前言
簡單寫一下Android部份的解題思路。
第三題：Android初級題
明顯的xxtea特徵。
解密後直接得到flag
第四題：Android中級題
目標是找到秘鑰。
Java層關鍵邏輯如下，調用了Check函數來檢查密鑰。
是個native函數。
嘗試直接hook RegisterNatives，發現Check果然是動態注冊的，在0xe8c54。
Check一開始是一些反調試邏輯。
先看anti1，它調用decrypt_str解密字符串，但奇怪的是解密出來的字符串不是以\x00結尾，導致opendir直接失敗，使得後面的反調試邏輯形同虛設？( 不知是故意的還是不小心的 )
anti2、do_something1也同理，皆因為decrypt_str的問題導致後續的邏輯失效。
繼續向下跟，看到它動態計算出一個函數地址，大概率就是加密函數，最後與密文進行對比。
一開始以為動態計算的那個函數地址是固定的，後來才發現有兩個不同的地址，會隨著上面anti1、anti2、do_something1、getenv等函數返回的結果而改變。
類似蜜罐的概念，當觸發anti邏輯後，不主動殺死APP，而是改變程序的執行流，導向錯誤的分支。
func1、func2如下，前者是錯誤的分支，後者是正確的，我的環境默認會走func1。
可以看到兩者的加密方式都是相同的異或加密，不同的只有異或的值。
經測試發現，手動hook getenv、do_something1修改其參數、返回值後，程序才會走向func2。這時再hook encrypt，將正確的異或值dump下來。
function hook_dlopen(soName) {
Interceptor.attach(Module.findExportByName(null, "dlopen"),
{
onEnter: function (args) {
var pathptr = args[0];
if (pathptr !== undefined && pathptr != null) {
var path = ptr(pathptr).readCString();
if (path.indexOf(soName) >= 0) {
this.is_can_hook = true;
}
}
},
onLeave: function (retval) {
if (this.is_can_hook) {
console.log("hook start...");
hook_func(soName)
}
}
}
);
Interceptor.attach(Module.findExportByName(null, "android_dlopen_ext"),
{
onEnter: function (args) {
var pathptr = args[0];
if (pathptr !== undefined && pathptr != null) {
var path = ptr(pathptr).readCString();
if (path.indexOf(soName) >= 0) {
this.is_can_hook = true;
}
}
},
onLeave: function (retval) {
if (this.is_can_hook) {
console.log("hook start...");
hook_func(soName)
}
}
}
);
}
function hook_func(soName) {
function hook_xorkey(base) {
Interceptor.attach(base.add(0xE9954), {
onLeave: function(retval) {
console.log("[xor_key] ", hexdump(retval))
}
})
}
function hook_test2(base) {
Interceptor.attach(base.add(0xE98A0), {
onEnter: function(args) {
console.log("[call func2] ")
}
})
Interceptor.attach(base.add(0xE74E8), {
onEnter: function(args) {
console.log("[call dosomething1] ")
},
onLeave: function(retval) {
console.log("[dosomething1] retval: ", retval)
retval.replace(0);
console.log("[dosomething1] retval: ", retval)
}
})
Interceptor.attach(Module.findExportByName(null, "getenv"), {
onEnter: function(args) {
let a0 = args[0].readCString();
if (a0.indexOf("name") != -1) {
Memory.writeUtf8String(args[0], "name");
this.flag = true
console.log("[getenv] a0: ", args[0].readCString())
}
},
onLeave: function(retval) {
if (this.flag) {
console.log("retval: ", retval.readCString())
}
}
})
}
var base = Module.findBaseAddress(soName);
hook_xorkey(base);
hook_test2(base);
}
function main() {
hook_dlopen("libwuaipojie2025_game.so")
}
setImmediate(main)
最終解密腳本：
xor_key1 = [0x2E, 0x4B, 0xEE, 0xC8, 0xE0, 0x95, 0x88, 0x47, 0xB0, 0x72, 0x1B, 0x68, 0x40, 0xD0, 0x0A, 0x84]
xor_key2 = [0x77, 0x70, 0x8a]
xor_key_list = [xor_key1, xor_key2]
data1 = 0x72ECF89BAF8F2748
data2 = 0xB63AE26B0C720798
data3 = 0xF75942
enc = data1.to_bytes(8, 'little') + data2.to_bytes(8, 'little') + data3.to_bytes(3, 'little')
enc = bytearray(enc)
xor_keylist_idx = 0
xor_key_idx = 0
flag = ""
for i in range(len(enc)):
if (i & 0xf) == 0:
xor_key = xor_key_list[xor_keylist_idx]
xor_keylist_idx += 1
xor_key_idx = 0
flag += chr(xor_key[xor_key_idx] ^ enc[i])
xor_key_idx += 1
print("flag: ", flag)
輸出：flag: flag{md5(uid+2025)}
第六題：Windows & Android高級題
Java層分析
先看看題目描述，要幾個重點：
- flag格式為
flag{XXXXX-XXXXX-XXXXX-XXXXX}，其中X要麼是大寫字母，要麼是數字。
- 不同UID對應不同的Flag，可能有多個解。
- SISC中的S意為堆棧。
再看看APP，要求輸入UID和Flag。
用新版jeb查看Java層邏輯( Java層有混淆，jeb能忽略部份混淆，方便分析 )，發現調用check函數來檢查，參考分別是UID和Flag。
check是Native函數。
vm初始化
native層的check是靜態注冊的，能直接搜到。
繼續深入分析( 配合動調來遂一分析每個函數的作用 )。
