n00bzCTF 2024

news/2024/11/20 12:29:18/文章来源:https://www.cnblogs.com/lpppp/p/18346286

n00bzCTF 2024

Crypto

Vinegar

题目：

Encrypted flag: nmivrxbiaatjvvbcjsf
Key: secretkey

exp:

维吉尼亚密码

​image-20240803213429-3gxhu0y​

flag：n00bz{vigenerecipherisfun}

RSA

题目：

e = 3
n = 135112325288715136727832177735512070625083219670480717841817583343851445454356579794543601926517886432778754079508684454122465776544049537510760149616899986522216930847357907483054348419798542025184280105958211364798924985051999921354369017984140216806642244876998054533895072842602131552047667500910960834243
c = 13037717184940851534440408074902031173938827302834506159512256813794613267487160058287930781080450199371859916605839773796744179698270340378901298046506802163106509143441799583051647999737073025726173300915916758770511497524353491642840238968166849681827669150543335788616727518429916536945395813

思路：

低加密指数攻击

exp :

from Crypto.Util.number import *
import gmpy2e = 3
n = 135112325288715136727832177735512070625083219670480717841817583343851445454356579794543601926517886432778754079508684454122465776544049537510760149616899986522216930847357907483054348419798542025184280105958211364798924985051999921354369017984140216806642244876998054533895072842602131552047667500910960834243
c = 13037717184940851534440408074902031173938827302834506159512256813794613267487160058287930781080450199371859916605839773796744179698270340378901298046506802163106509143441799583051647999737073025726173300915916758770511497524353491642840238968166849681827669150543335788616727518429916536945395813m = gmpy2.iroot(c, e)[0]print(long_to_bytes(m))

Vinegar 2

题目：

  • chall.py

    alphanumerical = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890!@#$%^&*(){}_?'
matrix = []
for i in alphanumerical:matrix.append([i])idx=0
for i in alphanumerical:matrix[idx][0] = (alphanumerical[idx:len(alphanumerical)]+alphanumerical[0:idx])idx += 1flag=open('../src/flag.txt').read().strip()
key='5up3r_s3cr3t_k3y_f0r_1337h4x0rs_r1gh7?'
assert len(key)==len(flag)
flag_arr = []
key_arr = []
enc_arr=[]
for y in flag:for i in range(len(alphanumerical)):if matrix[i][0][0]==y:flag_arr.append(i)for y in key:for i in range(len(alphanumerical)):if matrix[i][0][0]==y:key_arr.append(i)for i in range(len(flag)):enc_arr.append(matrix[flag_arr[i]][0][key_arr[i]])
encrypted=''.join(enc_arr)
f = open('enc.txt','w')
f.write(encrypted)

  • enc.txt

    *fa4Q(}$ryHGswGPYhOC{C{1)&_vOpHpc2r0({

exp:

# 定义字符集
alphanumerical = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890!@#$%^&*(){}_?'# 构建矩阵
matrix = []
for i in alphanumerical:matrix.append([i])idx = 0
for i in alphanumerical:matrix[idx][0] = (alphanumerical[idx:len(alphanumerical)] + alphanumerical[0:idx])idx += 1# 读取加密后的文件
with open('enc.txt', 'r') as f:encrypted = f.read().strip()# 定义key
key = '5up3r_s3cr3t_k3y_f0r_1337h4x0rs_r1gh7?'# 将key转换为索引数组
key_arr = []
for y in key:for i in range(len(alphanumerical)):if matrix[i][0][0] == y:key_arr.append(i)break# 解密加密后的字符串
decrypted_arr = []
for i in range(len(encrypted)):enc_char = encrypted[i]key_idx = key_arr[i]for row in range(len(matrix)):if matrix[row][0][key_idx] == enc_char:decrypted_arr.append(matrix[row][0][0])break# 将解密后的字符数组转换为字符串
decrypted = ''.join(decrypted_arr)# 输出解密后的flag
print(decrypted)

Random

题目：

I hid my password behind an impressive sorting machine. The machine is very luck based, or is it?!?!?!?

我把密码藏在一台令人印象深刻的分拣机后面。这台机器非常靠运气，或者是吗?!?!?!?

