Cover the face of pain and straight ahead.

SWPUCTF2021新生赛Web刷题记录

post @ 2025-07-21

[SWPUCTF 2021 新生赛]gift_F12

Ctrl+U查看前端源码再Ctrl+F搜搜Flag即可

[SWPUCTF 2021 新生赛]jicao

<?php
highlight_file('index.php');
include("flag.php");
$id=$_POST['id'];
$json=json_decode($_GET['json'],true);
if ($id=="wllmNB"&&$json['x']=="wllm")
{echo $flag;}
?>

代码审计，以post方式传id=wllmNB和以get方式传json={"x":"wllm"}即可

[SWPUCTF 2021 新生赛]easy_md5

<?php 
 highlight_file(__FILE__);
 include 'flag2.php';
 
if (isset($_GET['name']) && isset($_POST['password'])){
    $name = $_GET['name'];
    $password = $_POST['password'];
    if ($name != $password && md5($name) == md5($password)){
        echo $flag;
    }
    else {
        echo "wrong!";
    }
 
}
else {
    echo 'wrong!';
}
?>
wrong!

要求：name的值不等于password的值且其md5后相等

两种方法：

1.数组漏洞绕过

Read More

攻防世界Crypto-babyFibo

post @ 2025-07-21

分析

附件内容：

import os
import libnum
from secret import flag

def fibo(n):
    assert n >= 0
    if n < 2:
        return n
    return fibo(n-1) + fibo(n-2)

s = fibo(1000)
m = libnum.s2n(flag+os.urandom((len(bin(s))-2)//8-len(flag)))
c = m^s
print(c)

  • s是斐波那契数的第1000个数
  • m=flag+随机数字然后再(String to Number)转为一个超大数字
  • c=m^s

那解题思路就是将其中设计的步骤反着来一遍

开搞

计算fibo(1000)

以下由三种方法计算fibo(1000)

由于直接递归1000不可行

方法一：动态规划：

Read More

攻防世界Crypto-初识RSA

post @ 2025-07-21

分析

附件内容：

from Crypto.Util.number import bytes_to_long,inverse,getPrime
from flag import flag

m = bytes_to_long(flag)

p = getPrime(1024)
q = getPrime(1024)
n = p*q
print(n)
e = 65537

c = pow(m,e,n)

pq = p*(q-1)
qp = q*(p-1)

print("c=",c)
print("n=",n)
print("pq=",pq)
print("qp=",qp)

'''
c= 8722269075970644434253339592758512788160408912707387632591552130175707843950684315083250494010055435391879036285103810263591951437829414438640307561645721347859659807138051841516634704123100270651976676182059252251162982609391666023674158274992400910869692389001622774140191223807887675081808561012755545464977015973615407965906513878979919700065923364884766974187303774330319143647840846354404070430118235352622445115153298578370521811697710289716188726587743282814946239856766713516166990341116198180068191759095913957606379780234116317390622824096667107736103270907349927467971817639795094030622157581511033950777
n= 10466186506773626671397261081802640650185744558208505628349249045496105597268556020207175016523119333667851114848452038431498926527983706092607207796937431312520131882751891731564121558651246025754915145600686076505962750195353958781726515647847167067621799990588328894365930423844435964506372428647802381074584935050067254029262890188260006596141011807724688556673520261743199388391094490191001701011230322653422314758778116196105077883955436582364267530633358016652912054880813710531145973799193443828969535902856467548523653920307742364119002349899553478815101092655897400295925170383678499125295006364960124859003
pq= 10466186506773626671397261081802640650185744558208505628349249045496105597268556020207175016523119333667851114848452038431498926527983706092607207796937431312520131882751891731564121558651246025754915145600686076505962750195353958781726515647847167067621799990588328894365930423844435964506372428647802381074488896197029704465200125337817646702009123916866455067019234171839614862660036737875747177391796376553159880972782837853473250804807544086701088829096838316550146794766718580877976153967582795248676367265069623900208276878140709691073369415161936376086988069213820933152601453587292943483693378833664901178324
qp= 10466186506773626671397261081802640650185744558208505628349249045496105597268556020207175016523119333667851114848452038431498926527983706092607207796937431312520131882751891731564121558651246025754915145600686076505962750195353958781726515647847167067621799990588328894365930423844435964506372428647802381074475956379708898904933143429835002718457573266164923043251954374464149976302585916538814746811455883837138715445492053610047383292461097590195481556557381952895539341802954749542143253491617052100969586396996063822508764438280468492894012685918249843558593322831683872737943676955669923498182824352081785243246
'''