init_some_data函數如下，結合後面的分析可以知道，這裡是在初始化vm虛擬機的opcodes，存放在a1[0xC000 ~ 0xC200]。
將a1記為vm_ctx，意指vm虛擬機的上下文空間。
start_vm
初始化完成後便會調用start_vm正式啟動虛擬機進行計算。
一开始会通过一些运算获取_opcode和arg，前者是操作碼、後者是一些固定的參數( 在不同的操作碼中都有不同的含義 )。
接著就是vm最經典的一大段switch，每個case對應不同的handler，實現了不同的功能。
每個handler裡基本上都會用到vm_ctx[0x10002]，一些參數、中間值、計算結果都會存放在vm_ctx[0x10002]指向的位置。
而且可以看到vm_ctx[0x10002] + 4、vm_ctx[0x10002] - 4等等的运算，再结合題目的描述，可以猜測vm_ctx[0x10002]相當於sp( 棧指針 )，該虛擬機的运算操作都会在它自己维护的栈中进行( 沒有寄存器的概念 )。
vm handler分析與還原
大部份handler的實現都比較簡單，配合動調很容易就可以分析出來。
記錄幾個沒那麼容易看出來的handler。
handler7：&v26[-arg]相當於&v26 - arg，這裡是在將棧頂元素與棧頂後arg個元素交換。
handler22：注意_pc += (char)arg，對應匯編是ADD W11, W11, W12,SXTB，其中SXTB是對W12的修飾符，表示將W12的最低8位進行符號擴展，在還原handler時要特別留意這一點。
花億點時間，還原所有handler，實現一個簡單的vm解釋器：
def write_mem_str(addr, content):
global vm_ctx
if type(content) == str:
for i in range(len(content)):
vm_ctx[addr + i] = ord(content[i])
else:
raise Exception("TODO")
return addr
def write_mem_word(addr, content):
global vm_ctx
for i in range(2):
vm_ctx[addr + i] = content & 0xFF
content >>= 8
def write_mem_arr(addr, arr):
global vm_ctx
for i in range(len(arr)):
vm_ctx[addr + i] = arr[i]
def write_mem_dword(addr, content):
global vm_ctx
for i in range(4):
vm_ctx[addr + i] = content & 0xFF
content >>= 8
def read_mem_dword(addr):
global vm_ctx
return vm_ctx[addr] | (vm_ctx[addr + 1] << 8) | (vm_ctx[addr + 2] << 16) | (vm_ctx[addr + 3] << 24)
def read_mem_word(addr):
global vm_ctx
return vm_ctx[addr] | (vm_ctx[addr + 1] << 8)
def read_mem_byte(addr):
global vm_ctx
return vm_ctx[addr]
def push_data(data):
global vm_ctx
sp = read_mem_word(0x10002)
tmp = sp + 4
write_mem_word(0x10002, tmp)
write_mem_dword(tmp, data)
def pop_data():
global vm_ctx
sp = read_mem_word(0x10002)
data = read_mem_dword(sp)
write_mem_word(0x10002, sp - 4)
return data
def read_sp_data():
sp = read_mem_word(0x10002)
data = read_mem_dword(sp)
return data
def set_sp_data(data):
sp = read_mem_word(0x10002)
write_mem_dword(sp, data)
def load_opcodes():
global vm_ctx
with open("./dump/opcodes", mode = "rb") as f:
opcodes = bytearray(f.read())
for i in range(len(opcodes)):
vm_ctx[0xC000 + i] = opcodes[i]
def hex_to_negative(value, bits = 8):
if value & (1 << (bits - 1)):
value = value - (1 << bits)
return value
def start_vm():
global vm_ctx, pc, arg, v13
pc = None
arg = None
v13 = None
def handler_0_xor():
n1 = pop_data()
n2 = read_sp_data()
res = n1 ^ n2
set_sp_data(res)
print(f"[h0_xor]\t pop, *sp = {hex(n2)} ^ {hex(n1)} = {hex(res)}")
def handler_1_opposite():
n = read_sp_data()
set_sp_data(-n)
print(f"[h1_opposite]\t *sp = -{hex(n)}")
def handler_2_subsp():
sp = read_mem_word(0x10002)
write_mem_word(0x10002, sp - 4 * arg)
print(f"[h2_subsp]\t sp -= {4 * arg}")
def handler_4_orr():
n1 = pop_data()
n2 = read_sp_data()
res = n1 | n2
set_sp_data(res)
print(f"[h4_orr]\t pop, *sp = {hex(n2)} | {hex(n1)} = {hex(res)}")
def handler_5_():
global pc
sp = read_mem_word(0x10002)
v23 = read_sp_data()
v24 = sp - 8 - 4 * arg + 4
pc = read_mem_dword(sp - 4)
write_mem_word(0x10002, v24)
write_mem_dword(v24, v23)
print(f"[h5_]\t sp = {hex(v24)}, [{hex(v24)}] = {hex(v23)}, pc = {hex(pc)}")
def handler_6_noeq():
n1 = pop_data()
n2 = read_sp_data()
res = n1 != n2
set_sp_data(res)
print(f"[h6_noeq]\t pop, *sp = {hex(n2)} != {hex(n1)} = {hex(res)}")
def handler_7_swap():
global arg
sp = read_mem_word(0x10002)
n1 = read_mem_dword(sp)
n2 = read_mem_dword(sp - 4 * arg)
write_mem_dword(sp, n2)
write_mem_dword(sp - 4 * arg, n1)
print(f"[h7_swap]\t swap(sp, sp - {arg}) -> swap({hex(n1), hex(n2)})")
def handler_8_and():
n1 = pop_data()
