WEB:gift_F12
没啥好说的 直接F12得了
NSSCTF{We1c0me_t0_WLLMCTF_Th1s_1s_th3_G1ft}
RE
简简单单的解密
import base64, urllib.parse
def e():
key = "HereIsFlagggg"
flag = "xxxxxxxxxxxxxxxxxxx"
s_box = list(range(256))
j = 0
# 打乱s_box 和flag无关
for i in range(256):
j = (j + s_box[i] + ord(key[i % len(key)])) % 256
s_box[i], s_box[j] = s_box[j], s_box[i]
res = []
i = j = 0
for s in flag:
i = i + 1
j = (j + s_box[i]) % 256
s_box[i], s_box[j] = s_box[j], s_box[i]
t = (s_box[i] + s_box[j]) % 256
k = s_box[t]
res.append(chr(ord(s) ^ k))
enc = "".join(res)
#原来这里有个b64加密再解密 就等于没变 所以改了一下
enc = urllib.parse.quote(enc)
print(enc)
def d():
enc = "%C2%A6n%C2%87Y%1Ag%3F%C2%A01.%C2%9C%C3%B7%C3%8A%02%C3%80%C2%92W%C3%8C%C3%BA"
enc = urllib.parse.unquote(enc)#倒着往回做呗
key = "HereIsFlagggg"
s_box = list(range(256))
j = 0
r=[]
# 获得s_box
for i in range(256):
j = (j + s_box[i] + ord(key[i % len(key)])) % 256
s_box[i], s_box[j] = s_box[j], s_box[i]
i,j=0,0
for s in enc:
i = i + 1
j = (j + s_box[i]) % 256
s_box[i], s_box[j] = s_box[j], s_box[i]
t = (s_box[i] + s_box[j]) % 256
k = s_box[t]
r.append(chr(ord(s)^k))#基本直接复制就行 异或的逆运算就是再异或一次
print("".join(r))
if __name__ == '__main__':
e()
d()
脚本运行得到结果
NSSCTF{REAL_EZ_RC4}
简简单单的逻辑
先上脚本
def e():
flag = 'xxxxxxxxxxxxxxxxxx'
list = [47, 138, 127, 57, 117, 188, 51, 143, 17, 84, 42, 135, 76, 105, 28, 169, 25]
result = ''
for i in range(len(list)):
key = (list[i] >> 4) + ((list[i] & 0xf) << 4)
a = hex(ord(flag[i]) ^ key)
result += str(a)[2:].zfill(2)
#分析得知 这是16进制补全两位 所以是一个flag字符对应两位
print(result)
def d():
list = [47, 138, 127, 57, 117, 188, 51, 143, 17, 84, 42, 135, 76, 105, 28, 169, 25]
result = "bcfba4d0038d48bd4b00f82796d393dfec"
nums = []
#截取 转换 得到原来十进制
for i in range(0,len(result),2):
nums.append("0x" + result[i:i+ 2])
for index in range(len(nums)):
nums[index]=int(nums[index],16)
#直接加密里面复制key的计算 然后再异或一次即可
for i in range(len(list)):
key = (list[i] >> 4) + ((list[i] & 0xf) << 4)
print(chr(nums[i]^key),end="")
if __name__ == '__main__':
# e()
d()
这看脚本就行了
NSSCTF{EZEZ_RERE}
非常简单的逻辑题
def e():
flag = 'xxxxxxxxxxxxxxxxxxxxx'
s = 'wesyvbniazxchjko1973652048@$+-&*<>'
result = ''
for i in range(len(flag)):
s1 = ord(flag[i]) // 17
s2 = ord(flag[i]) % 17
#就 没啥好讲的 一点点拆开打断点调试就能看懂计算过程
#还是flag一个字符转换对应result的两个字符
a = s[(s1 + i) % 34]
b = s[-(s2 + i + 1) % 34]
result += a + b
print(result)
def d():
result = 'v0b9n1nkajz@j0c4jjo3oi1h1i937b395i5y5e0e$i'
s = 'wesyvbniazxchjko1973652048@$+-&*<>'
#套这么多层循环 主要是我不知道 取余怎么更方便的算回去
for x in range(-5, 5):
for y in range(-5, 5):
r = ""
for j in range(0, len(result), 2):
a = result[j]
b = result[j + 1]
#根据数学知识可得(
s1 = int(s.index(a) + 34 * x - j / 2)
s2 = -1*int(s.index(b) + 34 * y + j / 2 + 1)
#去除没啥意义的结果
if 32 < s1 * 17 + s2 < 130:
r += chr(s1 * 17 + s2)
