BUUCTF Re部分wp（四）

[FlareOn5]Minesweeper Championship Registration

java逆向，我把jar解压成了一个class文件和一个.MF文件，把class扔进ida

得到flag

[RoarCTF2019]polyre

拖进ida

控制流平坦化，用deflat处理：

虚假控制流程，看着还是很恶心

这的jnz是必定跳转的，用idapython处理一下

addr = 0x401198
while(addr<0x4020cc):
    next_addr = NextHead(addr)
    if "ds:dword_603054" in GetDisasm(addr):
        while(True):
            addr = next_addr
            next_addr = NextHead(addr)
            if GetMnem(addr)=="jnz":
                dest = GetOperandValue(addr,0)
                PatchByte(addr,0xe9)
                PatchByte(next_addr-1,0x90) 
                PatchDword(addr+1,dest-next_addr+1)
                addr = next_addr
                break
    else:
        addr = next_addr

idapython用法参考了https://blog.csdn.net/oShuangYue12/article/details/85675751

处理之后

 

 

main已经可以看懂了，crc64，emm然而我逆不回去。。。

在看了官方wp后，emmm密码学真神奇

[FlareOn6]Overlong

把unk_1242008 解密后输出了，但程序只用了unk前1Ch长度

把push 1ch改为afh，得到flag

[2019红帽杯]childRE

64exe,拖进ida

输入经过变换后以以上方式进行比对，

a="1234567890-=!@#$%^&*()_+qwertyuiop[]QWERTYUIOP{}asdfghjkl;'ASDFGHJKL:"ZXCVBNM<>?zxcvbnm,./"
b="55565653255552225565565555243466334653663544426565555525555222"
c="(_@4620!08!6_0*0442!@186%%0@3=66!!974*3234=&0^3&1@=&0908!6_0*&"
f=""
for i in range(62):
    for j in range(128):
        if a[j//23]==b[i] and a[j%23]==c[i]:
            f+=chr(j)
print(f)
#private: char * __thiscall R0Pxx::My_Aut0_PWN(unsigned char *)

关键在于

通过C++ 编译器的函数名修饰规则

得到?My_Aut0_PWN@R0Pxx@@AAEPADPAE@Z

输入时经过了换位，输入了一串字符测试

c="?My_Aut0_PWN@R0Pxx@@AAEPADPAE@Z"
a="pqhrsidtujvwkebxylz1mf23n45ogca"
b="abcdefghijklmnopqrstuvwxyz12345"
f=[]
for i in range(31):
    f.append(0)
for i in range(31):
    for j in range(31):
        if b[j]==a[i]:
            f[j]=c[i]
            break
for i in range(31):
    print(f[i],end="")

md5得flag

findkey

根据字符串找到关键函数

创建函数失败，显示

.text:0040191F: The function has undefined instruction/data at the specified address.
Your request has been put in the autoanalysis queue.

往下看

 这里连续push了两个，nop掉下面那个

string1加密后 与v21加密后比较，加密v12的sub_401005是与v17异或

string1加密方式使用了CryptCreateHash的md5

a="0kk`d1a`55k222k2a776jbfgd`06cjjb"
b="S"
for i in a:
    print(chr(ord(i)^ord(b[0])),end="")

得到md5值，在线求得string1为123321

之后通过sub_401005又将string1与unk_423030异或得到flag

c="123321"
d="57 5E 52 54 49 5F 01 6D  69 46 02 6E 5F 02 6C 57 5B 54 4C"
d=d.split()
for i in range(len(d)):
    d[i]=eval("0x"+d[i])

for i in range(len(d)):
    print(chr(d[i]^ord(c[i%6])),end="")

[MRCTF2020]hello_world_go

go逆向，没去符号，找到main_main

 

 

找到flag

[MRCTF2020]PixelShooter

apk不太会，但这是用unity写的，把Assembly-CSharp.dll用dnspy打开

 

这有个GameOver

[FlareOn3]Challenge1

又是换表base64

import base64

c="x2dtJEOmyjacxDemx2eczT5cVS9fVUGvWTuZWjuexjRqy24rV29q"
flag=""

table="ZYXABCDEFGHIJKLMNOPQRSTUVWzyxabcdefghijklmnopqrstuvw0123456789+/"
table2="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
for i in c:
    for j in range(len(table)):
        if i==table[j]:
            flag+=table2[j]

flag=base64.b64decode(flag)
print(flag)