n2 = read_sp_data()
res = n1 & n2
set_sp_data(res)
print(f"[h8_and]\t pop, *sp = {hex(n2)} & {hex(n1)} = {hex(res)}")
def handler_9_lsl():
sp_data = read_sp_data()
set_sp_data(sp_data << arg)
print(f"[h9_lsl]\t *sp = *sp << arg = {hex(sp_data)} << {arg} = {hex(sp_data << arg)}")
def handler_10_not():
sp_data = read_sp_data()
set_sp_data(~sp_data)
print(f"[h10_not]\t *sp = ~(*sp) = ~{hex(sp_data)} = {hex(~sp_data & 0xffffffff)}")
def handler_12_add():
n1 = pop_data()
n2 = read_sp_data()
res = n1 + n2
set_sp_data(res)
print(f"[h12_add]\t pop, *sp = {hex(n2)} + {hex(n1)} = {hex(res)}")
def handler_14_():
global pc
pc += hex_to_negative(arg)
print(f"[h14_]\t pc += {hex_to_negative(arg)}")
def handler_15_():
write_mem_word(0x10004, 257)
print("[h15_]\t write_mem_word(0x10004, 257)")
def handler_17_lsr():
sp_data = read_sp_data()
set_sp_data(sp_data >> arg)
print(f"[h17_lsr]\t *sp = *sp >> arg = {hex(sp_data)} >> {arg} = {hex(sp_data >> arg)}")
def handler_18_mod():
n1 = pop_data()
n2 = read_sp_data()
res = n2 % n1
set_sp_data(res)
print(f"[h18_mod]\t pop, *sp = {hex(n2)} % {hex(n1)} = {hex(res)}")
def handler_20_dword2byte():
sp = read_mem_word(0x10002)
sp_data = read_mem_byte(sp)
set_sp_data(sp_data)
print(f"[h20_dword2byte]\t *(dword*)sp = *(byte*)sp = {hex(sp_data)}")
def handler_21_mul():
n1 = pop_data()
n2 = read_sp_data()
res = n1 * n2
set_sp_data(res)
print(f"[h21_mul]\t pop, *sp = {hex(n2)} * {hex(n1)} = {hex(res)}")
def handler_22_pushpc():
global pc
sp = read_mem_word(0x10002)
pc_ = pc
pc += hex_to_negative(arg)
v34 = sp + 4
write_mem_word(0x10002, v34)
write_mem_dword(v34, pc_)
print(f"[h22_pushpc]\t push(pc) -> push({hex(pc_)}), pc += {hex_to_negative(arg)}")
def handler_23_eq():
global pc
sp = read_mem_word(0x10002)
v16 = sp - 4
v15 = sp - 8
n1 = read_mem_dword(sp)
n2 = read_mem_dword(sp - 4)
write_mem_word(0x10002, v15)
if (v13 == 25) == (n1 == n2):
print(f"[h23_eq]\t sp = sp - 8")
return
if arg & 0xFFFFFF00 != 0:
raise Exception("TODO")
pc += hex_to_negative(arg)
print(f"[h23_eq]\t sp = sp - 8, pc += {hex_to_negative(arg)} ({hex(arg)})")
def handler_26_getinput():
n1 = pop_data()
n2 = read_sp_data()
res = read_mem_byte(n1 + n2)
set_sp_data(res)
print(f"[h26_getinput]\t pop, *sp = vm_ctx[{hex(n2)} + {hex(n1)}] = {hex(res)}")
def handler_27_pusharg():
global arg
sp = read_mem_word(0x10002)
orig_arg = arg
arg = read_mem_dword(sp - 4 * arg)
push_data(arg)
print(f"[h27_pusharg]\t push({hex(arg)}) arg == [sp - 4 * {orig_arg}]")
def handler_29_pusharg2():
push_data(arg)
print(f"[h29_pusharg2]\t push({hex(arg)})")
def handler_30_sub1():
sp_data = read_sp_data()
set_sp_data(sp_data - 1)
print(f"[h30_sub1]\t *sp = *sp - 1 = {hex(sp_data)} - 1 = {hex(sp_data - 1)}")
pc = read_mem_word(0x10000)
while True:
pc_1 = pc + 1
cur_opcode = read_mem_byte(pc)
arg = cur_opcode & 7
if arg != 7:
pc += 1
v13 = cur_opcode >> 3
_opcode = v13 - 1
else:
pc += 2
arg = read_mem_byte(pc_1)
v13 = cur_opcode >> 3
_opcode = v13 - 1
if v13 - 1 > 0x1E:
raise Exception("TODO")
break
if _opcode == 0:
handler_0_xor()
elif _opcode == 1:
handler_1_opposite()
elif _opcode == 2:
handler_2_subsp()
elif _opcode == 3 or _opcode == 25:
continue
elif _opcode == 4:
handler_4_orr()
elif _opcode == 5:
handler_5_()
elif _opcode == 6:
handler_6_noeq()
elif _opcode == 7:
handler_7_swap()
elif _opcode == 8:
handler_8_and()
elif _opcode == 9:
handler_9_lsl()
elif _opcode == 10:
handler_10_not()
elif _opcode == 12:
handler_12_add()
elif _opcode == 14:
handler_14_()
elif _opcode == 15:
handler_15_()
break
elif _opcode == 17:
handler_17_lsr()
