新手区部分题解：

1，easy_RSA

2，Normal_RSA

3, 幂数加密

4,easy_ECC

高手进阶区部分题题解

5, ENC

6,告诉你个秘密

7，Easy-one

8，说我作弊需要证据

9.x_xor_md5

10，easy_crypto

11，cr2-many-time-secrets

12，oldDevier

13，wtc_rsa_bbq

14，cr4-poor-rsa

15,flag_in_your_hand1

16,cr3-what-is-this-encryption

17，safer_than_rot13

18，flag_in_your_hand

19，Decode_The_File

新手区部分题解：

1，easy_RSA

摘别人的图

或者用脚本 python2：（装了 gmpy2库的可以用）

import gmpy2
p = 473398607161
q = 4511491
e = 17
s = (p-1)*(q-1)d = gmpy2.invert(e,s)print d

2，Normal_RSA

解压出来两个文件：flag.enc 和 pubkey.pem

pem文件：以前是利用公钥加密进行邮件安全的一个协议，而现在PEM这个协议仅仅在使用的就是.pem这种文件
格式，这种文件里面保存了keys和X.509证书，看到了吗数字证书和key都能保存。
而现在一般表示用Base64 encoded DER编码的证书，
这种证书的内容打开能够看到在”—–BEGIN CERTIFICATE—–” and “—–END CERTIFICATE—–”之间

enc文件：该.enc文件扩展用于通过在UUenconded格式文件，它们是加密的文件。
这也意味着这些ENC文件包含受保护的数据和保存在该格式中，所以未经授权的收视数据和复制可以被防止

本题考查两个工具的使用 openssl （linux自带）和 rsatool

这两个工具我是在 windows 下安装使用的，当然可以在 linux中使用，个人喜好

分享：openssl： https://pan.baidu.com/s/1wP73GPJZo-HITH50UA61YQ 提取码: v2ti

rsatools ： https://pan.baidu.com/s/16Hl_p1zv_ajZadkW5GFnzQ 提取码: 8qqr

其中 openssl 是exe文件直接执行安装，默认安装到：C:\Program Files\OpenSSL-Win64

然后把 C:\Program Files\OpenSSL-Win64\bin 添加到环境变量中

我的电脑右键》属性》高级系统设置》环境变量》系统变量中双击 path 》新建》把复制的路径添加进去确定

一个是加密过的的flag.enc，另一个是公钥pubkey.pem。我们需要用四个步骤拿到flag

把 flag.enc 和 pubkey.pem 复制到 rsatools 文件夹中，打开cmd

1，提取 pubkey.pem 中的参数：

openssl rsa -pubin -text -modulus -in warmup -in pubkey.pem

对得到的Modulus（16）进制的转化为十进制：

87924348264132406875276140514499937145050893665602592992418171647042491658461

然后把这个数分解为两个互质数就是 q和 p

这里需要用到 yafu 中的 factor (kali自带)，我偏向于在windows中玩

分享源码，python脚本 linux也可以用，

https://pan.baidu.com/s/1TQqM8naEyVKhh101kz_igQ 提取码: rcb9

275127860351348928173285174381581152299
319576316814478949870590164193048041239

知道两个素数，随机定义大素数e,求出密钥文件 private.pem （python2）

python rsatool.py -o private.pem -e 65537 -p 275127860351348928173285174381581152299 -q 319576316814478949870590164193048041239