else:
break
if len(r)==len('xxxxxxxxxxxxxxxxxxxxx'):
print(r)
if __name__ == '__main__':
# e()
d()
NSSCTF{Fake_RERE_QAQ}
fakerondom
import random
def e():
flag = 'xxxxxxxxxxxxxxxxxxxx'
#别的都不重要 就是这个随机数种子
#同样的随机数种子生成的一定是相同的数字
random.seed(1)
l = []
for i in range(4):
l.append(random.getrandbits(8))
result = []
for i in range(len(l)):
random.seed(l[i])
for n in range(5):
result.append(ord(flag[i * 5 + n]) ^ random.getrandbits(8))
print(result)
def d():
result = [201, 8, 198, 68, 131, 152, 186, 136, 13, 130, 190, 112, 251, 93, 212, 1, 31, 214, 116, 244]
random.seed(1)
ran=[]
t=[]
for x in range(4):
ran.append(random.getrandbits(8))
l=ran[0:4]
for i in range(len(l)):
random.seed(l[i])
for n in range(5):
#异或题最好写了 基本原样复制代码就行
print(chr(result[i * 5 + n]^ random.getrandbits(8)),end="")
if __name__ == '__main__':
e()
d()
NSSCTF{FakeE_random}
fakebase
def e():
flag = 'xxxxxxxxxxxxxxxxxxx'
s_box = 'qwertyuiopasdfghjkzxcvb123456#$'
tmp = ''
for i in flag:
x = bin(ord(i))
tmp += str(x)[2:].zfill(8)
b1 = int(tmp, 2)
s = ''
while b1 // 31 != 0:
c = b1 % 31
s += s_box[c]
b1 = b1 // 31
print(b1)
print(s)
def d():
s = "u#k4ggia61egegzjuqz12jhfspfkay"
# s="j3d4h1$6ggrouxktrgky5sxv6bk6i5"#测试
s_box = 'qwertyuiopasdfghjkzxcvb123456#$'
l = []
for x in s:
l.append(s_box.index(x))
l=l[::-1]
for n in range(0,31):
for x in l:
n=x+31*n
t=str(bin(n))[2:]
for j in range(0,len(t),8):
#这题最让我迷惑 我觉得我写的是对的
#但是拿上面的加密结果再解密出来的ascii值总差一倍
#但最后一个字符的ascii值又是对应的 所以只能除以2看看
#所以就 现在挺迷茫的
print(chr(int(int(t[j:j+8],2)/2)),end="")
print()
if __name__ == '__main__':
# e()
d()
NSSCTF{WHAt_BASe31}
astjs
经过查资料知道 这是把js文件转换为ast 但是我不知道怎么转回去
解不出 我太菜了
简 单 的 逻 辑
我是 san兵
我真看不出
出题人说少给条件了 但可以琼剧
密码
带rsa的都不会 萌新还没学呢
crypto6
var="************************************"
flag='NSSCTF{' + base64.b16encode(base64.b32encode(base64.b64encode(var.encode()))) + '}'
print(flag)
小明不小心泄露了源码，输出结果为：4A5A4C564B36434E4B5241544B5432454E4E32465552324E47424758534D44594C4657564336534D4B5241584F574C4B4B463245365643424F35485649534C584A5A56454B4D4B5049354E47593D3D3D，你能还原出var的正确结果吗？
会用base64库就行
import base64
x="4A5A4C564B36434E4B5241544B5432454E4E32465552324E47424758534D44594C4657564336534D4B5241584F574C4B4B463245365643424F35485649534C584A5A56454B4D4B5049354E47593D3D3D"
a=base64.b16decode(x)
a=base64.b32decode(a)
a=base64.b64decode(a)
print(a)
NSSCTF{5e110989-dc43-1bd3-00b4-9009206158fe}
Crypto7
69f7906323b4f7d1e4e972acf4abfbfc
直接md5解密
NSSCTF{md5yyds}
crypto8
介绍pyhton库:ciphey库! 人工智能 yyds
直接一把梭
终端执行命令
ciphey -t "73E-30U1&>V-H965S95]I<U]P;W=E<GT`"
这个题用了uuencode
crypto9
题干脚本直接python执行
letter_list = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' # 字母表
# 根据输入的key生成key列表
def Get_KeyList(key):
key_list = []
for ch in key:
key_list.append(ord(ch.upper()) - 65)