elif _opcode == 18:
handler_18_mod()
elif _opcode == 20:
handler_20_dword2byte()
elif _opcode == 21:
handler_21_mul()
elif _opcode == 22:
handler_22_pushpc()
elif _opcode == 23 or _opcode == 24:
handler_23_eq()
elif _opcode == 26:
handler_26_getinput()
elif _opcode == 27:
handler_27_pusharg()
elif _opcode == 29:
handler_29_pusharg2()
elif _opcode == 30:
handler_30_sub1()
else:
print("else _opcode: ", _opcode)
raise Exception("TODO")
break
write_mem_word(0x10000, pc)
res = read_sp_data()
return res
vm_ctx = [0] * 0x10006
load_opcodes()
write_mem_dword(0x10000, 0x8000C000)
write_mem_word(0x10004, 0)
write_mem_arr(0x204 * 0x10, [0x00, 0x03, 0x0F, 0x20, 0x0D, 0x02, 0x23, 0x06, 0x1B, 0x14,0x0E, 0x01, 0x16, 0x19, 0x08, 0x12])
write_mem_arr(0x205 * 0x10, [0x1F, 0x17, 0x24, 0x0B, 0x1E, 0x07, 0x1A, 0x05, 0x18, 0x1D, 0x22, 0x00, 0x00, 0x00, 0x00, 0x00])
write_mem_arr(0x203 * 0x10, [0x09, 0x0A, 0x10, 0x15, 0x21, 0x13, 0x0C, 0x04, 0x11, 0x1C, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00])
write_mem_str(0x1000, "flag{44444-44444-44444-44444}")
push_data(1898208)
push_data(0x1000)
push_data(0x2000)
res = start_vm()
print("[res]: ", hex(res))
提醒：flag格式為flag{XXXXX-XXXXX-XXXXX-XXXXX}，其中X要麼是大寫字母，要麼是數字。
腳本中的測試flag要記得符合這個格式，腳本的輸出日志記為vm.log。
加密邏輯分析
前置：在動調的過程中發現handler26會獲取輸入的Flag，加密邏輯大概會在那附近。
在vm.log中搜h26_getinput定位到相關位置，首先判斷了input是否flag{ }的格式。
[h26_getinput] pop, *sp = vm_ctx[0x1000 + 0x0] = 0x66
[h0_xor] pop, *sp = 0x66 ^ 0x66 = 0x0
[h4_orr] pop, *sp = 0x0 | 0x0 = 0x0
[h29_pusharg2] push(0x6c)
[h27_pusharg] push(0x1000) arg == [sp - 4 * 3]
[h29_pusharg2] push(0x1)
[h26_getinput] pop, *sp = vm_ctx[0x1000 + 0x1] = 0x6c
[h0_xor] pop, *sp = 0x6c ^ 0x6c = 0x0
[h4_orr] pop, *sp = 0x0 | 0x0 = 0x0
[h29_pusharg2] push(0x61)
[h27_pusharg] push(0x1000) arg == [sp - 4 * 3]
[h29_pusharg2] push(0x2)
[h26_getinput] pop, *sp = vm_ctx[0x1000 + 0x2] = 0x61
[h0_xor] pop, *sp = 0x61 ^ 0x61 = 0x0
[h4_orr] pop, *sp = 0x0 | 0x0 = 0x0
[h29_pusharg2] push(0x67)
[h27_pusharg] push(0x1000) arg == [sp - 4 * 3]
[h29_pusharg2] push(0x3)
[h26_getinput] pop, *sp = vm_ctx[0x1000 + 0x3] = 0x67
[h0_xor] pop, *sp = 0x67 ^ 0x67 = 0x0
[h4_orr] pop, *sp = 0x0 | 0x0 = 0x0
[h29_pusharg2] push(0x7b)
[h27_pusharg] push(0x1000) arg == [sp - 4 * 3]
[h29_pusharg2] push(0x4)
[h26_getinput] pop, *sp = vm_ctx[0x1000 + 0x4] = 0x7b
[h0_xor] pop, *sp = 0x7b ^ 0x7b = 0x0
[h4_orr] pop, *sp = 0x0 | 0x0 = 0x0
[h29_pusharg2] push(0x7d)
[h27_pusharg] push(0x1000) arg == [sp - 4 * 3]
[h29_pusharg2] push(0x1c)
[h26_getinput] pop, *sp = vm_ctx[0x1000 + 0x1c] = 0x7d
[h0_xor] pop, *sp = 0x7d ^ 0x7d = 0x0
從input[5]開始才是真正的內容，對input[5~8]的运算可以总结为：查表、自減、乘0x24。
[h26_getinput] pop, *sp = vm_ctx[0x1005 + 0x0] = 0x34
[h26_getinput] pop, *sp = vm_ctx[0x2000 + 0x34] = 0x21
[h27_pusharg] push(0x21) arg == [sp - 4 * 0]
[h29_pusharg2] push(0x0)
[h23_eq] sp = sp - 8
[h12_add] pop, *sp = 0x0 + 0x21 = 0x21
[h30_sub1] *sp = *sp - 1 = 0x21 - 1 = 0x20
[h7_swap] swap(sp, sp - 1) -> swap(('0x20', '0x0'))
[h29_pusharg2] push(0x1)
[h12_add] pop, *sp = 0x0 + 0x1 = 0x1
[h14_] pc += -24
[h27_pusharg] push(0x1) arg == [sp - 4 * 0]
[h27_pusharg] push(0x5) arg == [sp - 4 * 6]
[h23_eq] sp = sp - 8
[h7_swap] swap(sp, sp - 1) -> swap(('0x1', '0x20'))
[h29_pusharg2] push(0x24)
[h21_mul] pop, *sp = 0x20 * 0x24 = 0x480
[h26_getinput] pop, *sp = vm_ctx[0x1005 + 0x1] = 0x34
[h26_getinput] pop, *sp = vm_ctx[0x2000 + 0x34] = 0x21
[h27_pusharg] push(0x21) arg == [sp - 4 * 0]
[h29_pusharg2] push(0x0)
[h23_eq] sp = sp - 8
[h12_add] pop, *sp = 0x480 + 0x21 = 0x4a1