最后解密：

openssl rsautl -decrypt -in flag.enc -inkey private.pem

openssl是一个功能强大的工具，https://www.cnblogs.com/yangxiaolan/p/6256838.html

3, 幂数加密

提示是幂函数解密：

幂函数加密的密文只有 01234 不可能有 8，所以表层不是幂函数加密，

是云影密码：

4,easy_ECC

直接上脚本：

import collections
import randomEllipticCurve = collections.namedtuple('EllipticCurve', 'name p a b g n h')curve = EllipticCurve('secp256k1',# Field characteristic.p=int(input('p=')),# Curve coefficients.a=int(input('a=')),b=int(input('b=')),# Base point.g=(int(input('Gx=')),int(input('Gy='))),# Subgroup order.n=int(input('k=')),# Subgroup cofactor.h=1,
)# Modular arithmetic ##########################################################def inverse_mod(k, p):"""Returns the inverse of k modulo p.This function returns the only integer x such that (x * k) % p == 1.k must be non-zero and p must be a prime."""if k == 0:raise ZeroDivisionError('division by zero')if k < 0:# k ** -1 = p - (-k) ** -1 (mod p)return p - inverse_mod(-k, p)# Extended Euclidean algorithm.s, old_s = 0, 1t, old_t = 1, 0r, old_r = p, kwhile r != 0:quotient = old_r // rold_r, r = r, old_r - quotient * rold_s, s = s, old_s - quotient * sold_t, t = t, old_t - quotient * tgcd, x, y = old_r, old_s, old_tassert gcd == 1assert (k * x) % p == 1return x % p# Functions that work on curve points #########################################def is_on_curve(point):"""Returns True if the given point lies on the elliptic curve."""if point is None:# None represents the point at infinity.return Truex, y = pointreturn (y * y - x * x * x - curve.a * x - curve.b) % curve.p == 0def point_neg(point):"""Returns -point."""assert is_on_curve(point)if point is None:# -0 = 0return Nonex, y = pointresult = (x, -y % curve.p)assert is_on_curve(result)return resultdef point_add(point1, point2):"""Returns the result of point1 + point2 according to the group law."""assert is_on_curve(point1)assert is_on_curve(point2)if point1 is None:# 0 + point2 = point2return point2if point2 is None:# point1 + 0 = point1return point1x1, y1 = point1x2, y2 = point2if x1 == x2 and y1 != y2:# point1 + (-point1) = 0return Noneif x1 == x2:# This is the case point1 == point2.m = (3 * x1 * x1 + curve.a) * inverse_mod(2 * y1, curve.p)else:# This is the case point1 != point2.m = (y1 - y2) * inverse_mod(x1 - x2, curve.p)x3 = m * m - x1 - x2y3 = y1 + m * (x3 - x1)result = (x3 % curve.p,-y3 % curve.p)assert is_on_curve(result)return resultdef scalar_mult(k, point):"""Returns k * point computed using the double and point_add algorithm."""assert is_on_curve(point)if k < 0:# k * point = -k * (-point)return scalar_mult(-k, point_neg(point))result = Noneaddend = pointwhile k:if k & 1:# Add.result = point_add(result, addend)# Double.addend = point_add(addend, addend)k >>= 1assert is_on_curve(result)return result# Keypair generation and ECDHE ################################################def make_keypair():"""Generates a random private-public key pair."""private_key = curve.npublic_key = scalar_mult(private_key, curve.g)return private_key, public_keyprivate_key, public_key = make_keypair()
print("private key:", hex(private_key))
print("public key: (0x{:x}, 0x{:x})".format(*public_key))

cyberpeace{19477226185390}

高手进阶区部分题题解

5, ENC

是一个没有后缀的文件，用winhex打开，发现只有两种字符串 ZERO 和 ONE

复制下来保存为 123.txt，然后用脚本将ZERO 替换为 0 ONE替换为 1：


f = open("123.txt","r")
chars = f.read()
char = chars.split(" ")
string = ""
for c in char:if(c == "ZERO"):string += "0"elif(c == "ONE"):string +="1"
print(string)
f.close()

0100110001101001001100000110011101001100011010010011000001110101010011000110100101000001011101010100100101000011001100000111010101001100011010010011000001100111010011000101001100110100011101000100110001101001010000010111010001001001010000110011010001110101010011000101001100110100011001110100110001010011010000010111010101001100011010010011010001110101010010010100001100110100011101000100110001010011001100000111010001001001010000110011010001110101010011000110100100110100011101010100100101000011001100000111010001001100010100110100000101110101010011000101001100110000011101000100110001010011010000010111010101001100011010010011010001100111010011000101001100110000011101000100100101000011001101000111010101001100011010010011010001110101010010010100001100110100011101010100110001010011010000010111010101001100010100110011000001110101010010010100001100110100011101010100110001101001001100000111010001001001010000110011010001110100010011000110100101000001011101000100110001010011001100000110011101001100011010010011010001110101010011000110100100110100011001110100110001101001010000010111010001001100011010010011000001110101010010010100001100110100011101000100110001101001010000010111010101001100011010010011010001110100010011000101001101000001011101000100100101000011001100000111010001001100010100110100000101110100010010010100001100110000011101010100110001101001001100000110011101001100010100010011110100111101