#include<chrono>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<string>
#include<fstream>
#include<thread>
#include<map>
using namespace std;bool amazingcustomsortingalgorithm(string s) {int n = s.size();for (int i = 0; i < 69; i++) {cout << s << endl;bool good = true;for (int i = 0; i < n - 1; i++)good &= s[i] <= s[i + 1];if (good)return true;random_shuffle(s.begin(), s.end());this_thread::sleep_for(chrono::milliseconds(500));}return false;
}int main() {string s;getline(cin, s);map<char, int> counts;for (char c : s) {if (counts[c]) {cout << "no repeating letters allowed passed this machine" << endl;return 1;}counts[c]++;}if (s.size() < 10) {cout << "this machine will only process worthy strings" << endl;return 1;}if (s.size() == 69) {cout << "a very worthy string" << endl;cout << "i'll give you a clue'" << endl;cout << "just because something says it's random mean it actually is" << endl;return 69;}random_shuffle(s.begin(), s.end());if (amazingcustomsortingalgorithm(s)) {ifstream fin("flag.txt");string flag;fin >> flag;cout << flag << endl;}else {cout << "UNWORTHY USER DETECTED" << endl;}
}

‍

Misc

Addition

题目：

import time
import randomquestions = int(input("how many questions do you want to answer? "))for i in range(questions):a = random.randint(0, 10)b = random.randint(0, 10)yourans = int(input("what is " + str(a) + ' + ' + str(b) + ' = '))print("calculating")totaltime = pow(2, i)print('.')time.sleep(totaltime / 3)print('.')time.sleep(totaltime / 3)print('.')time.sleep(totaltime / 3)if yourans != a + b:print("You made my little brother cry")exit(69)f = open('/flag.txt', 'r')
flag = f.read()
print(flag[:questions])

思路： 很明显计算对一次就会读取一个 flag 的值，但是有个问题，延迟的时间呈指数级增长，做 8 个以上的问题很痛苦，我以开始没注意到，写了个脚本一直跑，出不来，最后看了一下题目，发现输入 -1 可以绕过这个循环， print(flag[:-1]) 也可以把 flag 打印出来

​image-20240805235532-qzmqkzx​

exp: 错误的

import pwn
import re
import timep = pwn.remote('24.199.110.35', 42189)# 接收
bytes = p.recv(1024)
print(bytes)# 发送
p.sendline(b'20')bytes = p.recv(1024)
print(bytes)
# b'what is 8 + 10 = ' 写个正则表达式匹配 8 + 10 = 
# eval_result = re.findall(r'\d+', bytes.decode())
# sum = int(eval_result[0]) + int(eval_result[1])
# print(sum)for i in range(20):eval_result = re.findall(r'\d+', bytes.decode())sum = int(eval_result[0]) + int(eval_result[1])print(sum)p.sendline(str(sum).encode())# time.sleep(1)bytes = p.recv(1024)print(bytes)

web

Passwordless

题目：

#!/usr/bin/env python3
from flask import Flask, request, redirect, render_template, render_template_string
import subprocess
import urllib
import uuid
global leetapp = Flask(__name__)
flag = open('/flag.txt').read()
leet=uuid.UUID('13371337-1337-1337-1337-133713371337')@app.route('/',methods=['GET','POST'])
def main():global usernameif request.method == 'GET':return render_template('index.html')elif request.method == 'POST':username = request.values['username']if username == 'admin123':return 'Stop trying to act like you are the admin!'uid = uuid.uuid5(leet,username) # super secure!return redirect(f'/{uid}')@app.route('/<uid>')
def user_page(uid):if uid != str(uuid.uuid5(leet,'admin123')):return f'Welcome! No flag for you :('else:return flagif __name__ == '__main__':app.run(host='0.0.0.0', port=1337)

思路：

看路由  **/&lt;uid&gt;**  ，当 uid = uuid.uuid5(leet,'admin123') 时可以 return flag，

学习:

[探索 Python 中的 uuid 模块：生成唯一标识符_python uuid-CSDN博客](https://blog.csdn.net/Trb601012/article/details/135744410)

exp：

所以我们本地可以生成这个 uid

import uuidleet=uuid.UUID('13371337-1337-1337-1337-133713371337')
uid = uuid.uuid5(leet,"admin123")
print(uid)

