BUUCTF reverse wp 11 - 20
Java逆向解密
java逆向, 用java-decompile逆
import java.util.ArrayList;
import java.util.Scanner;
public class Reverse {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
System.out.println("Please input the flag );
String str = s.next();
System.out.println("Your input is );
System.out.println(str);
char[] stringArr = str.toCharArray();
Encrypt(stringArr);
}
public static void Encrypt(char[] arr) {
ArrayList<Integer> Resultlist = new ArrayList<>();
for (int i = 0; i < arr.length; i++) {
int result = arr[i] + 64 ^ 0x20;
Resultlist.add(Integer.valueOf(result));
}
int[] KEY = {
180, 136, 137, 147, 191, 137, 147, 191, 148, 136,
133, 191, 134, 140, 129, 135, 191, 65 };
ArrayList<Integer> KEYList = new ArrayList<>();
for (int j = 0; j < KEY.length; j++)
KEYList.add(Integer.valueOf(KEY[j]));
System.out.println("Result:");
if (Resultlist.equals(KEYList)) {
System.out.println("Congratulations);
} else {
System.err.println("Error);
}
}
}
当Resultlist.equals(KEYList)时通过检测
Resultlist的逻辑
for (int i = 0; i < arr.length; i++) {
int result = arr[i] + 64 ^ 0x20;
Resultlist.add(Integer.valueOf(result));
}
逆回去就是先用Keylist ^ 0x20, 然后减去64(注意java中^的优先级是比+更低的,所以是先+再^
Keylist = [180, 136, 137, 147, 191, 137, 147, 191, 148, 136, \
133, 191, 134, 140, 129, 135, 191, 65]
flag = ''
for key in Keylist:
flag += chr((key ^ 0x20) - 64)
print(flag)
结果记得加flag{}
[GXYCTF2019]luck_guy
file看是x64文件, 没壳拖进IDA64
unsigned __int64 get_flag()
{
unsigned int v0; // eax
int i; // [rsp+4h] [rbp-3Ch]
int j; // [rsp+8h] [rbp-38h]
__int64 s; // [rsp+10h] [rbp-30h] BYREF
char v5; // [rsp+18h] [rbp-28h]
unsigned __int64 v6; // [rsp+38h] [rbp-8h]
v6 = __readfsqword(0x28u);
v0 = time(0LL);
srand(v0);
for ( i = 0; i <= 4; ++i )
{
switch ( rand() % 200 )
{
case 1:
puts("OK, it's flag:");
memset(&s, 0, 0x28uLL);
strcat((char *)&s, f1);
strcat((char *)&s, &f2);
printf("%s", (const char *)&s);
break;
case 2:
printf("Solar not like you");
break;
case 3:
printf("Solar want a girlfriend");
break;
case 4:
s = 0x7F666F6067756369LL;
v5 = 0;
strcat(&f2, (const char *)&s);
break;
case 5:
for ( j = 0; j <= 7; ++j )
{
if ( j % 2 == 1 )
*(&f2 + j) -= 2;
else
--*(&f2 + j);
}
break;
default:
puts("emmm,you can't find flag 23333");
break;
}
}
return __readfsqword(0x28u) ^ v6;
}
处理一下s字符串, 然后就是flag后半段
flag = "GXY{do_not_"
s = [0x7F, 0x66, 0x6F, 0x60, 0x67, 0x75, 0x63, 0x69] # 0x7F666F6067756369LL
s = s[::-1]
for i in range(8):
if i % 2 == 1: s[i] -= 2
else: s[i] -= 1
flag += chr(s[i])
print(flag)
GXY{do_not_hate_me}
提交时改成flag{}
刮开有奖
INT_PTR __stdcall DialogFunc(HWND hDlg, UINT a2, WPARAM a3, LPARAM a4)
{
const char *v4; // esi
const char *v5; // edi
int v7[2]; // [esp+8h] [ebp-20030h] BYREF
int v8; // [esp+10h] [ebp-20028h]
int v9; // [esp+14h] [ebp-20024h]
int v10; // [esp+18h] [ebp-20020h]
int v11; // [esp+1Ch] [ebp-2001Ch]
int v12; // [esp+20h] [ebp-20018h]
int v13; // [esp+24h] [ebp-20014h]
int v14; // [esp+28h] [ebp-20010h]