然后转为 16进制，再转为 asci，base64 摩斯密码

最后得到这个玩意，怎么提交都不对，TM的什么几把玩意，

ALEXCTFTH15O1SO5UP3RO5ECR3TOTXT

最后看了别人的writeup 说需要整理格式：

ALEXCTF{TH15_1S_5UP3R_5ECR3T_TXT}

6,告诉你个秘密

打开是两段十六进制，转为ascii 》 base64 》键盘密码（我自己命名的，几个字符包围的字符）

flag： TONGYUAN

7，Easy-one

这个题我不会，附上一个大神的链接：

https://blog.csdn.net/zz_Caleb/article/details/89575430

8，说我作弊需要证据

追踪流：

是base64 ，找几条解密一下：

SEQ = 14; DATA = 0x83afae83c1db7776751d56c3f09fL; SIG = 0x400a19b82a4700ffc8a7515d7599L;SEQ = 19; DATA = 0x75c1fbc28bb27b5d2db9601fb967L; SIG = 0x2b5b628bf8183400cdab7f5870b1L;SEQ = 18; DATA = 0x181901c059de3b0f2d4840ab3aebL; SIG = 0x1b8bdf9468f81ce33a0da2a8bfbeL;SEQ = 13; DATA = 0x3b04b26a0adada2f67326bb0c5d6L; SIG = 0x2e5ab24f9dc21df406a87de0b3b4L;

解码得到序列号，数据和签名（签名用来校验信息）

猜测应该是解密数据部分，便可以得到flag。

现在思路就清晰了，

题目中给了两个公钥，一个公钥对应加密的数据，另一个是用来加密校验码的保证数据的准确性

序列号，肯定是解密出字符的位置

首先求解两个公钥对应的私钥：（其中的 n = q * p ,n分解为 p 和q 需要用到 yafu（kali自带）里面的 factor命令，linux和win都有yafu这个工具，命令 linux： factor（n）；win： yafu.exe factor(n)）

from Crypto.PublicKey
import RSA import gmpy
n = long(3104649130901425335933838103517383)
e = long(65537)
p = 49662237675630289
q = 62515288803124247
d = long(gmpy.invert(e, (p-1)*(q-1)))
rsa = RSA.construct( (n, e, d) )

然后就可以拿私钥解密数据了：

给出完整的脚本：其中的 zero_one是保存的 tcp流数据

from Crypto.PublicKey import RSA
import gmpy2
import base64
#Alice's
A_n = 1696206139052948924304948333474767
A_p = 38456719616722997
A_q = 44106885765559411
#Bob's
B_n = 3104649130901425335933838103517383
B_p = 49662237675630289
B_q = 62515288803124247A_phi = (A_p - 1) * (A_q - 1)
B_phi = (B_p - 1) * (B_q - 1)e = 65537A_d = int(gmpy2.invert(e, A_phi))
B_d = int(gmpy2.invert(e, B_phi))A_rsa = RSA.construct( (A_n, e, A_d) )
B_rsa = RSA.construct( (B_n, e, B_d) )flag = {}
with open('zero_one') as f:for s in f.readlines():line = str(base64.b64decode(s), encoding = 'utf8')seq = int(line.split(';')[0].split(' ')[2])data = int(line.split('0x')[1].split('L;')[0], 16)sig = int(line.split('0x')[2].rstrip('L;\n'), 16)decry = B_rsa.decrypt(data)signcheck = A_rsa.sign(decry, '')[0]if signcheck == sig:flag[seq] = chr(decry)
dic = sorted(flag.items(), key = lambda item:item[0])  #对字典按键值进行排序，返回值为列表
print(dic)
f = ''
for i in dic:f += i[1]
print(f)

拿到flag：flag{n0th1ng_t0_533_h3r3_m0v3_0n}

9.x_xor_md5

下载下来是一个无后缀名的文件，用hex打开看到十六进制的内容中有几行是重复的，加上题目是异或 xor

怀疑重复的内容就是 xor key

用 xor key 去和每一行的数据进行异或运算，

['01', '78', '0C', '4C', '10', '9E', '32', '37', '12', '0C', 'FB', 'BA', 'CB', '8F', '6A', '53']

然后得到：

['68', '4d', '4d', '4d', '0c', '00', '47', '4f', '4f', '44', '00', '4a', '4f', '42', '0c', '2a',
'54', '48', '45', '00', '46', '4c', '41', '47', '00', '49', '53', '00', '4e', '4f', '54', '00',
'72', '63', '74', '66', '5b', '77', '45', '11', '4c', '7f', '44', '10', '4e', '13', '7f', '3c',
'55', '54', '7f', '57', '48', '14', '54', '7f', '49', '15', '7f', '0a', '4b', '45', '59', '20',
'5d', '2a', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00',
'00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00',
'00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00',
'00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '72', '6f', '69', '73']

