Some Internet Cup Competition Writeup

The competition practical only has 6 questions, involving web, misc, and crypto topics. This competition is organized for the client, but the questions are not difficult; otherwise, how could it be a small competition?

Question 1: 1-1#

Search for js, search keyword flag

dasctf{d6e9c56d7f078d298ed4695d899effbe}

Question 2: 2-1#

Open the pdf

There is a text box in the image

Question 3: 3-1#

The source code of the question is as follows, solved by a friend:

import libnum
import gmpy2
from Crypto.Util.number import *
import flag,e1,e2
# Generate prime numbers
p=libnum.generate_prime(1024)
q=libnum.generate_prime(1024)
ec1=pow(bytes_to_long(str(e1).ljust(20,"D").encode()),3,p*q)
ec2=pow(bytes_to_long(str(e2).ljust(20,"A").encode()),5,p*q)
m=libnum.s2n(flag)
n=p*q
c1=pow(m,e1,n)
c2=pow(m,e2,n)

print("n1=",n)
print("ec1=",ec1)
print("c1=",c1)
print("n2=",n)
print("ec2=",ec2)
print("c2=",c2)

# n1= 27929259512873502442719286790227037320417984116570178470037376373267390909685621247157535458203218619293705428397911754453556082799420494496904478215709219317542924547049229150153308059698341011305505985823374280465467094476511869541135508518055946815227085548571701115773386101962695795789178321155174729047033298389886321980592410739667139376075568555729949442873964097042006391886635957242436522435588904492484342259494858627609438654632887244523845583473711604632109405043439047289868784236481926074763997559971182741918345193506253460323445846136663027639802131457594564405906763806426256107923417802076262573737
# ec1= 24979839185643431898760549059477070141596292955202172081572583839065034831779499992829742773873064296311713734486020739853343887094398935731264
# c1= 17695186679431856780362905635257355413310120106982055323913669124182832151093921194946365178919380690844190324897933591567360925332869903671651849060145290581360223200011298757871213149464298017718829480721410479504940393501845624196721013966839230696831321482946841011452400364600924503121451272593970649100410603943321149604376033957124800064565646929720179239631538966228020882733213221035707244692798307971636126058586394357032072695921642665492919186476321028415907982915011972040971875733852055633796811898421692604356476773910338982400925245494707729878821466569634334862311750349321720627252469986162120031838
# n2= 27929259512873502442719286790227037320417984116570178470037376373267390909685621247157535458203218619293705428397911754453556082799420494496904478215709219317542924547049229150153308059698341011305505985823374280465467094476511869541135508518055946815227085548571701115773386101962695795789178321155174729047033298389886321980592410739667139376075568555729949442873964097042006391886635957242436522435588904492484342259494858627609438654632887244523845583473711604632109405043439047289868784236481926074763997559971182741918345193506253460323445846136663027639802131457594564405906763806426256107923417802076262573737
# ec2= 2838620519239658396968146844964839207179863729944843241951228382052657801460586137213053314019699976475855578055607417923815486109050614096157077528657405905877896929808094661904905136761365045387901486261011216958309860644255996588189249
# c2= 10770781309274554738409447671578241895686779262243081931452089039730277591151694112684863740412412713684216227740930573490322958500198235497947657939304932868457999239593145330718657422535271157683896034543125292529800047408131765376686654378173684648427311300423776510153307756388404568013401217965931456538849277670384454454507752525534110389604969437991055504757081225690155489265359117617764571537216746554060783131168749700810806387918510612057149583061938836035963175555630655718716139689761210220525955656039741684390906935720406757364893793459339618913268943282961044530062475057887777134887741597041684698119

Given values, public key

The script specifies the following parameters: n1, n2, ec1, ec2, c1, c2. Both ec1 and ec2 are obtained through small exponent encryption, so we can directly take the fifth root to get the encryption indices e1 and e2.

