2018SUSCTF-Mobile题

BabyAndroid

常规，JEB反编译

根据输入的字符串长度，第一位和97异或，后面跟v0的字符异或，把结果字符串转入a函数中。a函数首先把给出的常量字符串转为数组，把传入的字符串转化为bytes数组，再根据数组长度把每一位做for中&和左移计算，结果为3267347723651E492C1D7E117C1946325D02432D493B0B62067B则为真。所以逆推，由于一次计算拼接两位，所以按照两位为一个字符计算。其中含有字母，调试得知是对应的十六进制数。

#coding:utf-8

strs = "3267347723651E492C1D7E117C1946325D02432D493B0B62067B"

num = []
for i in range(0, len(strs)):
	num.append(strs[i])   #52

nums = []
id = 0
while id < 52:
	for i in range(0,200):
		if str((i & 0xF0) >> 4) == num[id]:
			if hex(i & 15).replace('0x','').upper() == num[id+1]:
				nums.append(i)
				break
	id = id + 2

flag = []
for i in range(0, len(nums)):
	if i == 0:
		first = nums[i] ^ 97
		flag.append(first)
	else:
		sc = nums[i] ^ nums[i-1]
		flag.append(sc)

text = ''
for i in flag:
	text = text + chr(i)
print(text)


>>> SUSCTF{We1come_to_Andr0id}

后来看了其他资料，才发现a函数下面的for循环就是找16进制对应的十进制数，如

>>> int('32',16)
50
其中是利用50找到对应的十进制32.

这个点卡了N久，看到一份别人的poc写的是相当简洁了。

CrackMe

又是一个native层的APP。

IDA打开伪代码

根据汇编得知，中间for循环处操作的是数组

其中v11是给出的数据块，按照int型，四位一个字段，正好28个字段。

按照Java层逻辑和伪代码写出c代码为

#include <stdio.h>
#include <stdlib.h>

int main(){
	int v11[28] = {0x571a9693,0x23a96034,0x6a943d8c,0x1f9222ed,0x73887c81,0x5c13a257,0x26407522,0x13646a3a,0x2139537f,0x35415c5f,0x321304b7,0x238a8c26,0xd7307f6,0x622d5268,0x7c3d2e04,0x72198f7e,0x7df76af2,0x4e8431aa,0x28650861,0xfd8e3e9,0x196c1f1a,0x5fe8ab3,0x1231495d,0x5359d998,0x35fcfde0,0x3b2d0dd4,0x61113e45,0x314c57b8};   //小端字节序
	int v0[28] = {0};
	int string[28] = {0};   //未定义,引入的参数变量
	int i;
	for (i=0; i<28; i++){
		if (i == 0){
			v0[i] = string[i] ^ 0xFF;
		}
		else{
			v0[i] = string[i] ^ string[i-1];
		}
	}

	srand(0x133ED6B);
	int v4[28] = {0};
	for (i=0; i<28; i++){
		v4[i] = v0[i] - rand();
		if (v11[i] != v4[i]){
			int v10 = 0;
			return v10;
		}
	int v10 = 1;
	printf("%c",v10);
	return v10;
	}

}

也就是我们需要反向求出string的值

#include <stdio.h>
#include <stdlib.h>

int main(){
	int v11[28] = {0x571a9693,0x23a96034,0x6a943d8c,0x1f9222ed,0x73887c81,0x5c13a257,0x26407522,0x13646a3a,0x2139537f,0x35415c5f,0x321304b7,0x238a8c26,0xd7307f6,0x622d5268,0x7c3d2e04,0x72198f7e,0x7df76af2,0x4e8431aa,0x28650861,0xfd8e3e9,0x196c1f1a,0x5fe8ab3,0x1231495d,0x5359d998,0x35fcfde0,0x3b2d0dd4,0x61113e45,0x314c57b8};
	int v0[28] = {0};
	srand(0x133ED6B);
	for (int i=0; i<28; i++){
		v0[i] = v11[i] - rand();
	}

	int string[28] = {0};
	for (int i=0; i<28; i++){
		if (i==0)
			string[i] = v0[i] ^ 0xFF;
		else{
			string[i] = v0[i] ^ v0[i-1];
		}
	}

	for (int i=0; i<28; i++){
		printf("%c", string[i]);
	}
	return 0;
}

但是尴尬的是算出来的是乱码

然后没有找到这题的wp….也没找到其他问题的存在点。就这样吧，毕竟是汇编渣渣。