obstacle的个人空间

These are the days we won't forget

发表于 更新于 分类于 本文字数： 3k 阅读时长 ≈ 5 分钟

前言：前段时间在写pta上面的代码题时遇到一个查找字符串字串的问题，一开始采取i，j循环遍历的方法暴力寻找，但是很可惜时间复杂度上出现问题，于是经过学习，初步尝试运用kmp算法来解决查找字符串字串中时间复杂度的问题。

对kmp的名称由来等就不展开描述了直接进入代码部分。

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#include <stdio.h>
#include <string.h>
#include<stdlib.h>
void computeLPSArray(char* pat, int M, int* lps) {
    int len = 0,i=1;  // 当前最长前缀后缀长度
    lps[0] = 0;   // lps[0] 总是为 0
    while (i < M)
    {
        if (pat[i] == pat[len]) 
        {
            len++;
            lps[i] = len;
            i++;
        }
         else 
        {
            if (len != 0)   // 不匹配时，根据 lps 数组更新 len
            {
                len = lps[len - 1];
            }
            else 
            {
                lps[i] = 0;
                i++;
            }
        }
    }
}
void KMPSearch(char* pat, char* txt) {
    int M = strlen(pat);
    int N = strlen(txt);
    int* lps = (int*)malloc(M * sizeof(int));  // 创建 lps 数组
    computeLPSArray(pat, M, lps);
    int i = 0; // txt 的索引
    int j = 0; // pat 的索引
    while (i < N)
    {
        if (pat[j] == txt[i])
        {
            i++;
            j++;
        }
        if (j == M)
        {
            printf("找到模式字符串 '%s' 的起始位置: %d\n", pat, i - j);
            j = lps[j - 1];
        } 
        else if(i < N && pat[j] != txt[i])
        {
            if (j != 0)   // 字符不匹配
            {
                j = lps[j - 1];
            } else
            {
                i++;
            }
        }
    }
}
int main(){
    char txt[] = "ABABDABACDABABCABAB";
    char pat[] = "ABABCABAB";
    KMPSearch(pat, txt);
    return 0;
}

个人理解，掌握kmp算法的关键就是掌握lps数组的构造（网上大部分也称next数组，就把lps换成next），lps数组也就是找字串中的最大公共前后缀，它可以帮助我们查找子串时优化“回溯”过程，因为用i，j单纯暴力循环每次都要回到最初会有很多浪费，利用lps数组的构造就可以优化这个过程。具体概念理解，以及构造思路还需要在网络上自己找视频理解。

接下来是pta里面的一道题，运用kmp算法解决。

7-11 删除字符串中的子串

分数 10

作者 白洪欢

单位 浙江大学

输入2个字符串S1和S2，要求删除字符串S1中出现的所有子串S2，即结果字符串中不能包含S2。

输入格式：

输入在2行中分别给出不超过80个字符长度的、以回车结束的2个非空字符串，对应S1和S2。

输出格式：

在一行中输出删除字符串S1中出现的所有子串S2后的结果字符串。

输入样例：

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Tomcat is a male ccatat
cat

输出样例：

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Tom is a male

下面是通过代码

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#include <stdio.h>
#include <string.h>
#include <stdlib.h>
// 计算LPS（最长前缀后缀）数组
void computeLPSArray(char* pat, int M, int* lps) {
    int len = 0; // 当前最长前缀长度
    lps[0] = 0; // lps[0] 总是为 0
    int i = 1; // 从 lps[1] 开始计算
    while (i < M) {
        if (pat[i] == pat[len]) {
            len++;
            lps[i] = len;
            i++;
        } else {
            if (len != 0) {
                len = lps[len - 1]; // 根据 lps 更新 len
            } else {
                lps[i] = 0;
                i++;
            }
        }
    }
}
// KMP搜索算法 - 查找并删除所有出现的子串
void KMPSearchAndDelete(char* txt, char* pat) {
    int M = strlen(pat);
    int N = strlen(txt);
    int* lps = (int*)malloc(M * sizeof(int));
    computeLPSArray(pat, M, lps);
    int i = 0; // 主串的索引
    int j = 0; // 模式串的索引
    while (i < N) {
        if (pat[j] == txt[i]) {
            i++;
            j++;
        }
        if (j == M) {
            // 找到一个匹配，删除子串
            memmove(&txt[i - j], &txt[i], N - i + 1); // 移动后面的字符
            N -= j; // 更新主串长度
            txt[N] = '\0'; // 确保以'\0'结束
            
