帮联队里别的师傅看的比赛，做了点简单题

Misc

ez-picture

伪加密改回去解开压缩包

image-20250713190109710

对于key.txt：base100 -> base64

image-20250713190221446
image-20250713190251979

用key对flag文件异或，得到一个png

image-20250713190344888

爆破png的宽高，修改为正确的值即可看到flag

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import binascii
import struct
crcbp = open("download.png", "rb").read()    #填入图片名
crc32frombp = int(crcbp[29:33].hex(),16)    
print(crc32frombp)
 
for i in range(10000):                       
    for j in range(10000):                    
        data = crcbp[12:16] + \
            struct.pack('>i', i)+struct.pack('>i', j)+crcbp[24:29]
        crc32 = binascii.crc32(data) & 0xffffffff
        # print(crc32)
        if(crc32 == crc32frombp):
            print(i, j)
            print('hex:', hex(i), hex(j))
            exit(0)
#3906191683
#1920 1080
#hex: 0x780 0x438
image-20250713190548509
download

Crypto

ez-learning

题目

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from Crypto.Util.number import getPrime, bytes_to_long, long_to_bytes, GCD
import gmpy2 
from secret import flag
def extended_gcd(a, b):
    if a == 0:
        return b, 0, 1
    d, x1, y1 = extended_gcd(b % a, a)
    x = y1 - (b // a) * x1
    y = x1
    return d, x, y

def main():
    print("RSA learning 挑戰 ")
    print("=" * 40)

    m1_val = bytes_to_long(flag_part1_str.encode('utf-8'))
    m2_val = bytes_to_long(flag_part2_str.encode('utf-8'))
    m3_val = bytes_to_long(flag_part3_str.encode('utf-8'))
    m4_val = bytes_to_long(flag_part4_str.encode('utf-8'))

    print("--- 第一部分 ---")
    p1_bits = 256
    q1_bits = 256
    p1 = getPrime(p1_bits)
    q1 = getPrime(q1_bits)
    while p1 == q1: 
        q1 = getPrime(q1_bits)   
    n1 = p1 * q1
    phi1 = (p1 - 1) * (q1 - 1) 
    e1 = 65537
    c1 = pow(m1_val, e1, n1)
    print(f"n1 = {n1}")
    print(f"e1 = {e1}")
    print(f"c1 = {c1}")
    print(f"p1_given = {p1}")
    print("-" * 40)

    print("--- 第二部分 ---")
    p2_bits = 256 
    p2 = getPrime(p2_bits)
    q2 = gmpy2.next_prime(p2)   
    n2 = p2 * q2
    phi2 = (p2 - 1) * (q2 - 1)
    e2 = 65537
    c2 = pow(m2_val, e2, n2)
    print(f"n2 = {n2}")
    print(f"e2 = {e2}")
    print(f"c2 = {c2}")
    print("-" * 40)

    print("--- 第三部分 ---")
    e3 = 3
    m3_bits = m3_val.bit_length()
    
    p3_q3_bits = 128 
    if m3_bits * e3 >= p3_q3_bits * 2 - 10 : 
        p3_q3_bits = (m3_bits * e3 // 2) + 20 

    p3 = getPrime(p3_q3_bits)
    while (p3 - 1) % e3 == 0: 
        p3 = getPrime(p3_q3_bits)
        
    q3 = getPrime(p3_q3_bits)
    while (q3 - 1) % e3 == 0 or p3 == q3:
        q3 = getPrime(p3_q3_bits)
    n3 = p3 * q3   
    phi3 = (p3 - 1) * (q3 - 1)
    c3 = pow(m3_val, e3, n3)
    print(f"n3 = {n3}")
    print(f"e3 = {e3}")
    print(f"c3 = {c3}")
    print("-" * 40)

    print("--- 第四部分 ---")
    p4_bits = 256
    q4_bits = 256
    p4 = getPrime(p4_bits)
    q4 = getPrime(q4_bits)
    while p4 == q4:
        q4 = getPrime(q4_bits)
    n4 = p4 * q4
    phi4 = (p4 - 1) * (q4 - 1)
    e4a = 65537
    e4b = 65539
    c4a = pow(m4_val, e4a, n4)
    c4b = pow(m4_val, e4b, n4)
    print(f"n4 = {n4}")
    print(f"e4a = {e4a}")
    print(f"c4a = {c4a}")
    print(f"e4b = {e4b}")
    print(f"c4b = {c4b}")
    print("=" * 40)
    print("生成完畢。")