return key_list
# 加密函数
def Encrypt(plaintext, key_list):
ciphertext = ""
i = 0
for ch in plaintext: # 遍历明文
if 0 == i % len(key_list):
i = 0
if ch.isalpha(): # 明文是否为字母,如果是,则判断大小写,分别进行加密
if ch.isupper():
ciphertext += letter_list[(ord(ch) - 65 + key_list[i]) % 26]
i += 1
else:
ciphertext += letter_list[(ord(ch) - 97 + key_list[i]) % 26].lower()
i += 1
else: # 如果密文不为字母,直接添加到密文字符串里
ciphertext += ch
return ciphertext
# 解密函数
def Decrypt(ciphertext, key):
plaintext = ""
i = 0
for ch in ciphertext: # 遍历密文
if 0 == i % len(key_list):
i = 0
if ch.isalpha(): # 密文为否为字母,如果是,则判断大小写,分别进行解密
if ch.isupper():
plaintext += letter_list[(ord(ch) - 65 - key_list[i]) % 26]
i += 1
else:
plaintext += letter_list[(ord(ch) - 97 - key_list[i]) % 26].lower()
i += 1
else: # 如果密文不为字母,直接添加到明文字符串里
plaintext += ch
return plaintext
if __name__ == '__main__':
print("加密请按D,解密请按E:")
user_input = input();
while (user_input != 'D' and user_input != 'E'): # 输入合法性判断
print("输入有误!请重新输入:")
user_input = input()
print("请输入密钥:")
key = input()
while (False == key.isalpha()): # 输入合法性判断
print("输入有误!密钥为字母,请重新输入:")
key = input()
key_list = Get_KeyList(key)
if user_input == 'D':
# 加密
print("请输入明文:")
plaintext = input()
ciphertext = Encrypt(plaintext, key_list)
print("密文为:\n%s" % ciphertext)
else:
# 解密
print("请输入密文:")
ciphertext = input()
plaintext = Decrypt(ciphertext, key_list)
print("明文为:\n%s" % plaintext)
试了试密码NSS解密成功
Crypto10
当然你可以接着用这个
ciphey -t "AFFPGS{pbatenghyngvbaf!!!}"
NSSCTF{congratulations!!!}
ez_caesar
def caesar(plaintext):
str_list = list(plaintext)
i = 0
while i < len(plaintext):
if not str_list[i].isalpha():
str_list[i] = str_list[i]
else:
a = "A" if str_list[i].isupper() else "a"
str_list[i] = chr((ord(str_list[i]) - ord(a) + 5) % 26 + ord(a) or 5)
i = i + 1
return ''.join(str_list)
if __name__ == '__main__':
small = [chr(i) for i in range(97, 123)]
big = [chr(i) for i in range(65, 91)]
alpha = small + big
dic = {}
for flag in alpha:
str = caesar(flag)
dic[str] = flag
str = "SXXHYK{dtzmfajpstbhfjxfw}"
for x in str:
if x.isalpha():
print(dic[x], end="")
else:
print(x, end="")
分析代码知道 这是个静态加密 所以直接先把所有字母对应的密文加到字典里面 然后一一对应
NSSCTF{youhaveknowcaesar}
ez_rsa
为数不多我会写的rsa了
工具算个d 然后再md5加密即可
d=104550CB8E2144921
pigpig
看名字就知道了 所以不想写
traditional
题目：
西方的二进制数学的发明者莱布尼茨，从中国的八卦图当中受到启发，演绎并推论出了数学矩
阵，
最后创造的二进制数学。二进制数学的诞生为计算机的发明奠定了理论基础。而计算机现在改
变
了我们整个世界，改变了我们生活，而他的源头却是来自于八卦图。现在，给你一组由八卦图
方位
组成的密文，你能破解出其中的含义吗？
震坤艮 震艮震 坤巽坤 坤巽震 震巽兑 震艮震 震离艮 震离艮
格式：NSSCTF{}
我想说 差不多得了(
八卦从内向外看 --代表0 —代表1
inp= "震坤艮 震艮震 坤巽坤 坤巽震 震巽兑 震艮震 震离艮 震离艮"
dic={"震":"100","离":"101","兑":"110","乾":"111","坤":"000","艮":"001","坎":"010","巽":"011"}
results= ""
for result in inp:
if result in dic.keys():
results+= dic[result][::-1]+""
else:
results+= " "
print(results)