            // 重置j为0，继续寻找进一步的匹配
            j = 0;
            i = 0; // 从头再继续检查
        } else if (i < N && pat[j] != txt[i]) {
            if (j != 0) {
                j = lps[j - 1]; // 根据lps调整j
            } else {
                i++;
            }
        }
    }
    free(lps); // 释放内存
}
int main() {
    char s1[256]; // 增大数组大小以容纳更长的字符串
    char s2[81];
    fgets(s1, sizeof(s1), stdin);
    s1[strcspn(s1, "\n")] = 0; // 去掉换行符
    fgets(s2, sizeof(s2), stdin);
    s2[strcspn(s2, "\n")] = 0; // 去掉换行符
    // 持续删除子串直到找不到为止
    while (strstr(s1, s2) != NULL) {
        KMPSearchAndDelete(s1, s2); // 查找并删除子串
    }
    printf("%s\n", s1); // 输出结果
    return 0;
}

发表于 更新于 本文字数： 439 阅读时长 ≈ 1 分钟

1.图片

往typora插入图片，图片文件夹一定要和创建的.md文件夹放在同一个文件夹下。不然文件移动位置后图片会因找不到地址而消失

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格式！[提示文字](图片地址)      即先！然后快捷键ctrl+k
例如：![伊蕾娜](D:\blog\source\images\Elaina.jpg)

问题

相信很多人尝试上面图片的插入方法会遇到在typora能正常显示图片，但在网页上图片却无法加载。这是因为上述地址如D:\blog\source\images\Elaina.jpg为本地地址，而hexo+github搭建的博客并不能访问本地地址所以自然无法成功。

解决

最简单粗暴的方法就是直接将图片拖到typora中，typora会直接创建博客文章同名的文件夹存放图片。但是就是感觉文章写多了有点乱乱的。

特别注意

说来也巧本以为早就解决了图片的问题结果又出现了，这次经过更长时间捣鼓得出结论上面例子里的【提示文字】不能为纯数字不然部署到GitHub上就被认定为width而不是title导致图片无法加载。

发表于 更新于 本文字数： 9 阅读时长 ≈ 1 分钟

这是我的第一个博客

发表于 更新于 本文字数： 573 阅读时长 ≈ 1 分钟

typora功能初步尝试

1.标题

一级标题：ctrl+1或#
二级标题：ctrl+2或##
剩下的三级四级等就可以以此类推了

2.文字

删除线：alt+shift+5 示例 ~~~~
加粗：ctrl+B 示例 加粗
斜体：ctrl+I 示例 斜体
下划线：ctrl+U 示例 下划线
高亮：==中间内容==

3.表情包

:smile: :100: :heart: 快捷键:windows+；

4.表格

week2 week3 week4

快捷键:ctrl+t

5.引用

一级应用

二级引用

瑞典厨师长

6.代码

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插入不确定代码，快捷键：ctrl+shift+k

7.分隔线

*** 然后回车

8.源代码模式

ctrl+/,退出一样

9.跳转

1.跳转到外部

bilibili
屑魔女的个人博客

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快捷键：ctrl+k      格式[提示文字](网址)

2.跳转到内部

[博客](#my first blog)

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快捷键：CTRL+k     [提示文字](#标题)

10.自动链接

使用<>然后括号里链接会自动转化为超链接

11.图片

往typora插入图片，图片文件夹一定要和创建的.md文件夹放在同一个文件夹下。不然文件移动位置后图片会因找不到地址而消失

1
2
格式！[提示文字](图片地址)      即先！然后快捷键ctrl+k
例如：![伊蕾娜](D:\blog\source\images\Elaina.jpg)