[FlareOn5]Ultimate Minesweeper

毫无游戏体验的扫雷小游戏，c#写的，扔进dnspy

public MainForm()
{
    this.InitializeComponent();
    this.MineField = new MineField(MainForm.VALLOC_NODE_LIMIT);
    this.AllocateMemory(this.MineField);
    this.mineFieldControl.DataSource = this.MineField;
    this.mineFieldControl.SquareRevealed += this.SquareRevealedCallback;
    this.mineFieldControl.FirstClick += this.FirstClickCallback;
    this.stopwatch = new Stopwatch();
    this.FlagsRemaining = this.MineField.TotalMines;
    this.mineFieldControl.MineFlagged += this.MineFlaggedCallback;
    this.RevealedCells = new List<uint>();
}

AllocateMemory创建雷区，SquareRevealedCallback判断成功失败

private void AllocateMemory(MineField mf)
        {
            for (uint num = 0U; num < MainForm.VALLOC_NODE_LIMIT; num += 1U)
            {
                for (uint num2 = 0U; num2 < MainForm.VALLOC_NODE_LIMIT; num2 += 1U)
                {
                    bool flag = true;
                    uint num3 = num + 1U;
                    uint num4 = num2 + 1U;
                    if (this.VALLOC_TYPES.Contains(this.DeriveVallocType(num3, num4)))
                    {
                        flag = false;
                    }
                    mf.GarbageCollect[(int)num2, (int)num] = flag;
                }
            }
        }

我本来想法是控制雷的生成

private void AllocateMemory(MineField mf)
        {
            for (uint num = 0U; num < MainForm.VALLOC_NODE_LIMIT; num += 1U)
            {
                for (uint num2 = 0U; num2 < MainForm.VALLOC_NODE_LIMIT; num2 += 1U)
                {
                    bool flag = true;
                    int num3 = (int)(num + 1U);
                    uint num4 = num2 + 1U;
                    if ((num3 == 1 && num4 == 1U) || (num3 == 2 && num4 == 2U) || (num3 == 3 && num4 == 3U))
                    {
                        flag = false;
                    }
                    mf.GarbageCollect[(int)num2, (int)num] = flag;
                }
            }
        }

但是flag是加密过的，需要正确位置解密，所以我改成了这样

private void AllocateMemory(MineField mf)
        {
            for (uint num = 0U; num < MainForm.VALLOC_NODE_LIMIT; num += 1U)
            {
                for (uint num2 = 0U; num2 < MainForm.VALLOC_NODE_LIMIT; num2 += 1U)
                {
                    bool flag = true;
                    uint num3 = num + 1U;
                    uint num4 = num2 + 1U;
                    if (this.VALLOC_TYPES.Contains(this.DeriveVallocType(num3, num4)))
                    {
                        flag = false;
                        Console.Write(num3);
                        Console.Write(num4);//断点下在这
                    }
                    mf.GarbageCollect[(int)num2, (int)num] = flag;
                }
            }
        }

这样就得到了坐标,（1D，8），（8，15），（19，1D）

点三个点得到flag

[安洵杯 2019]game

ollvm,用deflat去除，找到关键代码

 

D0g3是数独的盘，输入被转换后输入盘，结果已经在Sudoku给出来了

a="4693641762894685722843556137219876255986"
b="HJMGJHEKJFLMHJLIKFFLHGIIJEGKFEMLKJFIIMLJ"
#f.add((x[i]+0xf3|~x[i]-0xc) - 20 == eval(a[i]))
flag=[]

t1="@ABCDEFGHIJKLMNO"
t2="LMNOHIJKDEFG@ABC"
#x[i]+0xf3|~x[i]-0xc只改变了第67bit，所以可以换成这样的表

for i in range(len(b)):
    for j in range(len(t1)):
        if b[i]==t1[j]:
            flag.append(t2[j])
for i in range(0,len(flag),2):
    temp=flag[i]
    flag[i]=flag[i+1]
    flag[i+1]=temp
for i in flag:
    print(i,end="")

[Zer0pts2020]easy strcmp

strcmp被hook了，输入先进行了转换

a="42 09 4A 49 35 43 0A 41 F0 19 E6 0B F5 F2 0E 0B 2B 28 35 4A 06 3A 0A 4F"
a=a.split()
b="********CENSORED********"
for i in range(len(a)):
    print(chr((ord(b[i])+eval("0x"+a[i]))&0xFF),end="")

跑出来有几位不对，调试发现需要+1