再将其转为字符：得到有ctf字样的乱码，说明思路没有出错，还需要进一步操作

审查得到的 16 进制列表数据，发现不应该出现0x00，0x00是绝对意义上的空，而空格是 0x20

所以猜测应该是所有数据还要和 0x20 进行异或运算

得到，

['48', '6d', '6d', '6d', '2c', '20', '67', '6f', '6f', '64', '20', '6a', '6f', '62', '2c',
'a', '74', '68', '65', '20', '66', '6c', '61', '67', '20', '69', '73', '20', '6e', '6f',
'74', '20', '52', '43', '54', '46', '7b', '57', '65', '31', '6c', '5f', '64', '30', '6e',
'33', '5f', '1c', '75', '74', '5f', '77', '68', '34', '74', '5f', '69', '35', '5f', '2a',
'6b', '65', '79', '0', '7d', 'a', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '52', '4f', '49', '53']

转为字符：提交不对，看到 { }中间 u前面有一个空格，并且*key应该是 *key*

在 16 进制列表中先找到 key 前面的字符的16进制为 2a ，确定key后面的字符也应该是 2a

而，0x00与0x2a异或后就是 0x2a ，所以找到字符u 前面的字符 1c 与 2a进行异或运算得到 36

所以最后的结果就是将 1c 改为 36 ，将00 改为 2a

改完之后得到：

['48', '6d', '6d', '6d', '2c', '20', '67', '6f', '6f', '64', '20', '6a', '6f', '62', '2c', '0a',
'74', '68', '65', '20', '66', '6c', '61', '67', '20', '69', '73', '20', '6e', '6f', '74', '20',
'52', '43', '54', '46', '7b', '57', '65', '31', '6c', '5f', '64', '30', '6e', '33', '5f', '36',
'75', '74', '5f', '77', '68', '34', '74', '5f', '69', '35', '5f', '2a', '6b', '65', '79', '2a',
'7d', '0a', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20',
'20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '52', '4f', '49', '53']

转为字符串：

提交还是不对，

题目提示还有 md5

而且字符串最后一行是一串不能打印的字符，怀疑最后一行的16进制与md5有关，

将最后一行的 16 进制复制下来与 0x20 进行异或再进行md5解码得到 that

m = ['01','78','0C','4C','10','9E','32','37','12','0C','FB','BA','CB','8F','6A','53']
zz = []
for i in range(0,len(m)):x = int(m[i],16)zz.append(hex(x^32)[2:])
for i in zz:print(i,end="")

21582c6c30be1217322cdb9aebaf4a73

最后得到 flag：

RCTF{We1l_d0n3_6ut_wh4t_i5_that}

处理脚本：

file1 = "693541011C9E75785D48FBF084CD66795530494C56D273701245A8BA85C03E53731B782A4BE977265E73BFAA859C156F542C731B588A66485B1984B080CA33735C520C4C109E3237120CFBBACB8F6A5301780C4C109E3237120CFBBACB8F6A5301780C4C109E3237120CFBBACB8F6A5301780C4C109E3237120C89D5A2FC"file = []
#先将文件的16进制字符串两个字符一组制成列表
for i in range(0,len(file1),2):file.append(file1[i:i+2])
xor1 = "01780C4C109E3237120CFBBACB8F6A53"
xor = []
#将 异或的对象制成一个列表
for i in range(0,len(xor1),2):xor.append(xor1[i:i+2])
result = []
string = ""
print(xor)
for i in range(0,len(file)):s = int("0x" + file[i],16) ^ int("0x" + xor[i%16],16) ^ int("0x20",16)string += chr(s)result.append(hex(s)[2:])
print(result)
print(string)

修改数据后的处理脚本：

n = ['48', '6d', '6d', '6d', '2c', '20', '67', '6f', '6f', '64', '20', '6a', '6f', '62', '2c', '0a','74', '68', '65', '20', '66', '6c', '61', '67', '20', '69', '73', '20', '6e', '6f', '74', '20','52', '43', '54', '46', '7b', '57', '65', '31', '6c', '5f', '64', '30', '6e', '33', '5f', '36','75', '74', '5f', '77', '68', '34', '74', '5f', '69', '35', '5f', '2a', '6b', '65', '79', '2a','7d', '0a', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20','20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20','20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '20','20', '20', '20', '20', '20', '20', '20', '20', '20', '20', '52', '4f', '49', '53']
for i in n :print(chr(int(i,16)),end="")