伊蕾娜

发表于 更新于 本文字数： 14k 阅读时长 ≈ 25 分钟

总结一下hgame week2crypto部分。自己虽然学了一段时间，结果week1密码一道没做出来，还是非常遗憾的，所幸week2密码是ak了的

CRYPTO

1.ancient recall

题目：

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import random

Major_Arcana = ["The Fool", "The Magician", "The High Priestess","The Empress", "The Emperor", "The Hierophant","The Lovers", "The Chariot", "Strength","The Hermit", "Wheel of Fortune", "Justice","The Hanged Man", "Death", "Temperance","The Devil", "The Tower", "The Star","The Moon", "The Sun", "Judgement","The World"]
wands = ["Ace of Wands", "Two of Wands", "Three of Wands", "Four of Wands", "Five of Wands", "Six of Wands", "Seven of Wands", "Eight of Wands", "Nine of Wands", "Ten of Wands", "Page of Wands", "Knight of Wands", "Queen of Wands", "King of Wands"]
cups = ["Ace of Cups", "Two of Cups", "Three of Cups", "Four of Cups", "Five of Cups", "Six of Cups", "Seven of Cups", "Eight of Cups", "Nine of Cups", "Ten of Cups", "Page of Cups", "Knight of Cups", "Queen of Cups", "King of Cups"]
swords = ["Ace of Swords", "Two of Swords", "Three of Swords", "Four of Swords", "Five of Swords", "Six of Swords", "Seven of Swords", "Eight of Swords", "Nine of Swords", "Ten of Swords", "Page of Swords", "Knight of Swords", "Queen of Swords", "King of Swords"]
pentacles = ["Ace of Pentacles", "Two of Pentacles", "Three of Pentacles", "Four of Pentacles", "Five of Pentacles", "Six of Pentacles", "Seven of Pentacles", "Eight of Pentacles", "Nine of Pentacles", "Ten of Pentacles", "Page of Pentacles", "Knight of Pentacles", "Queen of Pentacles", "King of Pentacles"]
Minor_Arcana = wands + cups + swords + pentacles
tarot = Major_Arcana + Minor_Arcana
reversals = [0,-1]

Value = []
cards = []
YOUR_initial_FATE = []
while len(YOUR_initial_FATE)<5:
    card = random.choice(tarot)
    if card not in cards:
        cards.append(card)
        if card in Major_Arcana:
            k = random.choice(reversals)
            Value.append(tarot.index(card)^k)
            if k == -1:
                YOUR_initial_FATE.append("re-"+card)
            else:
                YOUR_initial_FATE.append(card)
        else:
            Value.append(tarot.index(card))
            YOUR_initial_FATE.append(card)
    else:
        continue
print("Oops!lets reverse 1T!")

FLAG=("hgame{"+"&".join(YOUR_initial_FATE)+"}").replace(" ","_")

YOUR_final_Value = Value
def Fortune_wheel(FATE):
    FATEd = [FATE[i]+FATE[(i+1)%5] for i in range(len(FATE))]
    return FATEd

for i in range(250):
    YOUR_final_Value = Fortune_wheel(YOUR_final_Value)
print(YOUR_final_Value)
YOUR_final_FATE = []
for i in YOUR_final_Value:
    YOUR_final_FATE.append(tarot[i%78])
print("Your destiny changed!\n",",".join(YOUR_final_FATE))
print("oh,now you GET th3 GOOd lU>k,^^")
"""
Oops!lets reverse 1T!
[2532951952066291774890498369114195917240794704918210520571067085311474675019, 2532951952066291774890327666074100357898023013105443178881294700381509795270, 2532951952066291774890554459287276604903130315859258544173068376967072335730, 2532951952066291774890865328241532885391510162611534514014409174284299139015, 2532951952066291774890830662608134156017946376309989934175833913921142609334]
Your destiny changed!
 Eight of Cups,Ace of Cups,Strength,The Chariot,Five of Swords
oh,now you GET th3 GOOd lU>k,^^
"""

