned __int64)(2155905153i64 * v7) >> 32) & 0x80000000) != 0i64)+ ((int)((unsigned __int64)(2155905153i64 * v7) >> 32) >> 7))) != *((_BYTE *)v9 + v6) ){print("Error???\n");exit(0);}++v5;++v6;}while ( v5 < v3 );}print("Right???\n");system("pause");return 0;
}

就是随机数v7与v9异或

随机数种子是通过time来取的，C语言中的srand(time)是伪随机，直接爆破，题目描述给出了flag头为"XYCTF"，根据这个信息去爆破随机数种子，即我们将v9的前5位与生成的前五位随机数做异或，如果结果与“XYCTF”相同，则那个随机数种子就是我们需要求的结果 

#include<iostream>
#include<cstdlib>
using namespace std;
int main()
{unsigned char str[5] = { 0x5D, 0x0C, 0x6C, 0xEA, 0x46 };unsigned char random[6] = { 0 };unsigned char flag[6] = { 'X', 'Y', 'C', 'T', 'F', '\0' };for (int i = 0xFFFF; i >= 0; i--) {srand(i);for (int j = 0; j < 5; j++) {random[j] = rand() % 0xFF;}bool found = true;for (int j = 0; j < 5; j++) {if ((random[j] ^ str[j]) != flag[j]) {found = false;break;}}if (found) {cout << "Found! It is: " << i << endl;break;}elsecout << "Not " << i << " Nope" << endl;}return 0;
}
//Found! It is: 21308

 爆破出随机数种子：21308

#include<iostream>
#include<cstdlib>
using namespace std;int main()
{srand(21308);for (int i = 0; i < 29; i++) {int num = rand();cout << num << ",";}return 0;
}
//得到随机数4085,19210,5147,22630,16830,25853,6039,15416,9400,1281,32764,16374,8177,18485,16126,29528,5590,4777,18044,26256,25694,24259,10836,5327,13701,7138,5244,22538,13308,

 

随机数是16位的： 可以int num = rand() % 0xFF ;

得到：16位key

v9=[ 0x5D, 0x0C, 0x6C, 0xEA, 0x46, 0x19, 0xFC, 0x34, 0xB2, 0x62,0x23, 0x07, 0x62, 0x22, 0x6E, 0xFB, 0xB4, 0xE8, 0xF2, 0xA9,0x91, 0x12, 0x21, 0x86, 0xDB, 0x8E, 0xE9, 0x43, 0x4D]
key=[5,85,47,190,0,98,174,116,220,6,124,54,17,125,61,203,235,187,194,246,194,34,126,227,186,253,144,98,48]
flag=''
for i in range(len(v9)):flag+=chr(v9[i]^key[i])
print(flag)
#XYCTF{R@nd_1s_S0_S0_S0_easy!}

为什么就是v12^v7!=v9(?)

三，ez_cube

__int64 sub_140012930()
{int i; // [rsp+44h] [rbp+24h]char v2; // [rsp+64h] [rbp+44h]int v3; // [rsp+84h] [rbp+64h]sub_140011384(&unk_1400240A2);for ( i = 0; i < 9; ++i ){top[i] = &unk_14001CC24;                    // redunder[i] = "Blue";right[i] = "Green";left[i] = "Orange";advance[i] = "Yellow";below[i] = "White";}under[1] = &unk_14001CC24;top[1] = "Green";right[1] = "Blue";while ( 1 ){dov2 = getchar();while ( v2 == 10 );switch ( v2 ){case 'R':sub_140011375();break;case 'U':sub_1400113BB();break;case 'r':sub_140011366();break;case 'u':sub_14001115E();break;}++dword_14001F1C0;v3 = j_check();if ( v3 == 1 )break;if ( v3 == 2 )goto LABEL_19;}print(aGreatYouAreAGo);
LABEL_19:system("pause");return 0i64;
}