既然题目已经提醒了是RSA的相关内容那就直接开搞即可

关键点：各个参数的值(即公钥 密钥)、欧拉定理：通用$phi=(p−1)(q−1)$

Exploit:

from Crypto.Util.number import *
from gmpy2 import *

e = 65537

c= 8722269075970644434253339592758512788160408912707387632591552130175707843950684315083250494010055435391879036285103810263591951437829414438640307561645721347859659807138051841516634704123100270651976676182059252251162982609391666023674158274992400910869692389001622774140191223807887675081808561012755545464977015973615407965906513878979919700065923364884766974187303774330319143647840846354404070430118235352622445115153298578370521811697710289716188726587743282814946239856766713516166990341116198180068191759095913957606379780234116317390622824096667107736103270907349927467971817639795094030622157581511033950777
n= 10466186506773626671397261081802640650185744558208505628349249045496105597268556020207175016523119333667851114848452038431498926527983706092607207796937431312520131882751891731564121558651246025754915145600686076505962750195353958781726515647847167067621799990588328894365930423844435964506372428647802381074584935050067254029262890188260006596141011807724688556673520261743199388391094490191001701011230322653422314758778116196105077883955436582364267530633358016652912054880813710531145973799193443828969535902856467548523653920307742364119002349899553478815101092655897400295925170383678499125295006364960124859003
pq= 10466186506773626671397261081802640650185744558208505628349249045496105597268556020207175016523119333667851114848452038431498926527983706092607207796937431312520131882751891731564121558651246025754915145600686076505962750195353958781726515647847167067621799990588328894365930423844435964506372428647802381074488896197029704465200125337817646702009123916866455067019234171839614862660036737875747177391796376553159880972782837853473250804807544086701088829096838316550146794766718580877976153967582795248676367265069623900208276878140709691073369415161936376086988069213820933152601453587292943483693378833664901178324
qp= 10466186506773626671397261081802640650185744558208505628349249045496105597268556020207175016523119333667851114848452038431498926527983706092607207796937431312520131882751891731564121558651246025754915145600686076505962750195353958781726515647847167067621799990588328894365930423844435964506372428647802381074475956379708898904933143429835002718457573266164923043251954374464149976302585916538814746811455883837138715445492053610047383292461097590195481556557381952895539341802954749542143253491617052100969586396996063822508764438280468492894012685918249843558593322831683872737943676955669923498182824352081785243246

phi = pq * qp // n

d = gmpy2.invert(e, phi)

m = pow(c, d, n)

flag = long_to_bytes(m)

print(flag)
Read More

攻防世界Crypto-二元一次方程组

post @ 2025-07-21

分析：

import libnum
from Crypto.Util import number
from secret import flag


size = 256
e = 65537
p = number.getPrime(size)
q = number.getPrime(size)
avg = (p+q)/2
n = p*q

m = libnum.s2n(flag)
c = pow(m, e, n)

print('n = %d' % n)
print('avg = %d' % avg)
print('c = %d' % c)

看上述代码感觉老朋友了，详情可以看上一篇的baigeiRSA

size = 256
e = 65537

size=256定义了生成的素数p和q的比特长度

e=65537是RSA公钥指数，是一个固定的值

p = number.getPrime(size)
q = number.getPrime(size)
avg = (p+q)/2
n = p*q

生成 RSA密钥

其中设计两个公式：

$$avg = \frac{p+q} {2}$$

Read More

攻防世界Crypto-baigeiRSA

post @ 2025-07-21

拿到附件后查看内容

import libnum
from Crypto.Util import number
from secret import flag

size = 128
e = 65537
p = number.getPrime(size)
q = number.getPrime(size)
n = p*q

m = libnum.s2n(flag)
c = pow(m, e, n)

print('n = %d' % n)
print('c = %d' % c)

分析

参数设置

size = 128
e = 65537

size处设置了生成的素数p和q的比特长度

e=65537：RSA的公钥指数，这是一个常用的固定值

生成RSA密钥

p = number.getPrime(size)
q = number.getPrime(size)
n = p*q

number.getPrime(size)生成一个size比特的随机素数

Read More

Web康复训练-攻防世界Web2

post @ 2025-07-12

打开题目，给了以下代码：

<?php
$miwen="a1zLbgQsCESEIqRLwuQAyMwLyq2L5VwBxqGA3RQAyumZ0tmMvSGM2ZwB4tws";

function encode($str){
    $_o=strrev($str);
    // echo $_o;
        
    for($_0=0;$_0<strlen($_o);$_0++){
       
        $_c=substr($_o,$_0,1);
        $__=ord($_c)+1;
        $_c=chr($__);
        $_=$_.$_c;   
    } 
    return str_rot13(strrev(base64_encode($_)));
}

highlight_file(__FILE__);
/*
   逆向加密算法，解密$miwen就是flag
*/
?>

分析：

  • strlen() 函数返回字符串的长度

  • strrev() 反转字符串顺序

This is an example:

<?php
echo strrev("Hello World!");
?>

输出反转字符串 “!dlroW olleH”：

  • substr() 函数返回字符串的一部分。substr(string,start,length)

    tips：如果 start 参数是负数且 length 小于或等于 start，则 length 为 0。

  • <?php echo substr("Hello world",6); ?>从字符串中返回 “world”：

  • ord() 函数返回字符串中第一个字符的 ASCII 值。

  • chr() 函数从指定 ASCII 值返回字符。并可以指定ASCII是十进制或是十六进制或是八进制。

  • “.” 是字符运算符，作用是把两个字符串连接起来，其使用方法：首先创建一个PHP示例文件；然后定义两个变量；最后通过“$result.$result”方式将两个变量连接起来即可

  • str_rot13() 函数对字符串执行 ROT13 编码。

    ROT13 编码是把每一个字母在字母表中向前移动 13 个字母得到。数字和非字母字符保持不变。

综上，题目提示我们逆向加密算法，也就是说第一步加密变为最后一步，以此类推，那可以构造出逆向后的代码

<?php
$miwen="a1zLbgQsCESEIqRLwuQAyMwLyq2L5VwBxqGA3RQAyumZ0tmMvSGM2ZwB4tws";
$a = base64_decode(strrev(str_rot13($miwen)));
 
 
for($i=0;$i<strlen($a);$i++){
	$_c=substr($a,$i,1);
	$d=ord($_c)-1;
	$_c=chr($d); 
	$b=$b.$_c;
	
}
$p= strrev($b);
echo $p;
?>
Read More

攻防世界Crypto-了不起的盖茨比

post @ 2025-07-10

Description:

小王读了盖茨比深受震撼，于是摘录本书并效仿原作者为你加密留了一段言，但他忘记给你iv了，请问你可以恢复出来原文吗
(得到的flag内容用CatCTF{}包上提交)

附件内容为python源代码程序，对其进行分析：

from Crypto.Cipher import AES  
from Crypto.Util.strxor import strxor as xor  
from Crypto.Util.Padding import pad  
from random import *  
from base64 import *  
from copy import copy  
from secret import data  
  
iv=bytes([randint(0,2**8-1) for i in range(16)])   
iva=copy(iv)  
key=bytes([randint(0,2**8-1) for i in range(16)])  
cipher=AES.new(key,mode=AES.MODE_ECB)  
data=pad(data,16)  
c=b""  
for i in range(0,len(data),16):  
    s=data[i:i+16].ljust(16,b"\n")  
    c+=cipher.encrypt(xor(s,iv))  
    iv=xor(s,c[-16:])  
  
  
key=b64encode(key)  
c=b64encode(c)  
  
print(key)  
print(c)  
  
  
  
"""  
b'+0zkhmid1PFjVdxSP09zSw=='  
b'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'  
  
"""

这段代码实现了一个基于AES加密的自定义加密方案，结合了ECB模式和CBC模式的某些特性，明文按16字节分组进行加密，打印出key和密文，iv(偏移量)未知，需要还原明文。可以发现本题是多次使用同样的流密码。
所以本题其实是利用利用MTP攻击来求解，比较好的两篇文章可以参考
[参考一](Many-Time-Pad 攻击)
参考二

exp:

# Python3  
import base64  
from Crypto.Cipher import AES  
from Crypto.Util.strxor import strxor as xor  
from Crypto.Util.number import *  
import Crypto.Util.strxor as xo  
import libnum, codecs, numpy as np  
  
  
def isChr(x):  
    if ord('a') <= x and x <= ord('z'): return True  
    if ord('A') <= x and x <= ord('Z'): return True  
    return False  
  
def infer(index, pos):  
    if msg[index, pos] != 0:  
        return  
    msg[index, pos] = ord(' ')  
    for x in range(len(c)):  
        if x != index:  
            msg[x][pos] = xo.strxor(c[x], c[index])[pos] ^ ord(' ')  
  
  
def know(index, pos, ch):  
    msg[index, pos] = ord(ch)  
    for x in range(len(c)):  
        if x != index:  
            msg[x][pos] = xo.strxor(c[x], c[index])[pos] ^ ord(ch)  
  
  
def getSpace():  
    for index, x in enumerate(c):  
        res = [xo.strxor(x, y) for y in c if x != y]  
        f = lambda pos: len(list(filter(isChr, [s[pos] for s in res])))  
        cnt = [f(pos) for pos in range(len(x))]  
        for pos in range(len(x)):  
            dat.append((f(pos), index, pos))  
  