emmmm,自己看不太懂，但题目应该比较简单，丢给ai直接解出来（

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Major_Arcana = ["The Fool", "The Magician", "The High Priestess","The Empress", "The Emperor", "The Hierophant","The Lovers", "The Chariot", "Strength","The Hermit", "Wheel of Fortune", "Justice","The Hanged Man", "Death", "Temperance","The Devil", "The Tower", "The Star","The Moon", "The Sun", "Judgement","The World"]
wands = ["Ace of Wands", "Two of Wands", "Three of Wands", "Four of Wands", "Five of Wands", "Six of Wands", "Seven of Wands", "Eight of Wands", "Nine of Wands", "Ten of Wands", "Page of Wands", "Knight of Wands", "Queen of Wands", "King of Wands"]
cups = ["Ace of Cups", "Two of Cups", "Three of Cups", "Four of Cups", "Five of Cups", "Six of Cups", "Seven of Cups", "Eight of Cups", "Nine of Cups", "Ten of Cups", "Page of Cups", "Knight of Cups", "Queen of Cups", "King of Cups"]
swords = ["Ace of Swords", "Two of Swords", "Three of Swords", "Four of Swords", "Five of Swords", "Six of Swords", "Seven of Swords", "Eight of Swords", "Nine of Swords", "Ten of Swords", "Page of Swords", "Knight of Swords", "Queen of Swords", "King of Swords"]
pentacles = ["Ace of Pentacles", "Two of Pentacles", "Three of Pentacles", "Four of Pentacles", "Five of Pentacles", "Six of Pentacles", "Seven of Pentacles", "Eight of Pentacles", "Nine of Pentacles", "Ten of Pentacles", "Page of Pentacles", "Knight of Pentacles", "Queen of Pentacles", "King of Pentacles"]
Minor_Arcana = wands + cups + swords + pentacles
tarot = Major_Arcana + Minor_Arcana
def reverse_step(B):
    B0, B1, B2, B3, B4 = B
    numerator = B0 + B1 + B3 - B2 - B4
    if numerator % 2 != 0:
        raise ValueError("奇数无法整除")
    A1 = numerator // 2
    A0 = B0 - A1
    A2 = B1 - A1
    A3 = B2 - A2
    A4 = B3 - A3
    if A4 + A0 != B4:
        raise ValueError("验证失败")
    return [A0, A1, A2, A3, A4]
final_values = [
    2532951952066291774890498369114195917240794704918210520571067085311474675019,
    2532951952066291774890327666074100357898023013105443178881294700381509795270,
    2532951952066291774890554459287276604903130315859258544173068376967072335730,
    2532951952066291774890865328241532885391510162611534514014409174284299139015,
    2532951952066291774890830662608134156017946376309989934175833913921142609334
]
current = final_values.copy()
for _ in range(250):
    current = reverse_step(current)
def get_card(v):
    for k in range(22):
        if k ^ -1 == v:
            return f"re-{Major_Arcana[k]}"
    index = v % 78
    card = tarot[index]
    if card in Major_Arcana and v == index:
        return card
    return card
cards = [get_card(v) for v in current]
flag = "hgame{" + "&".join(cards).replace(" ", "_") + "}"
print(flag)
#hgame{re-The_Moon&re-The_Sun&Judgement&re-Temperance&Six_of_Cups}

2.Intergalactic Bound

题目：

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from Crypto.Util.number import *
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
from random import randint
import hashlib
from secrets import flag

def add_THCurve(P, Q):
    if P == (0, 0):
        return Q
    if Q == (0, 0):
        return P
    x1, y1 = P
    x2, y2 = Q
    x3 = (x1 - y1 ** 2 * x2 * y2) * pow(a * x1 * y1 * x2 ** 2 - y2, -1, p) % p
    y3 = (y1 * y2 ** 2 - a * x1 ** 2 * x2) * pow(a * x1 * y1 * x2 ** 2 - y2, -1, p) % p
    return x3, y3


def mul_THCurve(n, P):
    R = (0, 0)
    while n > 0:
        if n % 2 == 1:
            R = add_THCurve(R, P)
        P = add_THCurve(P, P)
        n = n // 2
    return R


p = getPrime(96)
a = randint(1, p)
G = (randint(1,p), randint(1,p))
d = (a*G[0]^3+G[1]^3+1)%p*inverse(G[0]*G[1],p)%p
x = randint(1, p)
Q = mul_THCurve(x, G)
print(f"p = {p}")
print(f"G = {G}")
print(f"Q = {Q}")

key = hashlib.sha256(str(x).encode()).digest()
cipher = AES.new(key, AES.MODE_ECB)
flag = pad(flag,16)
ciphertext = cipher.encrypt(flag)
print(f"ciphertext={ciphertext}")

"""
p = 55099055368053948610276786301
G = (19663446762962927633037926740, 35074412430915656071777015320)
Q = (26805137673536635825884330180, 26376833112609309475951186883)
ciphertext=b"k\xe8\xbe\x94\x9e\xfc\xe2\x9e\x97\xe5\xf3\x04'\x8f\xb2\x01T\x06\x88\x04\xeb3Jl\xdd Pk$\x00:\xf5"
"""

add_THcurve部分符合符合https://www.hyperelliptic.org/EFD/g1p/auto-twistedhessian.html 的定义。 所以按照文章里套换元 x’=X/Z y’=Y/Z 得到 ax’^3+y’^3+z’^3=dx’y’z’这样构造出了齐次式子之后就可以构造椭圆曲线了。所以现在只需要求a的值即可代入脚本求解。因为

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d = (a*G[0]^3+G[1]^3+1)%p*inverse(G[0]*G[1],p)%p

