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[原创]强网杯MINIRE
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发表于: 2024-11-11 20:18
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有upx壳，并且把upx改为gcc了
两种解法 一种是hook运算操作，尝试还原算法，我比赛时就是用的这种，但是这题校验非常难找，导致比赛时一直没找到，另一种是单字节爆破
脱壳，f5发现没法反编译
但是ghidra是可以的，它没有栈限制
对应的找到各个运算的位置下断点，很容易就能得到加密逻辑
这里打印的时候可以格式化一下，便于后续写脚本，因为没想到能直接写z3（）
参考0xcafebabe师傅的题解给出的脚本格式，会非常方便搞成z3，但是这里需要看出来eax - 46 就是index
接着就是找密文或者校验逻辑了，但是比赛卡这地方卡了一下午ww
后来将多个点全hook，并根据已知正确的前五个字符'flag{'的加密结果，可以发现其读的方式
猜测应该是存于某一个连续内存的，hook方式如下
通过上述脚本可断于特定位置，跟以下发现了以下内存块存储了enc
然后即可用z3解了
脚本可参考https://blog.hxzzz.asia/，属实是懒得写z3了
比赛的时候原本是有想用的，但是看波动范围比较大就没用，下来试了下发现没问题
插桩点分别是handler与数据清理出
hook.js
brute.py
case 'X':
local_20 = 0;
while (local_1c = local_1c + 1, *(char *)(param_1 + local_1c) != 'x') {
(&stack0xffffffffffc24328)[local_20] = *(undefined *)(param_1 + local_1c);
local_20 = local_20 + 1;
}
(&stack0xffffffffffc24328)[local_20] = 0;
if (1 < local_20) {
if (in_stack_ffffffffffc24328 != 'V') {
for (local_24 = 1;
((local_24 < local_20 && ('/' < (char)(&stack0xffffffffffc24328)[local_24])) &&
((char)(&stack0xffffffffffc24328)[local_24] < ':')); local_24 = local_24 + 1) {
}
}
iVar7 = (int)in_stack_ffffffffffc24328;
if (iVar7 == 0x20) {
bVar1 = (&DAT_005175a0)[DAT_00768090];
bVar5 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = bVar1 ^ bVar5;
iVar6 = DAT_00768094;
}
else if (((0x1f < iVar7) && (iVar7 < 0x6b)) && (0x40 < iVar7)) {
switch(iVar7) {
case 0x41:
cVar2 = (&DAT_005175a0)[DAT_00768090];
cVar4 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = cVar2 * cVar4;
iVar6 = DAT_00768094;
break;
case 0x4a:
iVar7 = FUN_00405c10(&stack0xffffffffffc24329);
iVar6 = DAT_00768094;
if (iVar7 != 0) {
cVar2 = (&DAT_005175a0)[DAT_00768090];
cVar4 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = cVar2 + cVar4;
iVar6 = DAT_00768094;
}
break;
case 0x51:
cVar2 = (&DAT_005175a0)[DAT_00768090];
cVar4 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = cVar2 - cVar4;
iVar6 = DAT_00768094;
break;
case 0x54:
bVar1 = (&DAT_005175a0)[DAT_00768090];
bVar5 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = bVar1 | bVar5;
iVar6 = DAT_00768094;
break;
case 0x55:
uVar3 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = uVar3;
iVar6 = DAT_00768094;
break;
case 0x5f:
cVar2 = (&DAT_005175a0)[DAT_00768090];
cVar4 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = cVar2 + cVar4;
iVar6 = DAT_00768094;
break;
case 0x61:
iVar7 = FUN_00405c10(&stack0xffffffffffc24329);
iVar6 = DAT_00768094;
if (iVar7 != 0) {
bVar1 = (&DAT_005175a0)[DAT_00768090];
bVar5 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = bVar1 / bVar5;
iVar6 = DAT_00768094;
}
break;
case 0x6a:
bVar1 = (&DAT_005175a0)[DAT_00768090];
bVar5 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = bVar1 & bVar5;
iVar6 = DAT_00768094;
case 'X':
local_20 = 0;
while (local_1c = local_1c + 1, *(char *)(param_1 + local_1c) != 'x') {