int v15; // [esp+2Ch] [ebp-2000Ch]
int v16; // [esp+30h] [ebp-20008h]
CHAR String[65536]; // [esp+34h] [ebp-20004h] BYREF
char v18[65536]; // [esp+10034h] [ebp-10004h] BYREF
if ( a2 == 272 )
return 1;
if ( a2 != 273 )
return 0;
if ( a3 == 1001 )
{
memset(String, 0, 0xFFFFu);
GetDlgItemTextA(hDlg, 1000, String, 0xFFFF);
if ( strlen(String) == 8 )
{
v7[0] = 90;
v7[1] = 74;
v8 = 83;
v9 = 69;
v10 = 67;
v11 = 97;
v12 = 78;
v13 = 72;
v14 = 51;
v15 = 110;
v16 = 103;
sub_4010F0(v7, 0, 10);
memset(v18, 0, 0xFFFFu);
v18[0] = String[5];
v18[2] = String[7];
v18[1] = String[6];
v4 = sub_401000(v18, strlen(v18));
memset(v18, 0, 0xFFFFu);
v18[1] = String[3];
v18[0] = String[2];
v18[2] = String[4];
v5 = sub_401000(v18, strlen(v18));
if ( String[0] == v7[0] + 34
&& String[1] == v10
&& 4 * String[2] - 141 == 3 * v8
&& String[3] / 4 == 2 * (v13 / 9)
&& !strcmp(v4, "ak1w")
&& !strcmp(v5, "V1Ax") )
{
MessageBoxA(hDlg, "U g3t 1T!", "@_@", 0);
}
}
return 0;
}
if ( a3 != 1 && a3 != 2 )
return 0;
EndDialog(hDlg, a3);
return 1;
}
输入8字符的string, 通过if校验后为flag
if ( String[0] == v7[0] + 34
&& String[1] == v10
&& 4 * String[2] - 141 == 3 * v8
&& String[3] / 4 == 2 * (v13 / 9)
&& !strcmp(v4, "ak1w")
&& !strcmp(v5, "V1Ax") )
v7 到 v16会经过sub_4010F0函数, 这里懒得分析了, 直接修改为c语言run, 可以在IDA中修改函数的item type, a1修改为char[]类型
双击v7局部变量, 进入栈窗口, 右键修改为数组类型
伪代码中的4 *进行数组索引是不合理的, 去掉之后得到符合语法的C代码如下
#includeZJSECaNH3ng
3CEHJNSZagn
可以确定string的前四位字符
_BYTE *__cdecl sub_401000(int a1, int a2)
{
int v2; // eax
int v3; // esi
size_t v4; // ebx
_BYTE *v5; // eax
_BYTE *v6; // edi
int v7; // eax
_BYTE *v8; // ebx
int v9; // edi
int v10; // edx
int v11; // edi
int v12; // eax
int i; // esi
_BYTE *result; // eax
_BYTE *v15; // [esp+Ch] [ebp-10h]
_BYTE *v16; // [esp+10h] [ebp-Ch]
int v17; // [esp+14h] [ebp-8h]
int v18; // [esp+18h] [ebp-4h]
v2 = a2 / 3;
v3 = 0;
if ( a2 % 3 > 0 )
++v2;
v4 = 4 * v2 + 1;
v5 = malloc(v4);
v6 = v5;
v15 = v5;
if ( !v5 )
exit(0);
memset(v5, 0, v4);
v7 = a2;
v8 = v6;
v16 = v6;
if ( a2 > 0 )
{
while ( 1 )
{
v9 = 0;
v10 = 0;
v18 = 0;
do
{
if ( v3 >= v7 )
break;
++v10;
v9 = *(v3 + a1) | (v9 << 8);
++v3;
}
while ( v10 < 3 );
v11 = v9 << (8 * (3 - v10));
v12 = 0;
v17 = v3;
for ( i = 18; i > -6; i -= 6 )
{
if ( v10 >= v12 )
{
*(&v18 + v12) = (v11 >> i) & 0x3F;
v8 = v16;
}
else
{
*(&v18 + v12) = 64;
}
*v8++ = byte_407830[*(&v18 + v12++)];
v16 = v8;
}
v3 = v17;
if ( v17 >= a2 )
break;
v7 = a2;
}
v6 = v15;
}
result = v6;
*v8 = 0;
return result;
}
分析sub_401000函数, 看到byte_407830数组
所以直接猜测为Base64加密, (因为传入3字节, 输出4字节
import base64
"""
if ( String[0] == a1[0] + 34
&& String[1] == a1[4]
&& 4 * String[2] - 141 == 3 * a1[2]
&& String[3] / 4 == 2 * (a1[7] / 9)
&& !strcmp(v4, "ak1w")
&& !strcmp(v5, "V1Ax") )
"""
s = "3CEHJNSZagn"
arr = [ord(_) for _ in s]
print(arr)
strings = [arr[0] + 34, arr[4], (3 * arr[2] + 141) // 4, int(2 * (arr[7] / 9) * 4)]
print(strings)
flag = "".join([chr(_) for _ in strings])
# print(flag)
b64str1 = "ak1w"
b64str2 = "V1Ax"
