浙江省省赛2024初赛wp

Re

个人感觉今年的初赛re没签到题，但是难题没有前几年的难

ezRe

010发现是pyc文件

nnd，pycdc反编译不了，可能是pyc文件被改了

只能用pycdas，先对key进行异或

类似rc4但是只生成了密钥流，并没有异或密文

后续异或key和51

最后是base64

逆的时候异或密文可以放在最后，不能反编译真恶心

import base64

encoded = 'w53Cj3HDgzTCsSM5wrg6FMKcw58Qw7RZSFLCljRxwrxbwrVdw4AEwqMjw7/DkMKTw4/Cv8Onw4NGw7jDmSdcwq4GGg=='
ciphertext = base64.b64decode(encoded).decode()
key = '7e021a7dd49e4bd0837e22129682551b'

s = list(range(256))
j = 0
key = [ord(c) ^ 102 for c in key] 

for i in range(256):
    j = (j + s[i] + key[i % len(key)]) % 256
    s[i], s[j] = s[j], s[i]

i = j = 0
data = []

for _ in range(50):
    i = (i + 1) % 256
    j = (j + s[i]) % 256
    s[i], s[j] = s[j], s[i]
    k = s[(s[i] + s[j]) % 256]
    data.append(k)


result = ''
for c, k in zip(ciphertext, data):
    result += chr(ord(c) ^ k ^ 51)

print( result)

Midmath

修改特征码DAS为UPX

然后脱壳

发现是找最大路径，用动态规划做

#include <bits/stdc++.h>
using namespace std;

#define MAX_SUM 6668912
int a[510][510], f[510][510], path[510];

int main() {
    srand(time(NULL));
    int x = 1, y = 1;
    FILE *fp = fopen("out", "rb");
    

    while (fscanf(fp, "%d", &a[x][y]) != EOF) {
        if (x == y) {
            y = 1;
            x++;
            continue;
        }
        y++;
    }
    fclose(fp);
    x--; 


    f[1][1] = a[1][1];
    for (int i = 2; i <= x; i++) {
        for (int j = 1; j <= i; j++) {

            f[i][j] = f[i - 1][j] + a[i][j];
            if (j > 1) {
                f[i][j] = max(f[i][j], f[i - 1][j - 1] + a[i][j]); 
            }
        }
    }

    for (int i = 1; i <= x; i++) {
        if (f[x][i] > MAX_SUM) {
            int cx = x, cy = i;
            while (cx > 1) {
                if (f[cx][cy] == f[cx - 1][cy] + a[cx][cy]) {
                    path[cx] = 1; 
                } else {
                    path[cx] = 2; 
                    cy--;
                }
                cx--;
            }
            break;
        }
    }

    for (int i = 2; i <= x; i++) {
        printf("%d", path[i]);
    }
    
    return 0;
}


***
2121111112222111112111111111112121211221122121111121122222222222212222122122212121111121122111121111111112211121221111222222222212111122221111211111221122111111121121212112211222121222222222112222111122222121222222222212211111111211112112212112222111211222122222122222222212221222212122211211222122221121222121212112112212121122111121122111221221221111121211211112122222211111212122221222222112111112122221111121221111111111111112111211111122122111111111111211212122212222111222212212111121112221211
***

最后输出16进制小写

import hashlib
data = "2121111112222111112111111111112121211221122121111121122222222222212222122122212121111121122111121111111112211121221111222222222212111122221111211111221122111111121121212112211222121222222222112222111122222121222222222212211111111211112112212112222111211222122222122222222212221222212122211211222122221121222121212112112212121122111121122111221221221111121211211112122222211111212122221222222112111112122221111121221111111111111112111211111122122111111111111211212122212222111222212212111121112221211"

md5_full = hashlib.md5(data.encode()).hexdigest()
md5_16 = md5_full[8:24]

print(md5_16)
#f4135424cb2f161c

MidRe-1

有花指令

中间一段始终无法处理，使得congratulatuion显示在main中

一开始对输入有循环异或

往上翻找到两个字符串

第一个是key，第二个是iv

看上去有点像aes，findcrypt一下果然是

当时解出来是这个，没有多想为什么前几位是正常的，赛后发现是0xb看成0x8了🥲

from cryptography.hazmat.primitives.ciphers import Cipher, algorithms, modes

ciphertext = bytes([
    0x52, 0x06, 0xC4, 0x9D, 0x28, 0x71, 0x26, 0x04, 0xBA, 0x98,
    0x4D, 0x20, 0x03, 0x81, 0x39, 0x39, 0x8C, 0x6E, 0x14, 0x8C,
    0x7E, 0xBF, 0x44, 0x5A, 0x67, 0xF5, 0x0A, 0x7F, 0x61, 0x7F,
    0xCE, 0x72
])

key = "5855eab53a2275d3".encode('utf-8')
iv = "b051a57d6d05b393".encode('utf-8')


cipher = Cipher(algorithms.AES(key), modes.CBC(iv))
decryptor = cipher.decryptor()


plaintext = decryptor.update(ciphertext) + decryptor.finalize()


plaintext_list = list(plaintext)


a=[0x77, 0x68, 0x61, 0x74, 0x27, 0x73, 0x20, 0x74, 0x68, 0x69,0x73]

for i in range(len(plaintext_list)):
    print(chr(plaintext_list[i]^a[i%len(a)]), end="")