'R' 'U' 'r' 'u'操作每一步

_QWORD *sub_1400117F0()
{_QWORD *result; // rax__int64 v1; // [rsp+28h] [rbp+8h]__int64 v2; // [rsp+48h] [rbp+28h]__int64 v3; // [rsp+68h] [rbp+48h]__int64 v4; // [rsp+88h] [rbp+68h]__int64 v5; // [rsp+A8h] [rbp+88h]sub_140011384(&unk_1400240A2);v1 = top[2];v2 = top[5];v3 = top[8];top[2] = below[2];top[5] = below[5];top[8] = below[8];below[2] = left[6];below[5] = left[3];below[8] = left[0];left[0] = advance[8];left[3] = advance[5];left[6] = advance[2];advance[2] = v1;advance[5] = v2;advance[8] = v3;v4 = right[1];right[1] = right[3];right[3] = right[7];right[7] = right[5];right[5] = v4;v5 = right[0];right[0] = right[6];right[6] = right[8];right[8] = right[2];result = right;right[2] = v5;return result;
}

嗯，自己玩分析可能有点麻烦，想想应该可以直接写脚本。（自己用c++写了一下，不知道怎么搞四个字符操作那里，有点麻烦）先借一下别人的脚本吧

爆破的脚本还是需要再学一下。

四，What's this

Lua bytecode

可以找一个lua在线反编译网站。Lua 工具箱 (luatool.cn)

应该先变字符然后base64解密 ，发现不对，前面应该还有一些操作

function Xor(num1, num2)local tmp1 = num1local tmp2 = num2local str = ""repeatlocal s1 = tmp1 % 2local s2 = tmp2 % 2if s1 == s2 thenstr = "0" .. strelsestr = "1" .. strendtmp1 = math.modf(tmp1 / 2)tmp2 = math.modf(tmp2 / 2)until tmp1 == 0 and tmp2 == 0return tonumber(str, 2)
endvalue = ""
output = ""
i = 1
while true dolocal temp = string.byte(flag, i)temp = string.char(Xor(temp, 8) % 256)value = value .. tempi = i + 1if i > string.len(flag) thenbreakend
end
for _ = 1, 1000 dox = 3y = x * 3z = y / 4w = z - 5if w == 0 thenprint("This line will never be executed")end
end
for i = 1, string.len(flag) dotemp = string.byte(value, i)temp = string.char(temp + 3)output = output .. temp
end
result = output:rep(10)
invalid_list = {1,2,3
}
for _ = 1, 20 dotable.insert(invalid_list, 4)
end
for _ = 1, 50 doresult = result .. "A"table.insert(invalid_list, 4)
end
for i = 1, string.len(output) dotemp = string.byte(output, i)temp = string.char(temp - 1)
end
for _ = 1, 30 doresult = result .. string.lower(output)
end
for _ = 1, 950 dox = 3y = x * 3z = y / 4w = z - 5if w == 0 thenprint("This line will never be executed")end
end
for _ = 1, 50 dox = -1y = x * 4z = y / 2w = z - 3if w == 0 thenprint("This line will also never be executed")end
end
require("base64")
obfuscated_output = to_base64(output)
obfuscated_output = string.reverse(obfuscated_output)
obfuscated_output = string.gsub(obfuscated_output, "g", "3")
obfuscated_output = string.gsub(obfuscated_output, "H", "4")
obfuscated_output = string.gsub(obfuscated_output, "W", "6")
invalid_variable = obfuscated_output:rep(5)
if obfuscated_output == "==AeuFEcwxGPuJ0PBNzbC16ctFnPB5DPzI0bwx6bu9GQ2F1XOR1U" thenprint("You get the flag.")
elseprint("F**k!")
end

先异或8后+3

import  base64
enc='==AeuFEcwxGPuJ0PBNzbC16ctFnPB5DPzI0bwx6bu9GQ2F1XOR1U'
print(enc[::-1])
str=list(enc[::-1])
for i in range(len(str)):if str[i]=='3':str[i]='g'elif str[i]=='4':str[i]='H'elif str[i]=='6':str[i]="W"
print(''.join(str))
ant='U1ROX1F2QG9ubWxwb0IzPD5BPnFtcW1CbzNBP0JuPGxwcEFueA=='
date=base64.b64decode(ant)
flag = ""
for i in date:flag += chr((i - 3) ^ 8)
print(flag)
#XYCTF{5dcbaed781363fbfb7d8647c1aee6c}