print(base64.b64decode(b64str1))
print(base64.b64decode(b64str2))
flag += "1jMp"
print("flag{" + flag + "}")
findit
安卓逆向
package com.example.findit;
import android.os.Bundle;
import android.support.v7.app.ActionBarActivity;
import android.view.MenuItem;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
public class MainActivity extends ActionBarActivity {
/* access modifiers changed from: protected */
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView((int) R.layout.activity_main);
final EditText edit = (EditText) findViewById(R.id.widget2);
final TextView text = (TextView) findViewById(R.id.widget1);
final char[] a = {'T', 'h', 'i', 's', 'I', 's', 'T', 'h', 'e', 'F', 'l', 'a', 'g', 'H', 'o', 'm', 'e'};
final char[] b = {'p', 'v', 'k', 'q', '{', 'm', '1', '6', '4', '6', '7', '5', '2', '6', '2', '0', '3', '3', 'l', '4', 'm', '4', '9', 'l', 'n', 'p', '7', 'p', '9', 'm', 'n', 'k', '2', '8', 'k', '7', '5', '}'};
((Button) findViewById(R.id.widget3)).setOnClickListener(new View.OnClickListener() {
public void onClick(View v) {
char[] x = new char[17];
char[] y = new char[38];
for (int i = 0; i < 17; i++) {
if ((a[i] < 'I' && a[i] >= 'A') || (a[i] < 'i' && a[i] >= 'a')) {
x[i] = (char) (a[i] + 18);
} else if ((a[i] < 'A' || a[i] > 'Z') && (a[i] < 'a' || a[i] > 'z')) {
x[i] = a[i];
} else {
x[i] = (char) (a[i] - 8);
}
}
if (String.valueOf(x).equals(edit.getText().toString())) {
for (int i2 = 0; i2 < 38; i2++) {
if ((b[i2] < 'A' || b[i2] > 'Z') && (b[i2] < 'a' || b[i2] > 'z')) {
y[i2] = b[i2];
} else {
y[i2] = (char) (b[i2] + 16);
if ((y[i2] > 'Z' && y[i2] < 'a') || y[i2] >= 'z') {
y[i2] = (char) (y[i2] - 26);
}
}
}
text.setText(String.valueOf(y));
return;
}
text.setText("答案错了肿么办。。。不给你又不好意思。。。哎呀好纠结啊~~~");
}
});
}
public boolean onOptionsItemSelected(MenuItem item) {
if (item.getItemId() == R.id.action_settings) {
return true;
}
return super.onOptionsItemSelected(item);
}
}
输入与x相同则输出flag, 图省事, java转cpp比较方便, 所以直接用cpp写exp
exp.cpp
#include
[BJDCTF2020]JustRE
shift + F12, 找到flag字符串
INT_PTR __stdcall DialogFunc(HWND hWnd, UINT a2, WPARAM a3, LPARAM a4)
{
CHAR String[100]; // [esp+0h] [ebp-64h] BYREF
if ( a2 != 272 )
{
if ( a2 != 273 )
return 0;
if ( (_WORD)a3 != 1 && (_WORD)a3 != 2 )
{
sprintf(String, Format, ++dword_4099F0);
if ( dword_4099F0 == 19999 )
{
sprintf(String, " BJD{%d%d2069a45792d233ac}", 19999, 0);
SetWindowTextA(hWnd, String);
return 0;
}
SetWindowTextA(hWnd, String);
return 0;
}
EndDialog(hWnd, (unsigned __int16)a3);
}
return 1;
}
记得将BJD{} 改为flag{}
简单注册器
package com.example.flag;
import android.os.Bundle;
import android.support.v4.app.Fragment;
import android.support.v7.app.ActionBarActivity;
import android.view.LayoutInflater;
import android.view.Menu;
import android.view.MenuItem;
import android.view.View;
import android.view.ViewGroup;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
public class MainActivity extends ActionBarActivity {
/* access modifiers changed from: protected */