​image-20240805234610-o4a24xu​

Rev

Brain

题目：

+++++++++++[<++++++++++>-]<[-]>++++++++[<++++++>-]<[-]>++++++++[<++++++>-]<[-]>++++++++++++++[<+++++++>-]<[-]>+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++[<++>-]<[-]>+++++++++++++++++++++++++++++++++++++++++[<+++>-]<[-]>+++++++[<+++++++>-]<[-]>+++++++++++++++++++[<+++++>-]<[-]>+++++++++++[<+++++++++>-]<[-]>+++++++++++++[<++++>-]<[-]>+++++++++++[<++++++++++>-]<[-]>+++++++++++++++++++[<+++++>-]<[-]>+++++++++++[<+++++++++>-]<[-]>++++++++[<++++++>-]<[-]>++++++++++[<++++++++++>-]<[-]>+++++++++++++++++[<+++>-]<[-]>+++++++++++++++++++[<+++++>-]<[-]>+++++++[<+++++++>-]<[-]>+++++++++++[<++++++++++>-]<[-]>+++++++++++++++++++[<+++++>-]<[-]>++++++++++++++[<+++++++>-]<[-]>+++++++++++++++++++[<++++++>-]<[-]>+++++++++++++[<++++>-]<[-]>+++++++[<+++++++>-]<[-]>+++++++++++[<++++++++++>-]<[-]>+++++++++++++++++[<++++++>-]<[-]>+++++++[<++++++>-]<[-]>+++++++++++[<+++++++++>-]<[-]>+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++[<+>-]<[-]>+++++++++++[<+++>-]<[-]>+++++++++++++++++++++++++[<+++++>-]<[-]

思路：

Brainfuck 解密，解密失败，说明需要将给的BF转成python代码，

学习：​记一次brainfuck的解密思路 - 『脱壳破解区』 - 吾爱破解 - LCG - LSG |安卓破解|病毒分析|www.52pojie.cn

exp:

def shrinkBFCode(code):cPos2Vars = {}   #位置对应的变量cPos2Change = {}  #位置中 + 号 增加的值varPos = 0nCode = []incVal = 0lc = NonedataChangeOp = set(['+', '-'])dataShiftOp = set(['>', '<'])for i in range(len(code)):c = code[i]if c not in dataChangeOp and lc in dataChangeOp:cPos2Change[len(nCode)] = incValcPos2Vars[len(nCode)] = varPosnCode.append('+')incVal = 0if c == '>':varPos += 1elif c == '<':varPos -= 1else:if c in dataChangeOp:incVal += 1 if c == '+' else -1else:#if lc == '>' or lc == '<':#    cPos2Vars[len(nCode)] = varPoscPos2Vars[len(nCode)] = varPosnCode.append(c)lc = creturn ''.join(nCode), cPos2Vars, cPos2Change
def generatePyCode(shellCode, pVars, pChange):pyCodes = []bStacks = []whileVarCache = {}for i, c in enumerate(shellCode):d_pos = i if i not in pVars else pVars[i]d_change = 1 if i not in pChange else pChange[i]indentLevel = len(bStacks)indentStr = ' '*(4*indentLevel)if c == '[':pyCodes.append('{}while data[{}] != 0:'.format(indentStr, d_pos))bStacks.append((c, i))whileVarCache[i] = {}elif c == ']':if bStacks[-1][0] != '[':raise Exception('miss match of {}] found between {} and {}'.format(bStacks[-1][0], bStacks[-1][1], i))cNum = i-bStacks[-1][1]if cNum == 2:del pyCodes[-1]del pyCodes[-1]d_pos_l = i-1 if i-1 not in pVars else pVars[i-1]pyCodes.append('{}data[{}] = 0'.format(' '*(4*(indentLevel-1)), d_pos_l))whileCode = shellCode[bStacks[-1][1]+1 : i]if cNum>2 and '[' not in whileCode and not '%' in whileCode:  # nested loop is a bit complicated, just skiploopCondvar = bStacks[-1][1]d_pos_l = loopCondvar if loopCondvar not in pVars else pVars[loopCondvar]whileVars = whileVarCache[bStacks[-1][1]]cVarChange = whileVars[d_pos_l]# remove statement of same indentwhile len(pyCodes)>0 and pyCodes[-1].startswith(indentStr) and pyCodes[-1][len(indentStr)]!=' ':pyCodes.pop()pyCodes.pop()#del pyCodes[bStacks[-1][1]-i:]for vPos, vChange in whileVars.items():if vPos == d_pos_l:continuectimes = abs(vChange / cVarChange)ctimesStr = '' if ctimes==1 else '{}*'.format(ctimes)cSign = '+' if vChange > 0 else '-'pyCodes.append('{}data[{}] {}= {}data[{}]'.format(' '*(4*(indentLevel-1)),vPos, cSign,  ctimesStr, d_pos_l))pyCodes.append('{}data[{}] = 0'.format(' '*(4*(indentLevel-1)), d_pos_l))del whileVarCache[bStacks[-1][1]]bStacks.pop()elif c == '.':pyCodes.append('{}print(data[{}])'.format(indentStr, d_pos))elif c == ',':pyCodes.append('{}data[{}] = ord(stdin.read(1))'.format(indentStr, d_pos))elif c == '+':opSign = '-=' if d_change < 0 else '+='if pyCodes and pyCodes[-1] == '{}data[{}] = 0'.format(indentStr, d_pos):pyCodes[-1] = '{}data[{}] = {}'.format(indentStr, d_pos, d_change)else:pyCodes.append('{}data[{}] {} {}'.format(indentStr, d_pos, opSign, abs(d_change)))if bStacks:whileVarCache[bStacks[-1][1]].setdefault(d_pos, 0)whileVarCache[bStacks[-1][1]][d_pos] += d_changeelif c == '-':opSign = '+=' if d_change < 0 else '-='if pyCodes and pyCodes[-1] == '{}data[{}] = 0'.format(indentStr, d_pos):pyCodes[-1] = '{}data[{}] = {}'.format(indentStr, d_pos, -d_change)else:pyCodes.append('{}data[{}] {} {}'.format(indentStr, d_pos, opSign, abs(d_change)))if bStacks:whileVarCache[bStacks[-1][1]].setdefault(d_pos, 0)whileVarCache[bStacks[-1][1]][d_pos] -= d_changeelif c == '%':pyCodes.append('{}data[{}] %= data[{}]'.format(indentStr, d_pos, d_pos+1))return '\n'.join(pyCodes)s = '>+++++++++++[<++++++++++>-]<[-]>++++++++[<++++++>-]<[-]>++++++++[<++++++>-]<[-]>++++++++++++++[<+++++++>-]<[-]>+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++[<++>-]<[-]>+++++++++++++++++++++++++++++++++++++++++[<+++>-]<[-]>+++++++[<+++++++>-]<[-]>+++++++++++++++++++[<+++++>-]<[-]>+++++++++++[<+++++++++>-]<[-]>+++++++++++++[<++++>-]<[-]>+++++++++++[<++++++++++>-]<[-]>+++++++++++++++++++[<+++++>-]<[-]>+++++++++++[<+++++++++>-]<[-]>++++++++[<++++++>-]<[-]>++++++++++[<++++++++++>-]<[-]>+++++++++++++++++[<+++>-]<[-]>+++++++++++++++++++[<+++++>-]<[-]>+++++++[<+++++++>-]<[-]>+++++++++++[<++++++++++>-]<[-]>+++++++++++++++++++[<+++++>-]<[-]>++++++++++++++[<+++++++>-]<[-]>+++++++++++++++++++[<++++++>-]<[-]>+++++++++++++[<++++>-]<[-]>+++++++[<+++++++>-]<[-]>+++++++++++[<++++++++++>-]<[-]>+++++++++++++++++[<++++++>-]<[-]>+++++++[<++++++>-]<[-]>+++++++++++[<+++++++++>-]<[-]>+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++[<+>-]<[-]>+++++++++++[<+++>-]<[-]>+++++++++++++++++++++++++[<+++++>-]<[-]'shrinkCode, pVars, pChange = shrinkBFCode(s)
res = generatePyCode(shrinkCode, pVars, pChange)# 将结果写入文件，
with open('flag.txt', 'w') as f:f.write(res)print(res)