[h30_sub1] *sp = *sp - 1 = 0x4a1 - 1 = 0x4a0
對input[9]有特別的處理，查表、自減操作仍舊保留，不同的是後面會判斷tmp >> 25是否不為0，若是則進行自加、取餘操作。
取餘操作中的模數，會根據輸入的UID不同而變化，即固定UID對應固定的模數。
( 注：以-分隔的每組字串的最個一個元素都是這樣處理的 )
[h26_getinput] pop, *sp = vm_ctx[0x1005 + 0x4] = 0x34
[h26_getinput] pop, *sp = vm_ctx[0x2000 + 0x34] = 0x21
[h12_add] pop, *sp = 0x34b8e80 + 0x21 = 0x34b8ea1
[h30_sub1] *sp = *sp - 1 = 0x34b8ea1 - 1 = 0x34b8ea0
[h17_lsr] *sp = *sp >> arg = 0x34b8ea0 >> 25 = 0x1
[h12_add] pop, *sp = 0x34b8ea0 + 0x1 = 0x34b8ea1
[h18_mod] pop, *sp = 0x34b8ea1 % 0xb05f17 = 0x8a1245
以-作為分隔符，每組處理完後會以|來融合。
[h4_orr] pop, *sp = 0x1fc3d5 | 0x8a1245 = 0x9fd3d5
最後會自減、異或0xc15303fb，這個值是固定的。
[h30_sub1] *sp = *sp - 1 = 0x19fffff - 1 = 0x19ffffe
[h0_xor] pop, *sp = 0x19ffffe ^ 0xc15303fb = 0xc0ccfc05
[h5_] sp = 0x8014, [0x8014] = 0xc0ccfc05, pc = 0xc088
[h15_] write_mem_word(0x10004, 257)
[res]: 0xc0ccfc05
綜合上述分析，可以大概用Python還原出加密邏輯：
tables = [0x09, 0x0A, 0x10, 0x15, 0x21, 0x13, 0x0C, 0x04, 0x11, 0x1C, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x03, 0x0F, 0x20, 0x0D, 0x02, 0x23, 0x06, 0x1B, 0x14,0x0E, 0x01, 0x16, 0x19, 0x08, 0x12, 0x1F, 0x17, 0x24, 0x0B, 0x1E, 0x07, 0x1A, 0x05, 0x18, 0x1D, 0x22, 0x00, 0x00, 0x00, 0x00, 0x00]
mod_arr = [0xa91f91, 0xb66962, 0xf19ad9, 0xef305d]
def encrypt(input):
length = len(input)
res = 0
tmp = 0
i = 0
mi = 0
while True:
if i >= length:
res |= tmp
print("tmp res: ", hex(res))
break
if input[i] == '-':
res |= tmp
print("tmp res: ", hex(res))
tmp = 0
i += 1
continue
table_idx = ord(input[i]) - 0x30
if table_idx < 0 or table_idx >= 0x30:
sep = input.find("-", i)
if sep == -1:
break
i = sep
tmp |= 1
continue
tmp += tables[table_idx]
tmp -= 1
if i + 1 < length and input[i + 1] != '-':
tmp *= 0x24
else:
if (tmp >> 25) != 0:
tmp += 1
tmp %= mod_arr[mi]
mi += 1
print("tmp: ", hex(tmp))
i += 1
res -= 1
res ^= 0xc15303fb
print(res)
print("res: ", hex(res))
encrypt("44444-RRRRR-RRRRR-RRRRR")
最終解密
基於上述加密腳本，似乎無法直接反推出對應的解密邏輯，而且題目描述中提到有多個解也認證了這一點。
密文是0x3EACFC04，(0x3EACFC04 ^ 0xc15303fb) == 0xFFFFFFFF，而-1的16進制正是該值，因此只要在最終的自減前，res的值為0，即可滿足等式。
上面提到，以-分隔的每個字串的最個一個元素都會進行取餘的操作( 前提是>>25不為0 )，這一步就可以很方便讓tmp歸0。
以-分隔的每組數據計算過程如下，現在的目標是讓tmp等於0，因此d + input_[4]必須是target的整數倍。
此時問題轉化為如何讓d + input_[4] == n * target，其中n、target都是已知的。
( 注：input_[i]指input[i]查表後的結果、target是每組的模數 )
a = (input_[0] - 1) * 0x24
b = (a + input_[1] - 1) * 0x24
c = (b + input_[2] - 1) * 0x24
d = (c + input_[3] - 1) * 0x24
tmp = (d + input_[4]) % target
以下腳本用來求input_[0 ~ 4]這幾個未知量( 初始為0 )，原理如下：
- 先爆破
input_[0]，若input_[0]為i會使func函數返回值>0且input_[0]為i+1會使func函數返回值<0，則代表i就是input_[0]的最大值，也是input_[0]其中一個可能的值。
- 確定了
input_[0]後，用同樣方法確定input_[1 ~ 4]。
- 最終可以確定
input_[0 ~ 4]，由此反查tables來確定input[0 ~ 4]字符串。
注：當input_[j]被確定為0時，是不合理的，要將input_[j - 1] -= 1，然後再重新計算input_[j]的最大值。
tables = [0x09, 0x0A, 0x10, 0x15, 0x21, 0x13, 0x0C, 0x04, 0x11, 0x1C, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x03, 0x0F, 0x20, 0x0D, 0x02, 0x23, 0x06, 0x1B, 0x14,0x0E, 0x01, 0x16, 0x19, 0x08, 0x12, 0x1F, 0x17, 0x24, 0x0B, 0x1E, 0x07, 0x1A, 0x05, 0x18, 0x1D, 0x22, 0x00, 0x00, 0x00, 0x00, 0x00]
def func(target, input_):
a = (input_[0] - 1) * 0x24
b = (a + input_[1] - 1) * 0x24
c = (b + input_[2] - 1) * 0x24
d = (c + input_[3] - 1) * 0x24
res = target - d
return res
def func2(target):
input_ = [0] * 4
res = []
for j in range(4):
for i in range(0x25):
input_[j] = i
a = func(target, input_)
input_[j] = i + 1
b = func(target, input_)
if a > 0 and b < 0:
if i == 0:
input_[j - 1] -= 1
res[len(res) - 1] = chr(0x30 + tables.index(input_[j - 1]))
continue
input_[j] = i
t = tables.index(i)
if t == -1:
raise Exception("??")