#726974812f6d9e532f9d838924e6c05f

这里key和iv哪怕显示的是16进制依旧是utf-8格式

信创安全

sm4rev-1

附件先是sh脚本，后续是未知文件

看到dir和decompress，猜测可能是在解压，于是修改后缀为zip，解压得到elf

题目提示是sm4,发现key和iv

Crypto

myez_encode-1

在ECC上生成pq，其实感觉和ECC没什么关系

from Crypto.Util.number import bytes_to_long, getPrime
from sympy import isprime
import random
from flag import flag
def generate_ecc_parameters():
    x = random.randint(1, 1 << 512)
    y = random.randint(1, 1 << 512)
    return x, y

def find_prime_on_curve(x, y, a, b, ecc_p):
    p = x
    q = y
    while not (isprime(p) and isprime(q)):
        p = random.randint(2, ecc_p - 1)
        q = (p**3 + a * p + b) % ecc_p
    return p, q

def generate_rsa_parameters():
    a = getPrime(512) 
    b = getPrime(512)   
    ecc_p = getPrime(512)  
    x, y = generate_ecc_parameters()
    p, q = find_prime_on_curve(x, y, a, b, ecc_p)
    n = p * q
    print(f"p= {p}\nq= {q}\nn= {n}")
    print(f"a= {a}\nb= {b}")
    print(f"P= {ecc_p}")
    
if __name__ == "__main__":
    generate_rsa_parameters()

n = p*q
e = 9
m = bytes_to_long(flag)
c = pow(m,e,n)
print(c)

'''
n= 23298836191712395990541254600776262066247692725919114528027158820049802443474994576179738462067629079873633948850637889127452791527914591229415148712172587856497614285410824614070907847594399218298016379507879066220104597707859246179921731928508884947347652904142879813069359815823184922170241099916465722623
a= 7388665644223916915334064243181348811184637180763467245762518813757790945069068654378380490110607063038613823004593920489924786053478102905200169738195523
b= 11742940161647091720180482697980016011774828087234021441133595442949631197989696508358388255191793888646498553804646435609849154496274569000398776043150743
P= 11300086101709077144191286182913849072593185125745291892398153828719453495325025227858328617077648296782357912556752467026523366682963139253552060862229027
c= 9314530945343661153059846131608414257092556390479105017633636336832925597262814680689800448223193301814365726128618348603188219757245073917910487794768758461683644600756896595336654006282030911824869219015400826589122838492456940861634378619000373353637666835642505021355710338342048772713981673863167110471
'''

q = (p**3 + a * p + b) % ecc_p左右两边同乘以p，然后sage求解

后续e和phi不互素，用中国剩余定理求解

from libnum import *
from Crypto.Util.number import *
a = 7388665644223916915334064243181348811184637180763467245762518813757790945069068654378380490110607063038613823004593920489924786053478102905200169738195523
b = 11742940161647091720180482697980016011774828087234021441133595442949631197989696508358388255191793888646498553804646435609849154496274569000398776043150743
n = 23298836191712395990541254600776262066247692725919114528027158820049802443474994576179738462067629079873633948850637889127452791527914591229415148712172587856497614285410824614070907847594399218298016379507879066220104597707859246179921731928508884947347652904142879813069359815823184922170241099916465722623
P = 11300086101709077144191286182913849072593185125745291892398153828719453495325025227858328617077648296782357912556752467026523366682963139253552060862229027
e = 9
c= 9314530945343661153059846131608414257092556390479105017633636336832925597262814680689800448223193301814365726128618348603188219757245073917910487794768758461683644600756896595336654006282030911824869219015400826589122838492456940861634378619000373353637666835642505021355710338342048772713981673863167110471


# R.<x> = Zmod(P)[]
# f = x ^ 4 + a * x ^ 2 + b * x - n
# res = f.roots()
# print(res)#2925490712948356009205547798331037409204468852265154197929696123102317330847028997592576845375767951888373634075473448002921250636926630905567362014595493
p = 2925490712948356009205547798331037409204468852265154197929696123102317330847028997592576845375767951888373634075473448002921250636926630905567362014595493
q = n // p
R.<x> = Zmod(p)[]
f = x ^ e - c
f = f.monic()
res1 = f.roots()
print(res1)
R.<x> = Zmod(q)[]
f = x ^ e - c
f = f.monic()
res2 = f.roots()
print(res2)
for i in res1:
    for j in res2:
        m =solve_crt([int(i[0]),int(j[0])],[p,q])
        flag = long_to_bytes(m)
        print(flag)
        
#DASCTF{Very_easy_3NC0dE_Is_1t}