10，easy_crypto

给了一个密文 enc.txt 和一个加密的脚本

根据加密脚本，写出解密脚本

get buf unsign s[256]get buf t[256]we have key:hello worldwe have flag:????????????????????????????????for i:0 to 256set s[i]:ifor i:0 to 256set t[i]:key[(i)mod(key.lenth)]for i:0 to 256set j:(j+s[i]+t[i])mod(256)swap:s[i],s[j]for m:0 to 37set i:(i + 1)mod(256)set j:(j + S[i])mod(256)swap:s[i],s[j]set x:(s[i] + (s[j]mod(256))mod(256))set flag[m]:flag[m]^s[x]fprint flagx to file

解密脚本

# -*- coding: utf-8 -*-
f = open('C:/Users/HP/Desktop/enc.txt','r',encoding='ISO-8859-1')
c = f.read()
t = []
key = 'hello world'
ch = ''
j = 0 #初始化
s = list(range(256)) #创建有序列表
for i in range(256):j = (j + s[i] + ord(key[i % len(key)])) % 256s[i],s[j] = s[j],s[i]
i = 0 #初始化
j = 0 #初始化
for r in c:i = (i + 1)  % 256j = (j + s[i])  % 256s[i], s[j] = s[j], s[i]x = (s[i] + (s[j] % 256)) % 256ch += chr(ord(r) ^ s[x])
print(ch)

得到：

EIS{55a0a84f86a6ad40006f014619577ad3}

11，cr2-many-time-secrets

没思路，是多字节xor运算，

附上一篇大牛的博客：https://pequalsnp-team.github.io/writeups/CR2


import binasciidef dec(msg, key):'''Simple char-by-char XOR with a key (Vigenere, Vernam, OTP)'''m = ""for i in range(0, len(key)):m += chr(msg[i] ^ ord(key[i]))return m######################################lines = []with open("msg", "r") as f:# Read lines from file and decode Hexls = f.readlines()for l in ls:lines.append(binascii.unhexlify(l[:-1]))# Step 1: Decode each line with the known key
k = "ALEXCTF{"
mes = []
for l in lines:m = dec(l, k)mes.append(m)
print(mes)# Step 2: Guess some part of the first message 'Dear Fri'
k = "Dear Friend, "
m = dec(lines[0], k)
print(m)# Step 3: Decode each line with the new known key
k = "ALEXCTF{HERE_"
mes = []
for l in lines:m = dec(l, k)mes.append(m)
print(mes)# Step 4: Guess some part of the last message 'ncryption sc'
k = 'ncryption scheme '
m = dec(lines[-1], k)
print(m)# Step 5: Decode each line with the new known key
k = "ALEXCTF{HERE_GOES_"
mes = []
for l in lines:m = dec(l, k)mes.append(m)
print(mes)# Step 6: Guess all the second message 'sed One time pad e'
# the third message is 'n scheme, I heard '
# so we can retrive the complete key
k = 'sed One time pad encryptio'
m = dec(lines[2], k)
print(m)'''
['Dear Fri', 'nderstoo', 'sed One ', 'n scheme', 'is the o', 'hod that', ' proven ', 'ever if ', 'cure, Le', 'gree wit', 'ncryptio']
ALEXCTF{HERE_
['Dear Friend, ', 'nderstood my ', 'sed One time ', 'n scheme, I h', 'is the only e', 'hod that is m', ' proven to be', 'ever if the k', 'cure, Let Me ', 'gree with me ', 'ncryption sch']
ALEXCTF{HERE_GOES
['Dear Friend, This ', 'nderstood my mista', 'sed One time pad e', 'n scheme, I heard ', 'is the only encryp', 'hod that is mathem', ' proven to be not ', 'ever if the key is', 'cure, Let Me know ', 'gree with me to us', 'ncryption scheme a']
ALEXCTF{HERE_GOES_THE_KEY}
'''

12，oldDevier

以前遇到过这种类型的题，是 RSA攻击方法中的一种，RSA低加密指数广播攻击（模数n、密文c不同，加密指数e相同）

至于什么是 RSA低加密指数广播攻击，自行百度，

如果想更详细的的掌握 RSA 的各种攻击方法，可以看看这篇博客https://www.zhihu.com/people/ou-yang-shen-pai/posts