  
key = b'+0zkhmid1PFjVdxSP09zSw=='  
key = base64.b64decode(key)  
c = b'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'  
c = base64.b64decode(c)  
cipher = AES.new(key, mode=AES.MODE_ECB)  
t = []  
for i in range(0, len(c), 16):  
    if i == 0:  
        t.append(cipher.decrypt(c[i:i + 16]))  # s1 ^ iv  
    else:  
        t.append(xor(cipher.decrypt(c[i:i + 16]), c[i - 16:i]))  
tmp = []  
tmp.append(t[0])  
for i in range(1, len(t)):  
    tttt = t[i]  
    for j in range(0, i):  
        tttt = xor(tttt, t[j])  
    tmp.append(tttt)  # si ^ iv  
# MTP attack  
c = tmp  
dat = []  
msg = np.zeros([len(c), len(c[0])], dtype=int)  
getSpace()  
dat = sorted(dat)[::-1]  
for w, index, pos in dat:  
    infer(index, pos)  
  
print(''.join([''.join([chr(c) for c in x]) for x in msg]))
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Many Time Pad Attack

post @ 2025-06-22

一些基础知识

符号：

⊕ 代表异或

C1 代表密文

M1 代表明文

性质：

  1. 交换律
  2. 结合律 (a ⊕ b ) ⊕ c = a⊕ ( b ⊕ c)
  3. 任何数x x ⊕ x = 0 x ⊕ 0 = X
  4. 自反性 x ⊕ b ⊕ b = x ⊕ 0 = x

Description

Many-Time-Pad (多时间垫) 攻击是一种针对多次使用相同密钥的流密码（如一次性密码本，One-Time Pad）的密码分析技术。其核心原理是利用密钥重用导致的明文信息泄露，通过数学和统计方法恢复部分或全部明文。以下是其核心原理和步骤：

1. 一次性密码本（OTP）的安全前提
OTP的安全性是建立在：

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攻防世界Crypto-cat's gift

post @ 2025-06-22

Description:

如图所示，下面这只好心的猫猫给你送来了跨年礼物。由于礼物不好拿，所以猫猫把礼物平均分成了四份，但是其中一份不小心掉在地上散落成了无数片，变成了 1 - 1/3 + 1/5 - 1/7 + …

聪明的你能算出或猜出猫猫的礼物是什么吗？

flag示例: CatCTF{apple} CatCTF{banana}

关键：Gregory-Leibniz 公式
德国数学家莱布尼茨（Leibniz）于 1674 年曾提出 Gregory-Leibniz 公式来计算π：
$π / 4 = 1 − 1 / 3 + 1 / 5 − 1 / 7 + 1 / 9 + . . . π/4 = 1 - 1/3 +1/5 - 1/7 + 1/9+ …π/4=1−1/3+1/5−1/7+1/9+…$
得到flag：CatCTF{pie

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upload-labs靶场小计

post @ 2025-06-17

Introduction：

  • Repo Address: https://github.com/c0ny1/upload-labs
  • Description：upload-labs是一个使用php语言编写的，专门收集渗透测试和CTF中遇到的各种上传漏洞的靶场。旨在帮助大家对上传漏洞有一个全面的了解。目前一共20关，每一关都包含着不同上传方式。

正文开始

Pass-01

源码分析:

function checkFile() {
    var file = document.getElementsByName('upload_file')[0].value;
    if (file == null || file == "") {
        alert("请选择要上传的文件!");
        return false;
    }
    //定义允许上传的文件类型
    var allow_ext = ".jpg|.png|.gif";
    //提取上传文件的类型
    var ext_name = file.substring(file.lastIndexOf("."));
    //判断上传文件类型是否允许上传
    if (allow_ext.indexOf(ext_name + "|") == -1) {
        var errMsg = "该文件不允许上传，请上传" + allow_ext + "类型的文件,当前文件类型为：" + ext_name;
        alert(errMsg);
        return false;
    }
}

在源码中我们可以看到其在客户端使用了js对不合法图片进行了检查，因此只需要在Firefox浏览器中禁用js即可绕过

Pass-02

源码分析:

$is_upload = false;
$msg = null;
if (isset($_POST['submit'])) {
    if (file_exists(UPLOAD_PATH)) {
        if (($_FILES['upload_file']['type'] == 'image/jpeg') || ($_FILES['upload_file']['type'] == 'image/png') || ($_FILES['upload_file']['type'] == 'image/gif')) {
            $temp_file = $_FILES['upload_file']['tmp_name'];
            $img_path = UPLOAD_PATH . '/' . $_FILES['upload_file']['name']            
            if (move_uploaded_file($temp_file, $img_path)) {
                $is_upload = true;
            } else {
                $msg = '上传出错！';
            }
        } else {
            $msg = '文件类型不正确，请重新上传！';
        }
    } else {
        $msg = UPLOAD_PATH.'文件夹不存在,请手工创建！';
    }
}

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