利用G和Q构造方程解出a

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p = 55099055368053948610276786301
Gx = 19663446762962927633037926740
Gy = 35074412430915656071777015320
Qx = 26805137673536635825884330180
Qy = 26376833112609309475951186883
# 计算 Gy^3 + 1 mod p
Gy_cubed = pow(Gy, 3, p)
Gy_cubed_plus_1 = (Gy_cubed + 1) % p
# 计算 Qy^3 + 1 mod p
Qy_cubed = pow(Qy, 3, p)
Qy_cubed_plus_1 = (Qy_cubed + 1) % p
# 计算分子：(Gy^3+1)*Qx*Qy - (Qy^3+1)*Gx*Gy mod p
term1 = (Gy_cubed_plus_1 * Qx) % p
term1 = (term1 * Qy) % p
term2 = (Qy_cubed_plus_1 * Gx) % p
term2 = (term2 * Gy) % p
numerator = (term1 - term2) % p
# 计算分母：Qx^3*Gx*Gy - Gx^3*Qx*Qy mod p
Qx_cubed = pow(Qx, 3, p)
term3 = (Qx_cubed * Gx) % p
term3 = (term3 * Gy) % p
Gx_cubed = pow(Gx, 3, p)
term4 = (Gx_cubed * Qx) % p
term4 = (term4 * Qy) % p
denominator = (term3 - term4) % p
# 计算逆元
inv_denominator = pow(denominator, -1, p)
a = (numerator * inv_denominator) % p
print(a)
#a=39081810733380615260725035189

求得a的值构建出椭圆曲线后使用 Pohlig Hellman 即可解出 Q = xG 中的 x

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from Crypto.Util.number import *
a = 39081810733380615260725035189
p = 55099055368053948610276786301
P = (19663446762962927633037926740, 35074412430915656071777015320)
Q = (26805137673536635825884330180, 26376833112609309475951186883)
d = (a * Q[0] ** 3 + Q[1] ** 3 + 1) * inverse(Q[0] * Q[1], p) % p
# construct ECC to get a solution of aX^3+Y^3+Z^3=dXYZ
R.<x,y,z> = Zmod(p)[]
cubic = a * x^3 + y^3 + z^3 - d*x*y*z
E = EllipticCurve_from_cubic(cubic,morphism=True)
P = E(P)
Q = E(Q)
P_ord = P.order()
def Pohlig_Hellman(n, P, Q):
    return discrete_log(Q, P, ord=n, operation='+')
x = Pohlig_Hellman(P_ord,P,Q)
print(x)
#x=2633177798829352921583206736
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import hashlib
from Crypto.Cipher import AES
from Crypto.Util.Padding import unpad
x = 2633177798829352921583206736
key = hashlib.sha256(str(x).encode()).digest()
cipher = AES.new(key, AES.MODE_ECB)
ciphertext = b"k\xe8\xbe\x94\x9e\xfc\xe2\x9e\x97\xe5\xf3\x04'\x8f\xb2\x01T\x06\x88\x04\xeb3Jl\xdd Pk$\x00:\xf5"
decrypted_flag = unpad(cipher.decrypt(ciphertext), 16)
print(f"解密后的数据: {decrypted_flag}")
#解密后的数据: b'hgame{N0th1ng_bu7_up_Up_UP!}'

3.Spica

题目：

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from Crypto.Util.number import getPrime, long_to_bytes,bytes_to_long
from secrets import flag
from sage.all import *

def derive_M(n):
    iota=0.035
    Mbits=int(2 * iota * n^2 + n * log(n,2))
    M = random_prime(2^Mbits, proof = False, lbound = 2^(Mbits - 1))
    return Integer(M)

m = bytes_to_long(flag).bit_length()
n = 70
p = derive_M(n)


F = GF(p)
x = random_matrix(F, 1, n)
A = random_matrix(ZZ, n, m, x=0, y=2)
A[randint(0, n-1)] = vector(ZZ, list(bin(bytes_to_long(flag))[2:]))
h = x*A

with open("data.txt", "w") as file:
    file.write(str(m) + "\n")
    file.write(str(p) + "\n")
    for item in h:
        file.write(str(item) + "\n")

隐子集和问题（HSSP / Hidden Subset Sum Problem）。解题参考：https://yanmo312.github.io/2022/11/26/gemima_6/#%E4%B8%89%E3%80%81%E9%9A%90%E5%AD%90%E9%9B%86%E5%92%8C%E9%97%AE%E9%A2%98%EF%BC%88HSSP-Hidden-Subset-Sum-Problem%EF%BC%89