将输出结果分析一下，就是一直在计算 d[0] 的值，处理一下就好

# 初始化data列表
data = [0, 0]# 存储每次data[0]的值
data_values = []data[1] = 0
data[0] = 0
data[1] += 11
data[0] += 10.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 8
data[0] += 6.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 8
data[0] += 6.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 14
data[0] += 7.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 61
data[0] += 2.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 41
data[0] += 3.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 7
data[0] += 7.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 19
data[0] += 5.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 11
data[0] += 9.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 13
data[0] += 4.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 11
data[0] += 10.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 19
data[0] += 5.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 11
data[0] += 9.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 8
data[0] += 6.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 10
data[0] += 10.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 17
data[0] += 3.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 19
data[0] += 5.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 7
data[0] += 7.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 11
data[0] += 10.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 19
data[0] += 5.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 14
data[0] += 7.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 19
data[0] += 6.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 13
data[0] += 4.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 7
data[0] += 7.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 11
data[0] += 10.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 17
data[0] += 6.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 7
data[0] += 6.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 11
data[0] += 9.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 107
data[0] += data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 11
data[0] += 3.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0
data[1] += 25
data[0] += 5.0*data[1]
data_values.append(data[0])data[1] = 0
data[0] = 0flag = ""
for i in range(len(data_values)):flag += chr(int(data_values[i]))print(flag)

flag： n00bz{1_c4n_c0d3_1n_br41nf*ck!}

‍

Forensics

Plane

题目：

So many plane-related challenges! Why not another one? The flag is the latitude, longitude of the place this picture is taken from, rounded upto two decimal places. Example: n00bz{55.51,-20.27}. Author: NoobMaster Author: NoobMaster

exp：

先下载图片：

curl https://static.n00bzunit3d.xyz/Forensics/Plane/plane.jpg -o plane.jpg

在查看图片信息

 exiftool plane.jpg

​image-20240806004346-55ar9na​

Wave

题目：

The Wave is not audible, perhaps corrupted? Note: Wrap the flag in n00bz{}. There are no spaces in the flag and it is all lowercase.

学习：​【音频处理】WAV 文件格式分析 ( 逐个字节解析文件头 | 相关字段的计算公式 )_52 49 46 46-CSDN博客

exp:

先下载这个音频

curl https://static.n00bzunit3d.xyz/Forensics/Wave/chall.wav -o wave.wav

我们打开时发现现在被损坏了

​image-20240806004937-62ujmug​

010 查看一下 16进制，不难发现，标志位都被换成了 00

​image-20240806161816-n8zfbva​

修改后，然后播放可以明显听到是摩斯电码，使用在线网站解即可

​image-20240806162436-5ec1aj5​

​image-20240806163319-2og40bd​

flag： n00bz{beepbopmorsecode}

Disk Golf

题目：

Let's play some disk golf!

思路：本来想使用 volatility2.6 ，但是无用，想到 fls 工具，就搜索到 flag

fls disk.img | grep "flag"

-r：递归地列出所有文件和目录。

-p：显示文件的路径。

​image-20240807023220-f9uz68x​

在使用 icat 命令读取数据

​image-20240807023612-uodhfgu​

​image-20240807024431-k2n9sjh​

flag： n00bz{7h3_l0ng_4w41t3d_d15k_f0r3ns1c5}

‍

本文来自互联网用户投稿，该文观点仅代表作者本人，不代表本站立场。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如若转载，请注明出处：http://www.hqwc.cn/news/779048.html

如若内容造成侵权/违法违规/事实不符，请联系编程知识网进行投诉反馈email:809451989@qq.com，一经查实，立即删除！

相关文章

两个coca略有不同词频文件 比较

两个coca略有不同词频文件 比较

coca20000xlsxFuzhi.a应该也是这样弄的.txt COCA60000.txt确实有一些词顺序不同，不知道为什么。。。
阅读更多...
【第2期】2024 搜索客 Meetup | Elasticsearch 的代码结构和写入查询流程的解读

【第2期】2024 搜索客 Meetup | Elasticsearch 的代码结构和写入查询流程的解读

本次活动由 搜索客社区、极限科技（INFINI Labs）联合举办，活动主题将深入探讨 Elasticsearch 的两个核心方面：代码结构以及写入和查询的关键流程。本次活动将为 Elasticsearch 初学者和有经验的用户提供宝贵的见解，欢迎大家报名参加、交流学习。 活动主题：Elasticsearch 的…
阅读更多...
【第2期】2024 搜索客 Meetup | Elasticsearch 的代码结构和写入查询流程的解读.md

