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香山杯2023初赛RE做题记录.md

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香山杯2023初赛个人做题记录
2023-10-15 14:08:36 -0700
解题报告
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香山杯2023初赛个人做题记录

签到

basecrack --magic一把梭，凯撒解码

URL从哪儿来

丢ida，发现在windows的temp文件夹中创建文件，提取ou_XXXX.tmp改为ou.exe，丢ida，找到szUrl为http://how.did.i.decrypted?id=，发现v3->base64->作为密钥rc4加密得url，但是动调发现密钥被篡改，怀疑原密钥为flag，exp:

import base64
v3=[0 for i in range(56)]
v3[0]=120
v3[1]=139
v3[2]=150
v3[3]=134
v3[4]=120
v3[5]=81
v3[6]=145
v3[7]=80
v3[8]=108
v3[9]=98
v3[10]=119
v3[11]=83
v3[12]=108
v3[13]=136
v3[14]=99
v3[15]=80
v3[16]=120
v3[17]=113
v3[18]=78
v3[19]=80
v3[20]=107
v3[21]=152
v3[22]=119
v3[23]=83
v3[24]=106
v3[25]=114
v3[26]=119
v3[27]=151
v3[28]=108
v3[29]=139
v3[30]=119
v3[31]=146
v3[32]=108
v3[33]=152
v3[34]=99
v3[35]=80
v3[36]=109
v3[37]=113
v3[38]=78
v3[39]=81
v3[40]=108
v3[41]=98
v3[42]=119
v3[43]=150
v3[44]=108
v3[45]=152
v3[46]=95
v3[47]=80
v3[48]=107
v3[49]=114
v3[50]=129
v3[51]=81
v3[52]=108
v3[53]=136
v3[54]=100
v3[55]=87
for i in range(56):
    v3[i]-=30
ret=base64.b64encode(bytes(v3))
print(ret)

hello_py

把python捅到apk里去了...，发现引擎叫chaquopy，丢进jeb中把一堆.py和.pyc弄出来，有个hello.py很扎眼，发现是关键代码：转为小端序然后xxtea加密。 hello.py很丑，改变量名：

from java import jboolean ,jclass #line:1
import struct #line:3
import ctypes #line:4
def MX (v1 ,v2 ,v3 ,v4 ,v5 ,v6 ):#line:7
    v7 =(v1 .value >>5 ^v2 .value <<2 )+(v2 .value >>3 ^v1 .value <<4 )#line:8
    v8 =(v3 .value ^v2 .value )+(v4 [(v5 &3 )^v6 .value ]^v1 .value )#line:9
    return ctypes .c_uint32 (v7 ^v8 )#line:11
def encrypt (len ,org ,key ):#line:14
    delta =0x9e3779b9 #line:15
    v13 =6 +52 //len #line:16
    v14 =ctypes .c_uint32 (0 )#line:18
    v15 =ctypes .c_uint32 (org [len -1 ])#line:19
    v16 =ctypes .c_uint32 (0 )#line:20
    while v13 >0 :#line:22
        v14 .value +=delta #line:23
        v16 .value =(v14 .value >>2 )&3 #line:24
        for v17 in range (len -1 ):#line:25
            v18 =ctypes .c_uint32 (org [v17 +1 ])#line:26
            org [v17 ]=ctypes .c_uint32 (org [v17 ]+MX (v15 ,v18 ,v14 ,key ,v17 ,v16 ).value ).value #line:27
            v15 .value =org [v17 ]#line:28
        v18 =ctypes .c_uint32 (org [0 ])#line:29
        org [len -1 ]=ctypes .c_uint32 (org [len -1 ]+MX (v15 ,v18 ,v14 ,key ,len -1 ,v16 ).value ).value #line:30
        v15 .value =org [len -1 ]#line:31
        v13 -=1 #line:32
    return org #line:34

def check (input_str ):#line:63
    print ("checking~~~: "+input_str )#line:64
    input_str =str (input_str )#line:65
    if len (input_str )!=36 :#line:66
        return jboolean (False )#line:67
    tmp1 =[]#line:69
    for i in range (0 ,36 ,4 ):#line:70
        encrypted =input_str [i :i +4 ].encode ('latin-1')#line:71
        tmp1 .append (encrypted )#line:72
    tmp2 =[]#line:73
    for i in tmp1 :#line:74
        tmp2 .append (struct .unpack ("<I",i )[0 ])#line:75
    print (tmp2 )#line:77
    encrypted =encrypt (9 ,tmp2 ,[12345678 ,12398712 ,91283904 ,12378192 ])#line:78
    ans =[689085350 ,626885696 ,1894439255 ,1204672445 ,1869189675 ,475967424 ,1932042439 ,1280104741 ,2808893494 ]#line:85
    for i in range (9 ):#line:86
        if ans [i ]!=encrypted [i ]:#line:87
            return jboolean (False )#line:88
    return jboolean (True )#line:90
def sayHello ():#line:92
    print ("hello from py")#line:93

