文件是两个python源码

我们来分析一下：

challenge.py：

```from N1ES import N1ES //导入另一个文件类

import base64 //导入关于base64的包

key = "wxy191iss00000000000cute" //给key赋值

n1es = N1ES(key)****** //将key带入N1ES的函数中

flag = "N1CTF{*****************************************}" //将N1CTF集合给flag变量

cipher = n1es.encrypt(flag)****** //对flag变量使用n1es类的encrypt(加密)方法

print base64.b64encode(cipher) # HRlgC2ReHW1/WRk2DikfNBo1dl1XZBJrRR9qECMNOjNHDktBJSxcI1hZIz07YjVx //输出经过base64编码的密文，也就是经过了两轮加密。

N1ES.py：

```# -*- coding: utf-8 -*-

def round_add(a, b):

f = lambda x, y: x + y - 2 * (x & y)

res = ''

for i in range(len(a)):

res += chr(f(ord(a[i]), ord(b[i])))

return res

def permutate(table, block): //将传参作为遍历table和block

return list(map(lambda x: block[x], table)) //

def string_to_bits(data): //data代表self.key参数

data = [ord(c) for c in data] //ord(c)返回对应的 ASCII 数值，for用c遍历key

l = len(data) * 8 //len根据data数据长度*8赋值给l

result = [0] * l //将列表【0】 * l

此时我已经看不懂要干嘛了，对数据元素全0操作？上面这张图基本就是目前对key1的操作。

pos = 0 //pos变量赋值为0

for ch in data: //用ch遍历data

for i in range(0,8): //用i遍历从下标0开始下标8结束

result[(pos<<3)+i] = (ch>>i) & 1 //<

pos += 1

return result

这是返回return result的值。我接下来可以又去看N1ES类。

```s_box = [54, 132, 138, 83, 16, 73, 187, 84, 146, 30, 95, 21, 148, 63, 65, 189, 188, 151, 72, 161, 116, 63, 161, 91, 37, 24, 126, 107, 87, 30, 117, 185, 98, 90, 0, 42, 140, 70, 86, 0, 42, 150, 54, 22, 144, 153, 36, 90, 149, 54, 156, 8, 59, 40, 110, 56,1, 84, 103, 22, 65, 17, 190, 41, 99, 151, 119, 124, 68, 17, 166, 125, 95, 65, 105, 133, 49, 19, 138, 29, 110, 7, 81, 134, 70, 87, 180, 78, 175, 108, 26, 121, 74, 29, 68, 162, 142, 177, 143, 86, 129, 101, 117, 41, 57, 34, 177, 103, 61, 135, 191, 74, 69, 147, 90, 49, 135, 124, 106, 19, 89, 38, 21, 41, 17, 155, 83, 38, 159, 179, 19, 157, 68, 105, 151, 166, 171, 122, 179, 114, 52, 183, 89, 107, 113, 65, 161, 141, 18, 121, 95, 4, 95, 101, 81, 156, 17, 190, 38, 84, 9, 171, 180, 59, 45, 15, 34, 89, 75, 164, 190, 140, 6, 41, 188, 77, 165, 105, 5, 107, 31, 183, 107, 141, 66, 63, 10, 9, 125, 50, 2, 153, 156, 162, 186, 76, 158, 153, 117, 9, 77, 156, 11, 145, 12, 169, 52, 57, 161, 7, 158, 110, 191, 43, 82, 186, 49, 102, 166, 31, 41, 5, 189, 27]

def generate(o):

k = permutate(s_box,o) //使用s_box,o作为参数，使用permutate方法

b = []

for i in range(0, len(k), 7):

b.append(k[i:i+7] + [1])

c = []

for i in range(32):

pos = 0

x = 0

for j in b[i]:

x += (j<

pos += 1

c.append((0x10001**x) % (0x7f))

return c

class N1ES:

def __init__(self, key): //初始化方法，传参key

if (len(key) != 24 or isinstance(key, bytes) == False ): //如果key的长度等于24或者，判断key是否是字节类型，并返回值

raise Exception("key must be 24 bytes long") //raise抛出异常，raise后面的语句将不能执行

self.key = key //外部传来的key参数转变为N1ES的变量self.key

self.gen_subkey() //参数返回值作为N1ES变量

def gen_subkey(self):

o = string_to_bits(self.key) //o变量等于string_to_bits方法的返回，self.key变量作为传参，我们转到string_to_bits方法查看

k = [] //空列表

for i in range(8): //range(8)也相当于range(0,8)，i遍历

o = generate(o) //将o当作参数，使用 generate方法

k.extend(o)

o = string_to_bits([chr(c) for c in o[0:24]])

self.Kn = []

for i in range(32):

self.Kn.append(map(chr, k[i * 8: i * 8 + 8]))

return

def encrypt(self, plaintext):

if (len(plaintext) % 16 != 0 or isinstance(plaintext, bytes) == False):

raise Exception("plaintext must be a multiple of 16 in length")

res = ''

for i in range(len(plaintext) / 16):

block = plaintext[i * 16:(i + 1) * 16]

L = block[:8]

R = block[8:]

for round_cnt in range(32):

L, R = R, (round_add(L, self.Kn[round_cnt]))

L, R = R, L

res += L + R

return res