【第2期】2024 搜索客 Meetup | Elasticsearch 的代码结构和写入查询流程的解读.md

本次活动由 搜索客社区、极限科技（INFINI Labs）联合举办，活动主题将深入探讨 Elasticsearch 的两个核心方面：代码结构以及写入和查询的关键流程。本次活动将为 Elasticsearch 初学者和有经验的用户提供宝贵的见解，欢迎大家报名参加、交流学习。 活动主题：Elasticsearch 的…
阅读更多...
使用黑群晖webdav服务同步obsidian笔记

使用黑群晖webdav服务同步obsidian笔记

前言 本文使用的黑群晖套件 webdav server进行黑群晖配置，接着使用obsidian插件将笔记同步到黑群晖上 使用的黑群晖已经经过内网穿透。 配置黑群晖 配置 webdav 并创建共享文件夹 首先下载套件，在套件中心搜索webdav server ，下载安装并打开进行配置此处已经安装好了，打开，…
阅读更多...
文件的读写

文件的读写

1.判断文件夹是否存在 2.文件的数据写入 3.文件内容的读取 4.文件的复制，移动，删除 5.log日志的记录 6.三种序列化器：
阅读更多...
Apache Answer Plugin 开发笔记

Apache Answer Plugin 开发笔记

0x00 引言 在前两天的 Apache CommunityOverCode Asia 2024 大会上认识到了一个很有意思的项目：Apache Answer。这是一个 Apache 开源问答社区系统，可以帮助开发者在社区中快速找到答案。尝试为这个项目做一些贡献，整理一下开发笔记。 领取的任务: incubator-answer-plugins…
阅读更多...
2024 暑假集训笔记

2024 暑假集训笔记

Day 1搜索 DFS： 基本思路：边界设置(if(...) return ;) 检查(for(....)) 标记结果(a[...]=...) 递归搜索(dfs(...)) 回溯例题： N皇后问题 [POJ 1321] 棋盘问题 [POJ 1011] Sticks BFS 略分治 基本思路将处理区间二分为两个区间(merge(l,mid) merge(mid+1,r)) 边界(if(l==r) r…
阅读更多...
macbook开机登录时输入正确的密码却提示密码错误

macbook开机登录时输入正确的密码却提示密码错误

背景 Macbook正常的账户登录，提示密码错误，但是我输入的密码确认了几次都是完全正确的，就是死活登不上去一直显示密码错误，一搜索原来是操作系统BUG，有对应解决方案。 原因 系统macOS Monterey 12.0.1设计的原因，导致有某些时刻可能忘记你的开机密码，导致你输入正确的密…
阅读更多...
中信证券抓取（页码范围）

中信证券抓取（页码范围）

中信证券抓取（页码范围） 创建时间：2024年8月5日 一、完整代码 import re import requests from lxml import etreeheaders = {User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/120.0.0.0 Safari/537.36, } url1 = ht…
阅读更多...
三国演义内容抓取（诗词名句网）

三国演义内容抓取（诗词名句网）

三国演义内容抓取（诗词名句网） 时间：2024-08-06 一、完整代码 import random import timeimport requests from lxml import etreefour_famous_novels = https://www.shicimingju.com/bookmark/sidamingzhu.html # 四大名著在线阅读地址 three_kingdoms = https://www.shic…
阅读更多...
表情党抓取（单页） (网站已转移)

表情党抓取（单页） (网站已转移)

表情党抓取（单页） (网站已转移) 创建时间：2024-08-06 一、完整代码 import requests from lxml import etree url = https://qq.yh31.com/xq/wq/ header = {User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/125.0.0.…
阅读更多...
腾牛网抓取（单页）

腾牛网抓取（单页）

腾牛网抓取（单页） 创建时间：2024-08-05 一、完整代码 import requests from lxml import etree url = https://www.qqtn.com/wm/meinvtp_1.html header = {User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/125.0.0.0…
阅读更多...
推荐文章
最新文章

Copyright @ 2022~2023