Use variable k to find a valid k value such that the expression ec2 + k * n is a perfect fifth power.

while 1:
    res=iroot(ec2+k*n,5) #print(res)

In the above loop, the script calculates the fifth root of the expression ec2 + k * n and checks if this fifth root is an integer. If it is an integer, it prints the corresponding string and breaks the loop. Otherwise, it increments the k value and continues the loop.

while True: 
    res = gmpy2.iroot(ec2 + k * n, 5) 
    if res[1] == True: 
        print(long_to_bytes(int(res[0])))

Next, the script uses the extended Euclidean algorithm to calculate the multiplicative inverses of the public keys e1 and e2, obtaining s1 and s2. Finally, the script calculates the s1-th power of c1 and the s2-th power of c2, multiplies these two results, and takes modulo n to obtain the plaintext m. These two indices are used to decrypt the ciphertext c1 and c2, thus obtaining the original message m.

s, s1, s2 = gmpy2.gcdext(e1, e2)
m = (pow(c1, s1, n) * pow(c2, s2, n)) % n  # m = (c1^s1) * (c2^s2)

The decryption script is as follows:

import gmpy2
from Crypto.Util.number import *

n = 27929259512873502442719286790227037320417984116570178470037376373267390909685621247157535458203218619293705428397911754453556082799420494496904478215709219317542924547049229150153308059698341011305505985823374280465467094476511869541135508518055946815227085548571701115773386101962695795789178321155174729047033298389886321980592410739667139376075568555729949442873964097042006391886635957242436522435588904492484342259494858627609438654632887244523845583473711604632109405043439047289868784236481926074763997559971182741918345193506253460323445846136663027639802131457594564405906763806426256107923417802076262573737

ec2 = 2838620519239658396968146844964839207179863729944843241951228382052657801460586137213053314019699976475855578055607417923815486109050614096157077528657405905877896929808094661904905136761365045387901486261011216958309860644255996588189249

c1 = 17695186679431856780362905635257355413310120106982055323913669124182832151093921194946365178919380690844190324897933591567360925332869903671651849060145290581360223200011298757871213149464298017718829480721410479504940393501845624196721013966839230696831321482946841011452400364600924503121451272593970649100410603943321149604376033957124800064565646929720179239631538966228020882733213221035707244692798307971636126058586394357032072695921642665492919186476321028415907982915011972040971875733852055633796811898421692604356476773910338982400925245494707729878821466569634334862311750349321720627252469986162120031838

c2 = 10770781309274554738409447671578241895686779262243081931452089039730277591151694112684863740412412713684216227740930573490322958500198235497947657939304932868457999239593145330718657422535271157683896034543125292529800047408131765376686654378173684648427311300423776510153307756388404568013401217965931456538849277670384454454507752525534110389604969437991055504757081225690155489265359117617764571537216746554060783131168749700810806387918510612057149583061938836035963175555630655718716139689761210220525955656039741684390906935720406757364893793459339618913268943282961044530062475057887777134887741597041684698119

e1 = 34967
e2 = 65535

# Find a valid k value
k = 0

while True:
    res = gmpy2.iroot(ec2 + k * n, 5)
    if res[1] == True:
        print(long_to_bytes(int(res[0])))  # Convert to string
        break
    k = k + 1

# Decrypt
s, s1, s2 = gmpy2.gcdext(e1, e2)
m = (pow(c1, s1, n) * pow(c2, s2, n)) % n  # m = (c1^s1) * (c2^s2)

# Print decryption result
print(long_to_bytes(m))

Question 4: 4-1#

The original code is as follows:

from uuid import *

from caser import something
from secret import flag
base_table="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
result =""

for i in range(len(flag)):
    if flag[i] in base_table:
        result +=base_table[(base_table.index(flag[i])+33)%64]
    else:
        result +=flag[i]
print(result)

#khzj0m{9caZ87VV-8X77-WW/+-78Wa-eZVdaYYe9/dZ}

Write a decryption script based on the given code

encrypted_flag = "khzj0m{9caZ87VV-8X77-WW/+-78Wa-eZVdaYYe9/dZ}"
base_table = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
flag = ""

for char in encrypted_flag:
    if char in base_table:
        index = (base_table.index(char) - 33) % 64
        flag += base_table[index]
    else:
        flag += char
print(flag)