故解密exp:

from ctypes import * 
 
def MX(z, y, sum1, k, p, e):
    return c_uint32(((z.value>>5^y.value<<2)+(y.value>>3^z.value<<4))^((sum1.value^y.value)+(k[(p&3)^e.value]^z.value)))
def btea(v,k,n,delta):
 
	if n>1:
		sum1=c_uint32(0)
		z=c_uint32(v[n-1])
		rounds=6+52//n
		e=c_uint32(0)
 
		while rounds>0:
			sum1.value+=delta
			e.value=((sum1.value>>2)&3)	#e都要32位哦
			for p in range(n-1):
				y=c_uint32(v[p+1])
				#v[p]=c_uint32(v[p]+c_uint32((((z.value>>5^y.value<<2)+(y.value>>3^z.value<<4))^((sum1.value^y.value)+(k[(p&3)^e.value]^z.value)))).value).value
				v[p] = c_uint32(v[p] + MX(z,y,sum1,k,p,e).value).value
				z.value=v[p]
 
			y=c_uint32(v[0])
			#v[n-1]=c_uint32(v[n-1]+c_uint32((((z.value>>5^y.value<<2)+(y.value>>3^z.value<<4))^((sum1.value^y.value)+(k[((n-1)&3)^e.value]^z.value)))).value).value		#这里tmd传入的是k[((n-1)&3)啊我草，找了半天！！！
			v[n-1] = c_uint32(v[n-1] + MX(z,y,sum1,k,n-1,e).value).value
			z.value=v[n-1]
			rounds-=1
 
	else:
		sum1=c_uint32(0)
		n=-n
		rounds=6+52//n
		sum1.value=rounds*delta
		y=c_uint32(v[0])
		e=c_uint32(0)
 
		while rounds>0:
			e.value=((sum1.value>>2)&3)	#e都要32位哦
			for p in range(n-1, 0, -1):
				z=c_uint32(v[p-1])
				#y[p]=c_uint32(v[p]-c_uint32((((z.value>>5^y.value<<2)+(y.value>>3^z.value<<4))^((sum1.value^y.value)+(k[(p&3)^e.value]^z.value)))).value).value
				v[p] = c_uint32(v[p] - MX(z,y,sum1,k,p,e).value).value
				y.value=v[p]
 
			z=c_uint32(v[n-1])
			#v[n-1]=c_uint32(v[n-1]-c_uint32((((z.value>>5^y.value<<2)+(y.value>>3^z.value<<4))^((sum1.value^y.value)+(k[((n-1)&3)^e.value]^z.value)))).value).value		#这里tmd传入的是k[((n-1)&3)啊我草，找了半天！！！
			v[0] = c_uint32(v[0] - MX(z,y,sum1,k,0,e).value).value
			y.value=v[0]
			sum1.value-=delta
			rounds-=1
 
	return v
 
 
 
 
if __name__=='__main__':
	a=[689085350 ,626885696 ,1894439255 ,1204672445 ,1869189675 ,475967424 ,1932042439 ,1280104741 ,2808893494]
	k=[12345678 ,12398712 ,91283904 ,12378192]
	delta=0x9e3779b9
	n=9
	# print("加密前数据:",a)
	# res=btea(a,k,n,delta)
	# print("加密后数据:",res)
	res=btea(a,k,-n,delta)
	print("解密后数据:",res)

解码exp：

import struct
enc=[946221411, 912614241, 878851117, 959720758, 1647128883, 761410866, 825241953, 959996208, 842610019]
for i in enc:
    print(str(hex(i))[2:])
tmp1=[]
tmp2=[]
for i in range(len(enc)):
    tmp1.append(struct.pack("<I",enc[i]))
    print(str(bytes(enc[i]).decode('latin-1')),end='')
    # print(str(tmp1[i]),end='')
print()
for i in range(len(tmp1)):
    print(str(tmp1[i].decode('latin-1')),end='')