| TermBreaker |
[Click to reveal]import random
kTarget = 0xB28
k_len = 8
kAlphabet = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
k_alpha_count = 36
def is_allowed(c):
return (c - 65 < 26) or (c - 48 < 10)
def calculate_checksum(s):
sum_value = 0
for i in range(k_len):
sum_value += (i + 1) * ord(s[i])
return sum_value
def generateKey():
for attemptCount in range(800):
key_string = ["0"] * k_len
for i in range(6):
key_string[i] = chr(ord("A") + random.randint(0, 25))
currentSum = 0
for i in range(6):
currentSum += (i + 1) * ord(key_string[i])
remaining_value = kTarget - currentSum
valid_pairs = []
for alphaIndex in range(k_alpha_count):
charValue = ord(kAlphabet[alphaIndex])
neededValue = remaining_value - 7 * charValue
if neededValue % 8 != 0:
continue
finalChar = neededValue // 8
if finalChar >= 0 and is_allowed(finalChar):
valid_pairs.append(
(kAlphabet[alphaIndex], chr(finalChar))
)
if not valid_pairs:
continue
sixthChar, seventh_char = random.choice(valid_pairs)
key_string[6] = sixthChar
key_string[7] = seventh_char
key_string = "".join(key_string)
if calculate_checksum(key_string) == kTarget:
return key_string
return "OOOPOOOP"
generated_keys = set()
printedCount = 0
for guard_count in range(400):
if printedCount >= 10:
break
generatedKeyValue = generateKey()
if generatedKeyValue in generated_keys:
continue
generated_keys.add(generatedKeyValue)
print(generatedKeyValue)
printedCount += 1
exit(0 if printedCount == 10 else 1)
example output:
[cooper@lenowo 6a9950e9cab6678aefe9dc90]$ python keygen.py
UNOCVENZ
VWFEZAZP
OAERSKYN
RIIZKHNW
UWAIKYOQ
XWIIOXPK
WHXGVIWK
UMUEOYKO
SORMRZIK
FSUGIEZT
|
2026-09-15 00:31 |
| FindLicenseKey |
[Click to reveal][user@hostname 6a2c61c06840520e01b21e2d]$ ./findlicensekey "hatenal"
Enter license key to continue:
dr2d1s1XlFNGqY9OXXP0AXX3
Key validated
[user@hostname 6a2c61c06840520e01b21e2d]$
code used:
#include <stdio.h>
void generate_license_key(const char *username, char *outKey)
{
static const char charset[] =
"QAZPLWSXOKMEYDCIJNRFVUHBTG"
"qpalzmwoeirutyskdjfhgxncbv"
"1750284369"; // 62 characters
for (int i = 0; i < 24; i++) {
outKey[i] = charset[(username[i] + i) % 62];
}
outKey[24] = '\0';
}
int main(int argc, char** argv) {
char pass[264];
generate_license_key(argv[1], pass);
printf(pass);
return 0;
}
|
2026-06-16 09:49 |
| ASCII_CRACK |
[Click to reveal]CTF{S3cR3T_AsSc1_Fl4g}{@_@}
found using this python script:
def decode(a:str, b:int):
return chr(ord(a)-b)
s = "IYJ~U4cQ1Q[<mL[(U;`'Ynk/M-i"
new_s = ""
for i in range(len(s)):
new_s += decode(s[i], 6-i)
print(new_s)
|
2026-04-05 09:18 |
| U Can't Pass |
[Click to reveal]replace assembly instruction:
0010117c 74 10 JZ LAB_0010118e
with a NOP
[user@hostname ~]$ ./main.out
Hello in my first programm for crackme.one
Success!
|
2026-04-05 08:23 |