（python 2 ）

#!/usr/bin/python
#coding:utf-8import gmpy2
import time
from Crypto.Util.number import long_to_bytesfile = [{"c": 7366067574741171461722065133242916080495505913663250330082747465383676893970411476550748394841437418105312353971095003424322679616940371123028982189502042, "e": 10, "n": 25162507052339714421839688873734596177751124036723831003300959761137811490715205742941738406548150240861779301784133652165908227917415483137585388986274803},
{"c": 21962825323300469151795920289886886562790942771546858500842179806566435767103803978885148772139305484319688249368999503784441507383476095946258011317951461, "e": 10, "n": 23976859589904419798320812097681858652325473791891232710431997202897819580634937070900625213218095330766877190212418023297341732808839488308551126409983193},
{"c": 6569689420274066957835983390583585286570087619048110141187700584193792695235405077811544355169290382357149374107076406086154103351897890793598997687053983, "e": 10, "n": 18503782836858540043974558035601654610948915505645219820150251062305120148745545906567548650191832090823482852604346478335353784501076761922605361848703623},
{"c": 4508246168044513518452493882713536390636741541551805821790338973797615971271867248584379813114125478195284692695928668946553625483179633266057122967547052, "e": 10, "n": 23383087478545512218713157932934746110721706819077423418060220083657713428503582801909807142802647367994289775015595100541168367083097506193809451365010723},
{"c": 22966105670291282335588843018244161552764486373117942865966904076191122337435542553276743938817686729554714315494818922753880198945897222422137268427611672, "e": 10, "n": 31775649089861428671057909076144152870796722528112580479442073365053916012507273433028451755436987054722496057749731758475958301164082755003195632005308493},
{"c": 17963313063405045742968136916219838352135561785389534381262979264585397896844470879023686508540355160998533122970239261072020689217153126649390825646712087, "e": 10, "n": 22246342022943432820696190444155665289928378653841172632283227888174495402248633061010615572642126584591103750338919213945646074833823905521643025879053949},
{"c": 1652417534709029450380570653973705320986117679597563873022683140800507482560482948310131540948227797045505390333146191586749269249548168247316404074014639, "e": 10, "n": 25395461142670631268156106136028325744393358436617528677967249347353524924655001151849544022201772500033280822372661344352607434738696051779095736547813043},
{"c": 15585771734488351039456631394040497759568679429510619219766191780807675361741859290490732451112648776648126779759368428205194684721516497026290981786239352, "e": 10, "n": 32056508892744184901289413287728039891303832311548608141088227876326753674154124775132776928481935378184756756785107540781632570295330486738268173167809047},
{"c": 8965123421637694050044216844523379163347478029124815032832813225050732558524239660648746284884140746788823681886010577342254841014594570067467905682359797, "e": 10, "n": 52849766269541827474228189428820648574162539595985395992261649809907435742263020551050064268890333392877173572811691599841253150460219986817964461970736553},
{"c": 13560945756543023008529388108446940847137853038437095244573035888531288577370829065666320069397898394848484847030321018915638381833935580958342719988978247, "e": 10, "n": 30415984800307578932946399987559088968355638354344823359397204419191241802721772499486615661699080998502439901585573950889047918537906687840725005496238621}]# 读入 e, n, c
c = []
n = []
for f in file:c.append(f["c"])n.append(f["n"])
e = 10def CRT(items):N = reduce(lambda x, y: x * y, (i[1] for i in items))result = 0for a, n in items:m = N / nd, r, s = gmpy2.gcdext(n, m)if d != 1: raise Exception("Input not pairwise co-prime")result += a * s * mreturn result % N, Ndata = zip(c, n)
x, n = CRT(data)
m = gmpy2.iroot(gmpy2.mpz(x), e)[0].digits()
print long_to_bytes(m)

13，wtc_rsa_bbq

给了个密文和公钥，直接用 RsaCtfTool 直接爆破就行

kali没有，需要自行去 github下载使用（需要先安装依赖包）

命令：

python RsaCtfTool.py --publickey key.pem --uncipherfile cipher.bin

得到：flag{how_d0_you_7urn_this_0n?}

14，cr4-poor-rsa

给了个无后缀的文件，先用winhex查看一下，发现是一个压缩包，后缀改为 zip解压

得到两个文件 flag.b64（密文）和 key.pub（公钥）

先处理flag.b64（将flag.b64中的内容进行解 base64操作）

使用 notepad++ 打开 flag.b64文件使用插件中的 MIME Tools 中的 base64 decode 将文件内容解密，然后另保存为 flag.enc 文件(这里不能直接复制粘贴保存，会丢失数据)