results=results.split()
for index in range(0,len(results)):
print(chr(int(results[index], 2)), end="")
写成二进制 然后转十进制 然后ascii码
NSSCTF{Da01sall}
misc
你喜欢osu吗？
我喜欢 但不至强行安利 真的(
OSU是一个windows的音游 要是想玩这个谱的话 可以去songs文件夹新建一个文件夹 然后把两个文件扔进去 你就能玩了
复制底下的铺面信息 对 就那一堆数字
224,224,0,5,0,0000
224,160,500,1,8,0000
224,160,1000,1,8,0000
224,160,1500,1,8,0000
224,160,2000,1,8,0000
224,224,2500,1,0,0000
224,160,3000,5,8,0000
224,160,3500,1,8,0000
224,224,4000,1,0,0000
224,224,4500,1,0,0000
224,160,5000,1,8,0000
224,224,5500,1,0,0000
224,224,6000,5,0,0000
224,224,6500,1,0,0000
224,224,7000,1,0,0000
224,160,7500,1,8,0000
224,224,8000,1,0,0000
224,160,8500,1,8,0000
224,160,9000,5,8,0000
224,160,9500,1,8,0000
224,224,10000,1,0,0000
224,160,10500,1,8,0000
224,224,11000,1,0,0000
224,160,11500,1,8,0000
224,224,12000,5,0,0000
224,160,12500,1,8,0000
224,160,13000,1,8,0000
224,160,13500,1,8,0000
224,224,14000,1,0,0000
224,224,14500,1,0,0000
224,160,15000,5,8,0000
224,160,15500,1,8,0000
224,224,16000,1,0,0000
224,160,16500,1,8,0000
224,160,17000,1,8,0000
224,224,17500,1,0,0000
224,160,18000,5,8,0000
224,160,18500,1,8,0000
224,160,19000,1,8,0000
224,160,19500,1,8,0000
224,224,20000,1,0,0000
224,160,20500,1,8,0000
224,224,21000,5,0,0000
224,160,21500,1,8,0000
224,160,22000,1,8,0000
224,160,22500,1,8,0000
224,160,23000,1,8,0000
224,160,23500,1,8,0000
224,224,24000,5,0,0000
224,160,24500,1,8,0000
224,160,25000,1,8,0000
224,224,25500,1,0,0000
224,224,26000,1,0,0000
224,160,26500,1,8,0000
224,224,27000,5,0,0000
224,160,27500,1,8,0000
224,224,28000,1,0,0000
224,160,28500,1,8,0000
224,160,29000,1,8,0000
224,224,29500,1,0,0000
224,160,30000,5,8,0000
224,224,30500,1,0,0000
224,160,31000,1,8,0000
224,160,31500,1,8,0000
224,224,32000,1,0,0000
224,160,32500,1,8,0000
224,160,33000,5,8,0000
224,224,33500,1,0,0000
224,160,34000,1,8,0000
224,224,34500,1,0,0000
224,224,35000,1,0,0000
224,160,35500,1,8,0000
224,224,36000,5,0,0000
224,160,36500,1,8,0000
224,160,37000,1,8,0000
224,224,37500,1,0,0000
224,160,38000,1,8,0000
224,160,38500,1,8,0000
224,224,39000,5,0,0000
224,224,39500,1,0,0000
224,224,40000,1,0,0000
224,160,40500,1,8,0000
224,224,41000,1,0,0000
224,160,41500,1,8,0000
224,160,42000,5,8,0000
224,160,42500,1,8,0000
224,160,43000,1,8,0000
224,160,43500,1,8,0000
224,224,44000,1,0,0000
224,160,44500,1,8,0000
224,160,45000,5,8,0000
224,160,45500,1,8,0000
224,224,46000,1,0,0000
224,160,46500,1,8,0000
224,224,47000,1,0,0000
224,160,47500,1,8,0000
224,224,48000,5,0,0000
224,160,48500,1,8,0000
224,160,49000,1,8,0000
224,224,49500,1,0,0000
224,160,50000,1,8,0000
224,160,50500,1,8,0000
224,160,51000,5,8,0000
224,160,51500,1,8,0000
224,224,52000,1,0,0000
224,160,52500,1,8,0000
224,160,53000,1,8,0000
224,160,53500,1,8,0000