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView((int) R.layout.activity_main);
if (savedInstanceState == null) {
getSupportFragmentManager().beginTransaction().add((int) R.id.container, (Fragment) new PlaceholderFragment()).commit();
}
final TextView textview = (TextView) findViewById(R.id.textView1);
final EditText editview = (EditText) findViewById(R.id.editText1);
((Button) findViewById(R.id.button1)).setOnClickListener(new View.OnClickListener() {
public void onClick(View v) {
int flag = 1;
String xx = editview.getText().toString();
if (!(xx.length() == 32 && xx.charAt(31) == 'a' && xx.charAt(1) == 'b' && (xx.charAt(0) + xx.charAt(2)) - 48 == 56)) {
flag = 0;
}
if (flag == 1) {
char[] x = "dd2940c04462b4dd7c450528835cca15".toCharArray();
x[2] = (char) ((x[2] + x[3]) - 50);
x[4] = (char) ((x[2] + x[5]) - 48);
x[30] = (char) ((x[31] + x[9]) - 48);
x[14] = (char) ((x[27] + x[28]) - 97);
for (int i = 0; i < 16; i++) {
char a = x[31 - i];
x[31 - i] = x[i];
x[i] = a;
}
textview.setText("flag{" + String.valueOf(x) + "}");
return;
}
textview.setText("输入注册码错误");
}
});
}
public boolean onCreateOptionsMenu(Menu menu) {
getMenuInflater().inflate(R.menu.main, menu);
return true;
}
public boolean onOptionsItemSelected(MenuItem item) {
if (item.getItemId() == R.id.action_settings) {
return true;
}
return super.onOptionsItemSelected(item);
}
public static class PlaceholderFragment extends Fragment {
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
return inflater.inflate(R.layout.fragment_main, container, false);
}
}
}
#include[GWCTF 2019]pyre
python逆向
python -m pip install uncompyle
uncompyle6 ./attachment.pyc > rev.py
# uncompyle6 version 3.8.0
# Python bytecode 2.7 (62211)
# Decompiled from: Python 3.8.10 (tags/v3.8.10:3d8993a, May 3 2021, 11:48:03) [MSC v.1928 64 bit (AMD64)]
# Embedded file name: encode.py
# Compiled at: 2019-08-19 21:01:57
print 'Welcome to Re World!'
print 'Your input1 is your flag~'
l = len(input1)
for i in range(l):
num = ((input1[i] + i) % 128 + 128) % 128
code += num
for i in range(l - 1):
code[i] = code[i] ^ code[(i + 1)]
print code
code = ['\x1f', '\x12', '\x1d', '(', '0', '4', '\x01', '\x06', '\x14', '4', ',', '\x1b', 'U', '?', 'o', '6', '*', ':', '\x01', 'D', ';', '%', '\x13']
# okay decompiling .\attachment.pyc
从后往前逆回去就行, code += num看起来不合理, 先不理它
exp
code = ['\x1f', '\x12', '\x1d', '(', '0', '4', '\x01', '\x06', '\x14', '4', ',', '\x1b', 'U', '?', 'o', '6', '*', ':', '\x01', 'D', ';', '%', '\x13']
l = len(code)
for i in range(l - 2, -1, -1):
code[i] = chr(ord(code[i]) ^ ord(code[i + 1]))
for i in range(l):
code[i] = chr((ord(code[i]) - i) % 128)
flag = ""
for i in range(l):
flag += code[i]
print(flag)
[ACTF新生赛2020]easyre
脱壳, 拖进IDA
upx -d easyre.exe
int __cdecl main(int argc, const char **argv, const char **envp)
{
_BYTE v4[12]; // [esp+12h] [ebp-2Eh] BYREF
_DWORD v5[3]; // [esp+1Eh] [ebp-22h]
_BYTE v6[5]; // [esp+2Ah] [ebp-16h] BYREF