(&stack0xffffffffffc24328)[local_20] = *(undefined *)(param_1 + local_1c);
local_20 = local_20 + 1;
}
(&stack0xffffffffffc24328)[local_20] = 0;
if (1 < local_20) {
if (in_stack_ffffffffffc24328 != 'V') {
for (local_24 = 1;
((local_24 < local_20 && ('/' < (char)(&stack0xffffffffffc24328)[local_24])) &&
((char)(&stack0xffffffffffc24328)[local_24] < ':')); local_24 = local_24 + 1) {
}
}
iVar7 = (int)in_stack_ffffffffffc24328;
if (iVar7 == 0x20) {
bVar1 = (&DAT_005175a0)[DAT_00768090];
bVar5 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = bVar1 ^ bVar5;
iVar6 = DAT_00768094;
}
else if (((0x1f < iVar7) && (iVar7 < 0x6b)) && (0x40 < iVar7)) {
switch(iVar7) {
case 0x41:
cVar2 = (&DAT_005175a0)[DAT_00768090];
cVar4 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = cVar2 * cVar4;
iVar6 = DAT_00768094;
break;
case 0x4a:
iVar7 = FUN_00405c10(&stack0xffffffffffc24329);
iVar6 = DAT_00768094;
if (iVar7 != 0) {
cVar2 = (&DAT_005175a0)[DAT_00768090];
cVar4 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = cVar2 + cVar4;
iVar6 = DAT_00768094;
}
break;
case 0x51:
cVar2 = (&DAT_005175a0)[DAT_00768090];
cVar4 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = cVar2 - cVar4;
iVar6 = DAT_00768094;
break;
case 0x54:
bVar1 = (&DAT_005175a0)[DAT_00768090];
bVar5 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = bVar1 | bVar5;
iVar6 = DAT_00768094;
break;
case 0x55:
uVar3 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = uVar3;
iVar6 = DAT_00768094;
break;
case 0x5f:
cVar2 = (&DAT_005175a0)[DAT_00768090];
cVar4 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = cVar2 + cVar4;
iVar6 = DAT_00768094;
break;
case 0x61:
iVar7 = FUN_00405c10(&stack0xffffffffffc24329);
iVar6 = DAT_00768094;
if (iVar7 != 0) {
bVar1 = (&DAT_005175a0)[DAT_00768090];
bVar5 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = bVar1 / bVar5;
iVar6 = DAT_00768094;
}
break;
case 0x6a:
bVar1 = (&DAT_005175a0)[DAT_00768090];
bVar5 = FUN_00405c10(&stack0xffffffffffc24329);
(&DAT_005175a0)[DAT_00768090] = bVar1 & bVar5;
iVar6 = DAT_00768094;
import ida_dbg
index = ida_dbg.get_reg_value('RAX') - 0x46
# 获取 R8D 寄存器的值EDX_value = ida_dbg.get_reg_val("EDX")
print(f"x[{value}]^= 0x{EDX_value:X}")
import ida_dbg
index = ida_dbg.get_reg_value('RAX') - 0x46
# 获取 R8D 寄存器的值EDX_value = ida_dbg.get_reg_val("EDX")
print(f"x[{value}]^= 0x{EDX_value:X}")
import ida_dbg
EAX_value = ida_dbg.get_reg_val("EAX")
print(f"case 69: 0x{EAX_value:X}")
import ida_dbg
EAX_value = ida_dbg.get_reg_val("EAX")
[培训]内核驱动高级班，冲击BAT一流互联网大厂工作，每周日13:00-18:00直播授课
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链接：https://pan.quark.cn/s/80c269695999 提取码：stPW 不好意思师傅 这题目可能是pcb的 那几天打的比赛比较多可能搞混了 您看看这个 |
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有了，多谢分享。
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果然比较变态，我找到了hash对比的地方，想着应该不可能就解个hash，还要脱壳。。算了，嫌麻烦，好久没学习脱壳的工具了
 继续学习一下楼主别的帖子
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