224,160,54000,5,8,0000
224,224,54500,1,0,0000
224,224,55000,1,0,0000
224,160,55500,1,8,0000
224,224,56000,1,0,0000
224,160,56500,1,8,0000
224,224,57000,5,0,0000
224,160,57500,1,8,0000
224,160,58000,1,8,0000
224,160,58500,1,8,0000
224,160,59000,1,8,0000
224,160,59500,1,8,0000
224,224,60000,5,0,0000
224,160,60500,1,8,0000
224,160,61000,1,8,0000
224,160,61500,1,8,0000
224,224,62000,1,0,0000
224,160,62500,1,8,0000
224,160,63000,5,8,0000
224,160,63500,1,8,0000
224,224,64000,1,0,0000
224,160,64500,1,8,0000
224,160,65000,1,8,0000
224,224,65500,1,0,0000
224,160,66000,5,8,0000
224,160,66500,1,8,0000
224,160,67000,1,8,0000
224,160,67500,1,8,0000
224,224,68000,1,0,0000
224,160,68500,1,8,0000
224,160,69000,5,8,0000
224,224,69500,1,0,0000
224,160,70000,1,8,0000
224,160,70500,1,8,0000
224,160,71000,1,8,0000
224,224,71500,1,0,0000
224,224,72000,5,0,0000
224,160,72500,1,8,0000
224,160,73000,1,8,0000
224,224,73500,1,0,0000
224,160,74000,1,8,0000
224,224,74500,1,0,0000
224,160,75000,5,8,0000
224,160,75500,1,8,0000
224,224,76000,1,0,0000
224,160,76500,1,8,0000
224,224,77000,1,0,0000
224,160,77500,1,8,0000
224,160,78000,5,8,0000
224,160,78500,1,8,0000
224,160,79000,1,8,0000
224,160,79500,1,8,0000
224,224,80000,1,0,0000
224,160,80500,1,8,0000
224,160,81000,5,8,0000
224,224,81500,1,0,0000
224,160,82000,1,8,0000
224,224,82500,1,0,0000
224,224,83000,1,0,0000
224,160,83500,1,8,0000
224,224,84000,5,0,0000
224,160,84500,1,8,0000
224,160,85000,1,8,0000
224,160,85500,1,8,0000
224,160,86000,1,8,0000
224,160,86500,1,8,0000
224,224,87000,5,0,0000
224,160,87500,1,8,0000
224,224,88000,1,0,0000
224,160,88500,1,8,0000
224,224,89000,1,0,0000
224,224,89500,1,0,0000
224,224,90000,5,0,0000
224,160,90500,1,8,0000
224,160,91000,1,8,0000
224,224,91500,1,0,0000
224,224,92000,1,0,0000
224,160,92500,1,8,0000
224,224,93000,5,0,0000
224,160,93500,1,8,0000
224,224,94000,1,0,0000
224,160,94500,1,8,0000
224,224,95000,1,0,0000
224,224,95500,1,0,0000
224,224,96000,5,0,0000
224,160,96500,1,8,0000
224,224,97000,1,0,0000
224,224,97500,1,0,0000
224,224,98000,1,0,0000
224,224,98500,1,0,0000
224,160,99000,5,8,0000
224,160,99500,1,8,0000
224,224,100000,1,0,0000
224,160,100500,1,8,0000
224,224,101000,1,0,0000
224,160,101500,1,8,0000
然后观察嘛 第二个数字 224转为0 160转为1 然后8位转换一个十进制 当ascii码用 最后来个翻转
import re
if __name__ == '__main__':
with open("a.txt", "r")as file:
s = file.read()
s = s.split("\n")
for index in range(0, len(s)):
s[index] = s[index].split(",")[1]
if s[index] == "224":
s[index] = "0"
else:
s[index] = "1"
s="".join(s)
zero_one=re.findall(".{8}",s)
r=[]
for x in zero_one:
r.append(chr(int(x,2)))
print("".join(r[::-1]))
here_is_a_bug
真的 我一定是非预期解
右键文件夹 使用杀毒软件杀毒(
打开得知flag
NSSCTF{oh_you_catch_the_bug}
原来你也玩原神
异世相遇 尽享美味!