int v7; // [esp+2Fh] [ebp-11h]
int v8; // [esp+33h] [ebp-Dh]
int v9; // [esp+37h] [ebp-9h]
char v10; // [esp+3Bh] [ebp-5h]
int i; // [esp+3Ch] [ebp-4h]
__main();
qmemcpy(v4, "*F'\"N,\"(I?+@", sizeof(v4));
printf("Please input:");
scanf("%s", v6);
if ( v6[0] != 65 || v6[1] != 67 || v6[2] != 84 || v6[3] != 70 || v6[4] != 123 || v10 != 125 )
return 0;
v5[0] = v7;
v5[1] = v8;
v5[2] = v9;
for ( i = 0; i <= 11; ++i )
{
if ( v4[i] != _data_start__[*((char *)v5 + i) - 1] )
return 0;
}
printf("You are correct!");
return 0;
}
强逆, 可从hex窗口拷贝string
teststr = "*F'\"N,\"(I?+@"
print(len(teststr))
data = "~}|{zyxwvutsrqponmlkjihgfedcba`_^]\[ZYXWVUTSRQPONMLKJIHGFEDCBA@?>="
data += "<;:9876543210/.-,+*)("
data += chr(0x27)
data += "&%$# !\""
indexs = []
for ch in teststr:
for i in range(len(data)):
if ch == data[i]:
indexs.append(chr(i + 1))
print(i + 1)
flag = "".join(indexs)
print("ACTF{" + flag + "}")
print("flag{" + flag + "}")
rsa
密码学题目, 怎么丢进reverse来了…
pub.key改成pem文件, 拖进kali用openssl
n = 86934482296048119190666062003494800588905656017203025617216654058378322103517
http://www.factordb.com/index.php?query=86934482296048119190666062003494800588905656017203025617216654058378322103517
分解得到p, q
p = 285960468890451637935629440372639283459
q = 304008741604601924494328155975272418463
解密exp
import gmpy2
import rsa
e = 65537
n = 86934482296048119190666062003494800588905656017203025617216654058378322103517
p = 285960468890451637935629440372639283459
q = 304008741604601924494328155975272418463
d = gmpy2.invert(e, (q-1)*(p-1))
key = rsa.PrivateKey(n, e, int(d), p, q)
with open("./flag.enc", "rb+") as f:
f = f.read()
print(rsa.decrypt(f, key))
CrackRTF
int __cdecl main_0(int argc, const char **argv, const char **envp)
{
DWORD v3; // eax
DWORD v4; // eax
char Str[260]; // [esp+4Ch] [ebp-310h] BYREF
int v7; // [esp+150h] [ebp-20Ch]
char String1[260]; // [esp+154h] [ebp-208h] BYREF
char Destination[260]; // [esp+258h] [ebp-104h] BYREF
memset(Destination, 0, sizeof(Destination));
memset(String1, 0, sizeof(String1));
v7 = 0;
printf("pls input the first passwd(1): ");
scanf("%s", Destination);
if ( strlen(Destination) != 6 )
{
printf("Must be 6 characters!\n");
ExitProcess(0);
}
v7 = atoi(Destination);
if ( v7 < 100000 )
ExitProcess(0);
strcat(Destination, "@DBApp");
v3 = strlen(Destination);
sub_40100A((BYTE *)Destination, v3, String1);
if ( !_strcmpi(String1, "6E32D0943418C2C33385BC35A1470250DD8923A9") )
{
printf("continue...\n\n");
printf("pls input the first passwd(2): ");
memset(Str, 0, sizeof(Str));
scanf("%s", Str);
if ( strlen(Str) != 6 )
{
printf("Must be 6 characters!\n");
ExitProcess(0);
}
strcat(Str, Destination);
memset(String1, 0, sizeof(String1));
v4 = strlen(Str);
sub_401019((BYTE *)Str, v4, String1);
if ( !_strcmpi("27019e688a4e62a649fd99cadaafdb4e", String1) )
{
if ( !(unsigned __int8)sub_40100F(Str) )
{
printf("Error!!\n");
ExitProcess(0);
}
printf("bye ~~\n");
}
}
return 0;
}