res.append(chr(0x30 + t))
break
res.append(chr(tables.index(func(target, input_)) + 0x30))
return "".join(res)
print(func2(0x2A47E44) + '-' + func2(0x2d9a588) + '-' + func2(0x2D4D08B) + '-' + func2(0x2CD9117))
得到一個可行的Flag為HB0P6-Y84V7-YSWDH-9RZPB：
第八題：Android高級題
直接hook RegisterNatives，看到flag驗證邏輯在lib52pojie.so!0x134d4。
[RegisterNatives] java_class: com.wuaipojie.crackme2025.MainActivity name: checkSn sig: (Ljava/lang/String;)Z
fnPtr: 0x7ebed554d4 fnOffset: 0x7ebed554d4 lib52pojie.so!0x134d4 callee: 0x7ebed553d8 lib52pojie.so!0x133d8
看到一堆~、^、|操作，但其實它們並非加密邏輯，而是類似ollvm裡的「指令替換」混淆，也叫MBA表達式。
簡單來說就是將一段很簡單的指令( 如a + b )，通過疊加~、^、|等操作符轉換成完全等價的複雜指令。
由於沒有解混淆的思路，因此只能直接動調慢慢看邏輯。
調用get_input_8取了input的一部份，然後傳入encrypt。
encrypt中主要分成3部份，先看encrypt_part1。
encrypt_part1
input.n128_u64[0]是低64位，代表傳入的flag，input.n128_u64[1]是高64位，用來存放結果。
只看與input有關的，hook發現input.n128_u64[0]每輪固定左移-1，即右移1。
由此得出input.n128_u64[0]的迭代方式：input = (input >> 1) & (2 ** 64 - 1)
input.n128_u64[1]只與tmp1有關。
frida stalker打印tmp1、input_1.n128_u64[1]+=的那個值，發現要將tmp1看成2進制位，每輪都會拼到input_1.n128_u64[1]的低位。
即input1 = (input1 << 1) | tmp1，而tmp1其實就是取input.n128_u64[0]的最低位。
[2] x26: 0x1 x27: 0x3332317b67616c66
[3] x8(tmp1): 0x0
[5] x8: 0x0
0 0 0 0 0 0 0 0
33 b6 b0 b3 bd 18 99 19
[2] x26: 0x1 x27: 0x199918bdb3b0b633
[3] x8(tmp1): 0x1
[5] x8: 0x1
0 0 0 0 0 0 0 0
19 5b d8 d9 5e 8c cc c
[2] x26: 0x1 x27: 0xccc8c5ed9d85b19
[3] x8(tmp1): 0x1
[5] x8: 0x3
1 0 0 0 0 0 0 0
8c 2d ec 6c 2f 46 66 6
[2] x26: 0x1 x27: 0x666462f6cec2d8c
[3] x8(tmp1): 0x0
[5] x8: 0x6
3 0 0 0 0 0 0 0
c6 16 76 b6 17 23 33 3
最終encrypt_part1可以簡化為：
def encrypt_part1(input):
input1 = 0
v18 = 1
for i in range(0x40):
tmp1 = input & v18
input = (input >> 1) & (2 ** 64 - 1)
input1 = (input1 << 1) | tmp1
print(tmp1)
return input1
encrypt_part2
encrypt_part2的邏輯比encrypt_part1複雜得多，繼續像上面那樣分析實在不太理智( 有心無力 )，本來都打算放棄了，結果當天晚上吾愛放出了提示：
2025.02.10 16:45 【春节】解题领红包之八 {Android 高级题} 对称算法，需要识别出算法类型，找出初始化后的密钥后反推即可，对应获取奖励也减半
對稱算法，結合分析過程中看到的一些表，嘗試直接搜看看表中的數據。
發現其實是DES算法。
而且根據提示，密鑰是初始化過的。
hook encrypt，打印args[0]，發現每個QWORD剛好都是6字節大小的數據，而DES算法的round key也是48位，因此這大概率就是提示所述的初始化過的密鑰。
算法分析
DES算法：https://blog.csdn.net/nicai_hualuo/article/details/123135670
基於原版DES，遂步分析，還原到最後發現其實是3DES。完整腳本如下：( 腳本是基於於上述文章改的 )
IP = [0x3A, 0x32, 0x2A, 0x22, 0x1A, 0x12, 0x0A, 0x02, 0x3C, 0x34,
0x2C, 0x24, 0x1C, 0x14, 0x0C, 0x04, 0x3E, 0x36, 0x2E, 0x26,
0x1E, 0x16, 0x0E, 0x06, 0x40, 0x38, 0x30, 0x28, 0x20, 0x18,
0x10, 0x08, 0x39, 0x31, 0x29, 0x21, 0x19, 0x11, 0x09, 0x01,
0x3B, 0x33, 0x2B, 0x23, 0x1B, 0x13, 0x0B, 0x03, 0x3D, 0x35,
0x2D, 0x25, 0x1D, 0x15, 0x0D, 0x05, 0x3F, 0x37, 0x2F, 0x27,
0x1F, 0x17, 0x0F, 0x07]
E = [32, 1, 2, 3, 4, 5,
4, 5, 6, 7, 8, 9,
8, 9, 10, 11, 12, 13,
12, 13, 14, 15, 16, 17,
16, 17, 18, 19, 20, 21,
20, 21, 22, 23, 24, 25,
24, 25, 26, 27, 28, 29,
28, 29, 30, 31, 32, 1]
P = [16, 7, 20, 21,
29, 12, 28, 17,
1, 15, 23, 26,
5, 18, 31, 10,
2, 8, 24, 14,
32, 27, 3, 9,
19, 13, 30, 6,
22, 11, 4, 25]
IPR = [40, 8, 48, 16, 56, 24, 64, 32,