file命令知道是个mp3
看了频谱文件 不是摩斯啥的
用MP3stego解密 密码留空就行
解密出一个txt文件 看内容有PK
后缀名改为.zip
里面一个txt
NSSCTF{So_you_also_play_Genshin_impact}
Mooooooooooorse
AU打开 你拿别的音频软件打开也行 敲个摩斯码
解密就行了
文件已经忘了 不想再敲了
Bill
看提示应该是个excel表格
后缀改为.xlsx 但是有密码
说不要被表象迷惑 猜测可能是有隐写或者伪加密
foremost分离得到压缩包
后缀名再改xlsx打开文件
然后修改一下excel表格 把所有的都换成中文大写数字之后 把脚本拿出来
(上次DAS也有算钱的题...那个表格还比这个长
import openpyxl
import re
def trad_to_int(money):
# 转换字典
trad_dict = {
'零':0,
'壹':1,
'贰':2,
'叁':3,
'肆':4,
'伍':5,
'陆':6,
'柒':7,
'捌':8,
'玖':9,
'拾':10,
'佰':100,
'仟':1000,
'万':10000,
'亿':100000000,
'角':0.1,
'分':0.01
}
trad = re.search(r"[零壹贰叁肆伍陆柒捌玖拾佰仟亿角分]+", money)
if trad is not None:
num = 0
add = 0
sum = 0
for i in money:
if i in ['零','壹','贰','叁','肆','伍','陆','柒','捌','玖']:
add = trad_dict[i]
sum = sum + add
elif i in ['拾','佰','仟','亿','角','分']:
num = add * trad_dict[i]
sum = sum - add
sum = sum + num
add = num
elif i == '万' or i == '亿':
sum = sum * trad_dict[i]
return float(sum)
else:
return money
def main():
submoneyCN = ["", "拾", "佰", "仟"]
numsCN = {"零": 0, "壹": 1, "贰": 2, "叁": 3, "肆": 4, "伍": 5, "陆": 6, "柒": 7, "捌": 8, "玖": 9}
w = openpyxl.open("00000025.xlsx")
ws = w.active
# 获取第一列 即单件商品金额
t = ws['A']
t1=[]
for x in range(1,len(t)):
t1.append(trad_to_int(t[x].value))
print(t1)
# 获取第二列并处理
t = ws['B']
t2=[]
for x in range(1,len(t)):
t2.append(trad_to_int(t[x].value))
print(t2)
r=0
for index in range(0,len(t2)):
r+=float(t1[index])*float(t2[index])
print(r)
if __name__ == '__main__':
main()
跑脚本
NSSCTF{5030782.26}
ZIPBOMB
我也不知道他为啥没炸 反正我直接解压了
010中得flag
NSSCTF{Z1p_B00m_d1sp0sal}
gif好像有点大
那确实挺大得 最开始以为是隐写
看了看gif发现他只是长而已...
拆分gif脚本
from PIL import Image
import os
pic=Image.open("CTF.gif")
try:
i=0
while True:
pic.seek(i)#搜寻文件帧数
pic.save(str(i)+".png")
i+=1
except:
pass
然后有一帧里面有个二维码 扫描即可