int __cdecl sub_401230(BYTE *pbData, DWORD dwDataLen, LPSTR lpString1)
{
int result; // eax
DWORD i; // [esp+4Ch] [ebp-28h]
CHAR String2[4]; // [esp+50h] [ebp-24h] BYREF
BYTE v6[20]; // [esp+54h] [ebp-20h] BYREF
DWORD pdwDataLen; // [esp+68h] [ebp-Ch] BYREF
HCRYPTHASH phHash; // [esp+6Ch] [ebp-8h] BYREF
HCRYPTPROV phProv; // [esp+70h] [ebp-4h] BYREF
if ( !CryptAcquireContextA(&phProv, 0, 0, 1u, 0xF0000000) )
return 0;
if ( CryptCreateHash(phProv, 0x8004u, 0, 0, &phHash) )
{
if ( CryptHashData(phHash, pbData, dwDataLen, 0) )
{
CryptGetHashParam(phHash, 2u, v6, &pdwDataLen, 0);
*lpString1 = 0;
for ( i = 0; i < pdwDataLen; ++i )
{
wsprintfA(String2, "%02X", v6[i]);
lstrcatA(lpString1, String2);
}
CryptDestroyHash(phHash);
CryptReleaseContext(phProv, 0);
result = 1;
}
else
{
CryptDestroyHash(phHash);
CryptReleaseContext(phProv, 0);
result = 0;
}
}
else
{
CryptReleaseContext(phProv, 0);
result = 0;
}
return result;
}
sub_401230用到了一个windows加密的加密库函数
观察发现6E32D0943418C2C33385BC35A1470250DD8923A9是40位的字符串
猜测可能是sha1加密
用sha1爆破一下6位密码
import hashlib
hash1 = "6E32D0943418C2C33385BC35A1470250DD8923A9".lower()
hash2 = "27019e688a4e62a649fd99cadaafdb4e".lower()
flag = "@DBApp"
for i in range(100000, 999999):
s = str(i) + flag
x = hashlib.sha1(s.encode())
if x.hexdigest() == hash1:
flag = s
print(flag)
break
123321@DBApp
验证通过
第二次hash是32位的字符串, 猜测是MD5, 但是直接全字符爆破不实际
分析其他线索
char __cdecl sub_4014D0(LPCSTR lpString)
{
LPCVOID lpBuffer; // [esp+50h] [ebp-1Ch]
DWORD NumberOfBytesWritten; // [esp+58h] [ebp-14h] BYREF
DWORD nNumberOfBytesToWrite; // [esp+5Ch] [ebp-10h]
HGLOBAL hResData; // [esp+60h] [ebp-Ch]
HRSRC hResInfo; // [esp+64h] [ebp-8h]
HANDLE hFile; // [esp+68h] [ebp-4h]
hFile = 0;
hResData = 0;
nNumberOfBytesToWrite = 0;
NumberOfBytesWritten = 0;
hResInfo = FindResourceA(0, (LPCSTR)0x65, "AAA");
if ( !hResInfo )
return 0;
nNumberOfBytesToWrite = SizeofResource(0, hResInfo);
hResData = LoadResource(0, hResInfo);
if ( !hResData )
return 0;
lpBuffer = LockResource(hResData);
sub_401005(lpString, (int)lpBuffer, nNumberOfBytesToWrite);
hFile = CreateFileA("dbapp.rtf", 0x10000000u, 0, 0, 2u, 0x80u, 0);
if ( hFile == (HANDLE)-1 )
return 0;
if ( !WriteFile(hFile, lpBuffer, nNumberOfBytesToWrite, &NumberOfBytesWritten, 0) )
return 0;
CloseHandle(hFile);
return 1;
}
unsigned int __cdecl sub_401420(LPCSTR lpString, int a2, int a3)
{
unsigned int result; // eax
unsigned int i; // [esp+4Ch] [ebp-Ch]
unsigned int v5; // [esp+54h] [ebp-4h]
v5 = lstrlenA(lpString);
for ( i = 0; ; ++i )
{
result = i;
if ( i >= a3 )
break;
*(_BYTE *)(i + a2) ^= lpString[i % v5];
}
return result;
}
从"AAA"resource中读取数据, 然后和输入的6个字符异或得到结果, 写入.rtf文件, 因为rtf文件头是确定的, 所以只要找到AAA中的数据就能得到第二个密钥
用resource-hacker
.rtf文件头
写exp解密第二个key, 这里rtf头有5字节, 往后多取一个字节, 为0x31
rtf_header = [0x7B, 0x5C, 0x72, 0x74, 0x66, 0x31]
arr = [0x05, 0x7D, 0x41, 0x15, 0x26, 0x01]
key2 = ''
for i in range(6):
key2 += chr(rtf_header[i] ^ arr[i])
print(key2)
~!3a@0
输入exe的第二个key, 生成rtf文件, 用wps打开得到flag