然后使用 RsaCtfTool 工具进行破解：

python RsaCtfTool.py --publickey key.pem --uncipherfile cipher.bin

得到flag:ALEXCTF{SMALL_PRIMES_ARE_BAD}

15,flag_in_your_hand1

给了两个文件 index.html 和一个js文件，考察js代码审计能力

首先借助浏览器来运行js 程序。用浏览器打开index.html，分析 js 代码: 首先无论在 token 输入框中输入什么字符串，getFlag() 都会算出一个 hash 值，

实际上是showFlag()函数中 ic 的值决定了 hash 值即 flag 是否正确。

那么在script-min.js中找到 ic 取值的函数 ck() ，找到一个 token 使得 ck()中ic =true即可。

token 是[118, 104, 102, 120, 117, 108, 119, 124, 48,123,101,120]每个数字减3 得到的ascii 码所对应的字符，

即security-xbu

可以利用浏览器的js 调试功能跟踪变量，逻辑梳理的会更快一些

在 token 处输入security-xbu ，点击Get flag!

出现You got the flag below!! ，

以及flag： RenIbyd8Fgg5hawvQm7TDQ

16,cr3-what-is-this-encryption

给了rsa解密的相关参数，直接用脚本解密就行了

import libnum
from Crypto.Util.number import long_to_bytesc = 0x7fe1a4f743675d1987d25d38111fae0f78bbea6852cba5beda47db76d119a3efe24cb04b9449f53becd43b0b46e269826a983f832abb53b7a7e24a43ad15378344ed5c20f51e268186d24c76050c1e73647523bd5f91d9b6ad3e86bbf9126588b1dee21e6997372e36c3e74284734748891829665086e0dc523ed23c386bb520e = int("0x6d1fdab4ce3217b3fc32c9ed480a31d067fd57d93a9ab52b472dc393ab7852fbcb11abbebfd6aaae8032db1316dc22d3f7c3d631e24df13ef23d3b381a1c3e04abcc745d402ee3a031ac2718fae63b240837b4f657f29ca4702da9af22a3a019d68904a969ddb01bcf941df70af042f4fae5cbeb9c2151b324f387e525094c41",16)q = int("0xa6055ec186de51800ddd6fcbf0192384ff42d707a55f57af4fcfb0d1dc7bd97055e8275cd4b78ec63c5d592f567c66393a061324aa2e6a8d8fc2a910cbee1ed9",16)
p = int("0xfa0f9463ea0a93b929c099320d31c277e0b0dbc65b189ed76124f5a1218f5d91fd0102a4c8de11f28be5e4d0ae91ab319f4537e97ed74bc663e972a4a9119307",16)
n = q*pd = libnum.invmod(e, (p - 1) * (q - 1))
m = pow(c, d, n)   # m 的十进制形式
string = long_to_bytes(m)  # m明文
print(string)  # 结果为 b‘ m ’ 的形式

flag： ALEXCTF{RS4_I5_E55ENT1AL_T0_D0_BY_H4ND}

17，safer_than_rot13

解压得到 cry100文件，复制文件内容尝试进行 rot13解密，没发现什么

在尝试词频分析，利用在线工具：https://quipqiup.com/

得到flag：no_this_is_not_crypto_my_dear

18，flag_in_your_hand

给了两个文件 index.html 和一个js文件，考察js代码审计能力

首先借助浏览器来运行js 程序。用浏览器打开index.html，分析 js 代码: 首先无论在 token 输入框中输入什么字符串，getFlag() 都会算出一个 hash 值，

实际上是showFlag()函数中 ic 的值决定了 hash 值即 flag 是否正确。

那么在script-min.js中找到 ic 取值的函数 ck() ，找到一个 token 使得 ck()中ic =true即可。

token 是[118, 104, 102, 120, 117, 108, 119, 124, 48,123,101,120]每个数字减3 得到的ascii 码所对应的字符，

即security-xbu

可以利用浏览器的js 调试功能跟踪变量，逻辑梳理的会更快一些

在 token 处输入security-xbu ，点击Get flag!

出现You got the flag below!! ，

以及flag： RenIbyd8Fgg5hawvQm7TDQ

19，Decode_The_File

这题跟攻防世界的 misc中的 base64stego 一样，都是利用 base64加密的特性进行数据的加密填充

利用base64 加密的特性，将flag进行编码分割，填充在每一行的最后

（简要原理，ascii码是用8位二进制表示一个字符的，而base64是用6位二进制表示一个字符，将明文字符转化为二进制后再每6位划分成一个 “字节”，然后将每个字节转化为一个字符，就变成了base64密文，而在base64的密文中加密是利用，每一段密文的最后4位二进制是不影响明文的，可以将flag转化为二进制后拆分隐藏在每一段的最后4位二进制中）