if __name__ == "__main__":
    main()
'''
--- 第一部分 ---
n1 = 6812896682529270617889699041268397231216344502100994418898483090399363050725238802825519610890413646309466494321918636827383749031784541498873387892731639
e1 = 65537
c1 = 4427729071087402462891221302870960264377546560852404414763265602889920597278631371895561554533422266459579128710764143355754615245149034250540901716118396
p1_given = 83018022170775357156881992679804004613671250493281300137495505566721083549673
----------------------------------------
--- 第二部分 ---
n2 = 8618941019390135762450560251440447449812344988348002904674567734387521342905027779518989812368861070593805164745239666403382823700616097172234834567736147
e2 = 65537
c2 = 5090157394401735030895991956180326655655690673641213577889910602055845449426923727333236685488219458241927620516575355892867356153466992699239192192910803
----------------------------------------
--- 第三部分 ---
n3 = 44935859354785283479720490973543160302824014776763758544522532443340696020267
e3 = 3
c3 = 345422409558921105091064923418692102360936093114398968000
----------------------------------------
--- 第四部分 ---
n4 = 8050202063335318202668477773676061807230884209991655693527143039730083343444420055444708024772406517257344636757261685775610919720030730312364281795306843
e4a = 65537
c4a = 5944697746898769084130690069563137465626689325180948580645355816012208942196895961186706314322689146225405184339425445878469418993890870252122256787094490
e4b = 65539
c4b = 590873621149745423995166880880817186889129377479311590030588253033344142558245476231076545393071246375504750217939956792516706707037717951120503693278372
========================================
'''

第一部分有了p直接算就行

第二部分p和q十分接近，所以可以通过开根去nextprime挨个试

第三部分低指数幂攻击

第四部分共模攻击

exp：

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from Crypto.Util.number import *
import gmpy2
n1 = 6812896682529270617889699041268397231216344502100994418898483090399363050725238802825519610890413646309466494321918636827383749031784541498873387892731639
e1 = 65537
c1 = 4427729071087402462891221302870960264377546560852404414763265602889920597278631371895561554533422266459579128710764143355754615245149034250540901716118396
p1 = 83018022170775357156881992679804004613671250493281300137495505566721083549673
q1=n1//p1
phi1=(p1-1)*(q1-1)
d1=gmpy2.invert(65537,phi1)
m1=long_to_bytes(pow(c1,d1,n1))

n2 = 8618941019390135762450560251440447449812344988348002904674567734387521342905027779518989812368861070593805164745239666403382823700616097172234834567736147
e2 = 65537
c2 = 5090157394401735030895991956180326655655690673641213577889910602055845449426923727333236685488219458241927620516575355892867356153466992699239192192910803
tmp2=gmpy2.iroot(n2,2)[0]
while n2%tmp2 !=0:
    tmp2=gmpy2.next_prime(tmp2)
p2=n2//tmp2
q2=tmp2
phi2=(p2-1)*(q2-1)
d2=gmpy2.invert(65537,phi2)
m2=long_to_bytes(pow(c2,d2,n2))

n3 = 44935859354785283479720490973543160302824014776763758544522532443340696020267
e3 = 3
c3 = 345422409558921105091064923418692102360936093114398968000
k=0
while True:
    mm=c3+n3*k
    m3,flag=gmpy2.iroot(mm,e3)
    if True==flag:
        break
    k+=1
m3=long_to_bytes(m3)

n4 = 8050202063335318202668477773676061807230884209991655693527143039730083343444420055444708024772406517257344636757261685775610919720030730312364281795306843
e4a = 65537
c4a = 5944697746898769084130690069563137465626689325180948580645355816012208942196895961186706314322689146225405184339425445878469418993890870252122256787094490
e4b = 65539
c4b = 590873621149745423995166880880817186889129377479311590030588253033344142558245476231076545393071246375504750217939956792516706707037717951120503693278372
_gcd,s1,s2=gmpy2.gcdext(e4a,e4b)
m4=long_to_bytes(pow(c4a,s1,n4)*pow(c4b,s2,n4)%n4)
print(m1+m2+m3+m4)
#b'MOCSCTF{y0U_AR3_a_gO0d_learN3er}'

OSINT

ez-college

一搜就是成都大学

image-20250713191236972

Forensics

easymem

在查看记事本的时候就发现了flag（

image-20250713191440783

easysword

看题目名字就盲猜某webshell工具流量，直接丢给Net-A梭一下看，找到有个压缩包

image-20250713191704829

里面的文件都特别小，所以爆破CRC

image-20250713191811263
image-20250713191903694

注意他是按这个顺序排的，得记得换回来（

image-20250713191938726

MOCSCTF{9fc0b9fd2d3f666c8e3631dce9fe2d4d}