39, 7, 47, 15, 55, 23, 63, 31,
38, 6, 46, 14, 54, 22, 62, 30,
37, 5, 45, 13, 53, 21, 61, 29,
36, 4, 44, 12, 52, 20, 60, 28,
35, 3, 43, 11, 51, 19, 59, 27,
34, 2, 42, 10, 50, 18, 58, 26,
33, 1, 41, 9, 49, 17, 57, 25]
SBOX = [
[
[14, 4, 13, 1, 2, 15, 11, 8, 3, 10, 6, 12, 5, 9, 0, 7],
[0, 15, 7, 4, 14, 2, 13, 1, 10, 6, 12, 11, 9, 5, 3, 8],
[4, 1, 14, 8, 13, 6, 2, 11, 15, 12, 9, 7, 3, 10, 5, 0],
[15, 12, 8, 2, 4, 9, 1, 7, 5, 11, 3, 14, 10, 0, 6, 13]
],
[
[15, 1, 8, 14, 6, 11, 3, 4, 9, 7, 2, 13, 12, 0, 5, 10],
[3, 13, 4, 7, 15, 2, 8, 14, 12, 0, 1, 10, 6, 9, 11, 5],
[0, 14, 7, 11, 10, 4, 13, 1, 5, 8, 12, 6, 9, 3, 2, 15],
[13, 8, 10, 1, 3, 15, 4, 2, 11, 6, 7, 12, 0, 5, 14, 9]
],
[
[10, 0, 9, 14, 6, 3, 15, 5, 1, 13, 12, 7, 11, 4, 2, 8],
[13, 7, 0, 9, 3, 4, 6, 10, 2, 8, 5, 14, 12, 11, 15, 1],
[13, 6, 4, 9, 8, 15, 3, 0, 11, 1, 2, 12, 5, 10, 14, 7],
[1, 10, 13, 0, 6, 9, 8, 7, 4, 15, 14, 3, 11, 5, 2, 12]
],
[
[7, 13, 14, 3, 0, 6, 9, 10, 1, 2, 8, 5, 11, 12, 4, 15],
[13, 8, 11, 5, 6, 15, 0, 3, 4, 7, 2, 12, 1, 10, 14, 9],
[10, 6, 9, 0, 12, 11, 7, 13, 15, 1, 3, 14, 5, 2, 8, 4],
[3, 15, 0, 6, 10, 1, 13, 8, 9, 4, 5, 11, 12, 7, 2, 14]
],
[
[2, 12, 4, 1, 7, 10, 11, 6, 8, 5, 3, 15, 13, 0, 14, 9],
[14, 11, 2, 12, 4, 7, 13, 1, 5, 0, 15, 10, 3, 9, 8, 6],
[4, 2, 1, 11, 10, 13, 7, 8, 15, 9, 12, 5, 6, 3, 0, 14],
[11, 8, 12, 7, 1, 14, 2, 13, 6, 15, 0, 9, 10, 4, 5, 3]
],
[
[12, 1, 10, 15, 9, 2, 6, 8, 0, 13, 3, 4, 14, 7, 5, 11],
[10, 15, 4, 2, 7, 12, 9, 5, 6, 1, 13, 14, 0, 11, 3, 8],
[9, 14, 15, 5, 2, 8, 12, 3, 7, 0, 4, 10, 1, 13, 11, 6],
[4, 3, 2, 12, 9, 5, 15, 10, 11, 14, 1, 7, 6, 0, 8, 13]
],
[
[4, 11, 2, 14, 15, 0, 8, 13, 3, 12, 9, 7, 5, 10, 6, 1],
[13, 0, 11, 7, 4, 9, 1, 10, 14, 3, 5, 12, 2, 15, 8, 6],
[1, 4, 11, 13, 12, 3, 7, 14, 10, 15, 6, 8, 0, 5, 9, 2],
[6, 11, 13, 8, 1, 4, 10, 7, 9, 5, 0, 15, 14, 2, 3, 12]
],
[
[13, 2, 8, 4, 6, 15, 11, 1, 10, 9, 3, 14, 5, 0, 12, 7],
[1, 15, 13, 8, 10, 3, 7, 4, 12, 5, 6, 11, 0, 14, 9, 2],
[7, 11, 4, 1, 9, 12, 14, 2, 0, 6, 10, 13, 15, 3, 5, 8],
[2, 1, 14, 7, 4, 10, 8, 13, 15, 12, 9, 0, 3, 5, 6, 11]
]
]
round_keys = [222483666014355, 34094049895368, 188087828272899, 30798344234022, 20170121439688, 154109401428571, 143409192342562, 80501118078826, 126994336798112, 150086336645229, 197956095638172, 182733681792953, 11125921617955, 224782889413428, 7453516311004, 200667612718677, 114350607846426, 27979304443188, 145503975706288, 90448879766041, 10827630596124, 1245263770020, 194907790650021, 89110378318487, 38106705338630, 149997549266822, 105755390509763, 75540444135499, 215007439009096, 119720110503264, 55615706156578, 143051418949992, 222483666014355, 34094049895368, 188087828272899, 30798344234022, 20170121439688, 154109401428571, 143409192342562, 80501118078826, 126994336798112, 150086336645229, 197956095638172, 182733681792953, 11125921617955, 224782889413428, 7453516311004, 200667612718677]
def dec2binary(dec):
res = bin(dec)[2:]
length = len(res)
if length < 4:
r = 4 - length
else:
r = length - 4 * (length // 4)
for i in range(r):
res = '0' + res
return res
def hex_to_binary_str(hex_val, n):
def byte2binary(val):
ret = "{:08b}".format(val)
for i in range(8 - len(ret)):
ret = '0' + ret
return ret
res = ""
arr = bytearray(int.to_bytes(hex_val, n, 'little'))
for i in range(n):
res += byte2binary(arr[i])
return res
def binary_str_to_hex(bin_str):
return hex(int(bin_str, 2))[2:]
def IPExchange(input):
res = ""
for i in range(64):
res += input[IP[i] - 1]
return res
def XOR(a, b):
if len(a) != len(b):
raise Exception("something wrong")
res = ""
for i in range(len(a)):
if a[i] == b[i]:
res += '0'
else:
res += '1'
return res
def EExchange(right):
res = ""
for i in range(48):
res += right[E[i] - 1]
return res
def SExchange(input):
res = ""
for i in range(0, 48, 6):
row = int(input[i]) * 2 + int(input[i + 5])
col = int(input[i + 1]) * 8 + int(input[i + 2]) * 4 + int(input[i + 3]) * 2 + int(input[i + 4])
res += dec2binary(SBOX[i // 6][row][col])
return res
def PExchange(input):
res = ""
for i in range(32):
res += input[P[i] - 1]
return res
def IPRExchange(input):
res = ""
for i in range(64):
res += input[IPR[i] - 1]
return res
def F(right, rk):
tmp = EExchange(right)
tmp = XOR(tmp, rk)
res = SExchange(tmp)
res = PExchange(res)
return res
def des_encrypt(input, key_start, mode):
tmp = IPExchange(input)
left = tmp[0:32]
right = tmp[32:]
for i in range(16):
middle = right
if mode == 0:
right = XOR(left, F(right, round_keys[key_start + i]))
else:
right = XOR(left, F(right, round_keys[key_start + 0xf - i]))
left = middle
cipher = right + left
res = IPRExchange(cipher)
return res
def convert(input):
res = ""
for i in range(0x40):
res += str((input & 1))
input >>= 1
return res
def convert_re(input):
res = ""
for i in range(0x40):
res = input[i] + res
return res
def convert2(input, bit):
res = ""
for i in range(bit):
res = str((input & 1)) + res
input >>= 1
return res
def convert_round_keys():
for i in range(len(round_keys)):
round_keys[i] = convert2(round_keys[i], 48)
def encrypt(input):
input = convert(input)
enc = des_encrypt(input, 0, 0)
enc = des_encrypt(enc, 16, 1)
enc = des_encrypt(enc, 32, 0)
print("enc: ", binary_str_to_hex(enc))
return enc
def decrypt(input):
input = convert2(input, 64)
enc = des_encrypt(input, 0, 1)
enc = des_encrypt(enc, 16, 0)
enc = des_encrypt(enc, 32, 1)
res = convert_re(enc)
return binary_str_to_hex(res)
def to_flag(input):
res = ""
for i in range(0, len(input), 2):
ch = chr(int(input[i: i + 2], 16))
res = ch + res
return res
if __name__ == "__main__":
convert_round_keys()
enc_data = [0x7C1A8B2E957A3115, 0x4B43E13562FC5DE6, 0x8346103AE93F945D]
flag = ""
for e in enc_data:
t = decrypt(e)
flag += to_flag(t)
print("flag: ", flag)
輸出flag：
flag: 52PojiEHaPpynEwY3ar2025!
发表于 2025-2-13 21:09
|
发表于 2025-2-13 17:51
|楼主
发表于 2025-2-13 11:09
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有用
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而且輪密鑰應該反推不了主密鑰吧