解密：

将文件重命名为 stego.txt

import base64b64chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
with open('stego.txt', 'rb') as f:   #stego.txt 为在base64密文中加密后的密文flag = ''bin_str = ''for line in f.readlines():stegb64 = str(line, "utf-8").strip("\n")rowb64 = str(base64.b64encode(base64.b64decode(stegb64)), "utf-8").strip("\n")offset = abs(b64chars.index(stegb64.replace('=', '')[-1]) - b64chars.index(rowb64.replace('=', '')[-1]))equalnum = stegb64.count('=')  # no equalnum no offsetif equalnum:bin_str += bin(offset)[2:].zfill(equalnum * 2)
for i in range(0, len(bin_str), 8):print(chr(int(bin_str[i:i + 8], 2)),end='')

刷不下去了，就到这儿吧

攻防世界 ——crypto相关推荐

[攻防世界]crypto新手练习区Caesar
[攻防世界]crypto新手练习区Caesar Caesar最佳Writeup由Um0 • Umo.提供难度系数: 1.0 题目来源: poxlove3 题目描述:你成功的解出了来了灯谜,小鱼一脸的 ...
攻防世界 Crypto高手进阶区 3分题 wtc_rsa_bbq
前言继续ctf的旅程攻防世界Crypto高手进阶区的3分题本篇是wtc_rsa_bbq的writeup 发现攻防世界的题目分数是动态的就仅以做题时的分数为准了解题过程得到一个无后缀文件扔 ...
攻防世界crypto高手题之wtc_rsa_bbq
攻防世界crypto高手题之wtc_rsa_bbq 照例下载附件,第一次解压后还是压缩包,所以要经过两次解压: 解压后的文件夹内是一个两个文件,因为题目提示是RSA,而这两个看上去像是密文和密钥文件: ...
攻防世界crypto高手题之best_rsa
攻防世界crypto高手题之best_rsa 继续开启全栈梦想之逆向之旅~ 这题是攻防世界crypto高手题的best_rsa . . 下载题目,是一个明文和密钥的4个附件: . . 其实一开始我并不 ...
攻防世界 Crypto高手进阶区 3分题你猜猜
前言继续ctf的旅程攻防世界Crypto高手进阶区的3分题本篇是你猜猜的writeup 发现攻防世界的题目分数是动态的就仅以做题时的分数为准了解题过程得到一串16进制 504B03040A ...
攻防世界crypto高手题之sherlock
攻防世界crypto高手题之sherlock 继续开启全栈梦想之逆向之旅~ 这题是攻防世界crypto高手题的sherlock . . (这里积累第一个经验) 下载附件,是一个txt文档,内容是一篇小 ...
攻防世界 Crypto sherlock
攻防世界 Crypto sherlock 1.打开文件 2.提取所有大写字母 2.得到以下内容 3.二进制转字符串 1.打开文件文件是一个大文本搜索了flag时候没有什么结果 2.提取所有大写字母 ...
攻防世界 Crypto高手进阶区 5分题简单流量分析
前言继续ctf的旅程攻防世界Crypto高手进阶区的5分题本篇是简单流量分析的writeup 发现攻防世界的题目分数是动态的就仅以做题时的分数为准了解题过程题目描述 binwalk无果查 ...
攻防世界 Crypto高手进阶区 6分题 xor_game
前言继续ctf的旅程攻防世界Crypto高手进阶区的6分题本篇是xor_game的writeup 发现攻防世界的题目分数是动态的就仅以做题时的分数为准了解题过程题目描述得到一段py和一个 ...
攻防世界 Crypto 幂数加密
攻防世界 Crypto 幂数加密 1.看到题目是幂数加密 1.看到题目是幂数加密先去了解一下什么是幂数加密二进制幂数加密法就是应用这个原理,由于英文字母只有26个字母,由公式可知,只要2的0.1. ...

攻防世界 ——crypto

新手区部分题解：

1，easy_RSA

2，Normal_RSA

3, 幂数加密

4,easy_ECC

高手进阶区部分题题解

5, ENC

6,告诉你个秘密

7，Easy-one

8，说我作弊需要证据

9.x_xor_md5

10，easy_crypto

11，cr2-many-time-secrets

12，oldDevier

13，wtc_rsa_bbq

14，cr4-poor-rsa

15,flag_in_your_hand1

16,cr3-what-is-this-encryption

17，safer_than_rot13

18，flag_in_your_hand

19，Decode_The_File

攻防世界 ——crypto相关推荐

最新文章

热门文章