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from Crypto.Util.number import *
from sage.all import *
import time
def read_data(filename):
    with open(filename, 'r') as f:
        m = int(f.readline().strip())
        n = 70
        p = int(f.readline().strip())
        h_line = f.readline().strip() 
        w = list(map(int, h_line[1:-1].split(', '))) 
    return m, n, p, w
# 生成 orthoLattice 的相关函数
def orthoLattice(b, x0):
    m = b.length()
    M = Matrix(ZZ, m, m)
    # 生成正交矩阵
    for i in range(1, m):
        M[i, i] = 1
    M[1:m, 0] = -b[1:m] * inverse_mod(b[0], x0)
    M[0, 0] = x0
    for i in range(1, m):
        M[i, 0] = mod(M[i, 0], x0)
    return M
def allpmones(v):
    return len([vj for vj in v if vj in [-1, 0, 1]]) == len(v)
def allones(v):
    if all(vj in (0, 1) for vj in v):
        return v
    if all(vj in (0, -1) for vj in v):
        return -v
    return None
# 恢复只包含 {0,1} 或 {-1,0,1} 的向量
def recoverBinary(M5):
    lv = [allones(vi) for vi in M5 if allones(vi)]
    n = M5.nrows()
    for v in lv:
        for i in range(n):
            nv = allones(M5[i] - v)
            if nv and nv not in lv:
                lv.append(nv)
            nv = allones(M5[i] + v)
            if nv and nv not in lv:
                lv.append(nv)
    return Matrix(lv)
def kernelLLL(M):
    n = M.nrows()
    m = M.ncols()
    if m < 2 * n:
        return M.right_kernel().matrix()
    K = 2 ^ (m // 2) * M.height()
    MB = Matrix(ZZ, m + n, m)
    MB[:n] = K * M
    MB[n:] = identity_matrix(m)
    MB2 = MB.T.LLL().T
    assert MB2[:n, : m - n] == 0
    Ke = MB2[n:, : m - n].T
    return Ke
def attack(m, n, p, w):
    print("n =", n, "m =", m)
    iota = 0.035
    nx0 = int(2 * iota * n^2 + n * log(n, 2))
    print("nx0 =", nx0)
    x0 = p
    b = vector(w)
    M = orthoLattice(b, x0)
    t = time.time()
    M2 = M.LLL()
    print("LLL step1: %.1f" % (time.time() - t))
    MOrtho = M2[: m - n]
    print("log(Height, 2) = ", int(log(MOrtho.height(), 2)))
    t2 = time.time()
    ke = kernelLLL(MOrtho)
    print("Kernel: %.1f" % (time.time() - t2))
    if n > 170:
        return
    beta = 2
    tbk = time.time()
    while beta < n:
        if beta == 2:
            M5 = ke.LLL()
        else:
            M5 = M5.BKZ(block_size=beta)
        if len([True for v in M5 if allpmones(v)]) == n:
            break
        if beta == 2:
            beta = 10
        else:
            beta += 10
    print("BKZ beta=%d: %.1f" % (beta, time.time() - tbk))
    t2 = time.time()
    MB = recoverBinary(M5)
    print("Recovery: %.1f" % (time.time() - t2))
    print("Number of recovered vector = ", MB.nrows())
    print("Number of recovered vector.T = ", MB.ncols())
    return MB
m, n, p, w = read_data('data.txt')
res = attack(m, n, p, w)
def bits_to_long(bits):
    return int(''.join(str(bit) for bit in bits), 2)
def extract_flags(MB):
    flags = []
    # 遍历 MB 的每一行，将每一行转换为一个二进制数字
    for row in MB:
        flag_bits = [int(element) for element in row]  # 获取每行的二进制位
        flag_long = bits_to_long(flag_bits)  # 转换为整数
        flag = long_to_bytes(flag_long)  # 转换为字节串
        flags.append(flag)    
    return flags
flags = extract_flags(res)
for flag in flags:
    print(f"Recovered flag: {flag}")
#Recovered flag: b'hgame{U_f0und_3he_5pec14l_0n3!}'

感觉代码最后加个对flag的处理，判断只有符合hgame{}格式的flag输出会好点（但数据不是很大，还是一眼就从输出里找到正确flag）。输出部分还是很好找的是吧（

最后说明这是第一次尝试写wp，肯定有很多不足之处，会继续在日后一点一点完善这个过程。

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