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[Sharif University CTF Quals 2013] one Line encoder For Specialist pRogrammer (crypto200)

CTF Events / Sharif University CTF Quals 2013

04/10/2013

[Sharif University CTF Quals 2013] one Line encoder For Specialist pRogrammer (crypto200)

>>> len(key)
9
>>> set(key)
set(['1', '0'])
>>>
>>>
>>> suctf_enc = lambda text, key: reduce(lambda x, y: (x[0] + str(int(y) ^ int(x[1][0])), x[1][1:] + str(int(x[1][4]) ^ int(x[1][0]))), ''.join(map(lambda x: bin(ord(x))[2:].rjust(8, '0'), text)), ('', key))[0]
>>>
>>>
>>> suctf_enc(text, key)
'10100101011010000001001011101110010000001000010111000001101000110010110110001010100000001110010001111011010101100110101001111111001110010110101111111011100000001011110101111100010101001111100011011111111011000110100010000011000110100000010111101100010110000000110010101001100001101110110111101101001001001101010100001101100101001000111110110110001111001000011000011000110000100100001001001100011101101111111101010101000010101110001100010011000111010110010101101011111100111000010100100001000001101010111111011101101110000111111000000000011111010011011010100010000101111100100111000001111001111010101001101110110110011001111101110011100111101100001101101011010110111011111011001100001111001000101101010100000000010011001001110000101100011101011010101100001011101001110010010110111011011011101001110110010000000010100100101000101001001000100110101110111111101111100010101001101110111110010010101011001001100000000101110001001110100010011010100110100001001010111011101111011000111100111010000011111001111010110010110011001000000100010001111001101111100101000110011111'
>>>
>>> # flag = md5(key)

>>> len(key)

>>> set(key)

set(['1', '0'])

>>>

>>> suctf_enc = lambda text, key: reduce(lambda x, y: (x[0] + str(int(y) ^ int(x[1][0])), x[1][1:] + str(int(x[1][4]) ^ int(x[1][0]))), ''.join(map(lambda x: bin(ord(x))[2:].rjust(8, '0'), text)), ('', key))[0]

>>>

>>> suctf_enc(text, key)

'10100101011010000001001011101110010000001000010111000001101000110010110110001010100000001110010001111011010101100110101001111111001110010110101111111011100000001011110101111100010101001111100011011111111011000110100010000011000110100000010111101100010110000000110010101001100001101110110111101101001001001101010100001101100101001000111110110110001111001000011000011000110000100100001001001100011101101111111101010101000010101110001100010011000111010110010101101011111100111000010100100001000001101010111111011101101110000111111000000000011111010011011010100010000101111100100111000001111001111010101001101110110110011001111101110011100111101100001101101011010110111011111011001100001111001000101101010100000000010011001001110000101100011101011010101100001011101001110010010110111011011011101001110110010000000010100100101000101001001000100110101110111111101111100010101001101110111110010010101011001001100000000101110001001110100010011010100110100001001010111011101111011000111100111010000011111001111010110010110011001000000100010001111001101111100101000110011111'

>>>

>>> # flag = md5(key)

Có thể để ý ngay những chữ viết hoa trong tên bài, nó là LFSR, nhưng kệ Nội dung chính Ở bài này, chúng ta có một hàm encode, đầu vào là...

[Sharif University CTF Quals 2013] too hard? (crypto300)

CTF Events / Sharif University CTF Quals 2013

03/10/2013

[Sharif University CTF Quals 2013] too hard? (crypto300)

Chúng ta cần tìm một số x sao cho: (11111111111111111111112 ^ x) % 123456790123456790123454320987654320987654321 = 1173805180904286755555543817503746512689 Nội dung chính Đây là một bài mà ngay khi đọc đề, trong trái tim nhân hậu và...

[No cON Name Facebook CTF Quals 2013] Writeups

CTF Events / No cON Name Facebook CTF Quals 2013

01/10/2013

[No cON Name Facebook CTF Quals 2013] Writeups

ACCESS LEVEL 1 Open crypto.js file, copy the parameter of eval() function and execute it in firebug, we will get:

function simpleHash(str) {
    var i, hash = 0;
    for (i = 0; i < str.length; i++) {
        hash += (str[i].charCodeAt() * (i + 1))
    }
    return Math.abs(hash) % 31337
}

function ascii_one(foo) {
    foo = foo.charAt(0);
    var i;
    for (i = 0; i < 256; ++i) {
        var hex_i = i.toString(16);
        if (hex_i.length == 1) hex_i = "0" + hex_i;
        hex_i = "%" + hex_i;
        hex_i = unescape(hex_i);
        if (hex_i == foo) break
    }
    return i
}

function numerical_value(str) {
    var i, a = 0,
        b;
    for (i = 0; i < str.length; ++i) {
        b = ascii_one(str.charAt(i));
        a += b * (i + 1)
    }
    return a
}

function encrypt(form) {
    var res;
    res = numerical_value(form.password.value);
    res = res * (3 + 1 + 3 + 3 + 7);
    res = res >>> 6;
    res = res / 4;
    res = res ^ 4153;
    if (res != 0) {
        alert('Invalid password!')
    } else {
        alert('Correct password :)')
    }
    form.key.value = numerical_value(form.password.value);
    form.verification.value = "yes" + simpleHash(form.password.value);
    return true
}

function simpleHash(str) {

var i, hash = 0;

for (i = 0; i < str.length; i++) {

hash += (str[i].charCodeAt() * (i + 1))

}

return Math.abs(hash) % 31337

}

function ascii_one(foo) {

foo = foo.charAt(0);

var i;

for (i = 0; i < 256; ++i) {

var hex_i = i.toString(16);

if (hex_i.length == 1) hex_i = "0" + hex_i;

hex_i = "%" + hex_i;

hex_i = unescape(hex_i);

if (hex_i == foo) break

}

return i

}

function numerical_value(str) {

var i, a = 0,

for (i = 0; i < str.length; ++i) {

b = ascii_one(str.charAt(i));

a += b * (i + 1)

}

return a

}

function encrypt(form) {

var res;

res = numerical_value(form.password.value);

res = res * (3 + 1 + 3 + 3 + 7);

res = res >>> 6;

res = res / 4;

res = res ^ 4153;

if (res != 0) {

alert('Invalid password!')

} else {

alert('Correct password :)')

}

form.key.value = numerical_value(form.password.value);

form.verification.value = "yes" + simpleHash(form.password.value);

return true

}

So, to get the goodboy, we must find a number res such that:...

CSAW CTF 2013 / CTF Events

24/09/2013

[CSAW CTF 2013] crackme (Re300)

crackme – 300 Points Solved by 174 teams. nc 128.238.66.218 54321 crackme File dự phòng: https://db.tt/Yw5prmr3

[CSAW CTF 2013] onlythisprogram (Crypto300)

CSAW CTF 2013 / CTF Events

23/09/2013

[CSAW CTF 2013] onlythisprogram (Crypto300)

onlythisprogram – 300 Points Solved by 127 teams. onlythisprogram.tgz File dự phòng: https://db.tt/3HpohDG2

CSAW CTF 2013 / CTF Events

23/09/2013

[CSAW CTF 2013] CSAWpad (Crypto100)

CSAWpad – 100 Points Solved by 141 teams. csawpad.py File dự phòng: https://db.tt/p6zWhqXR

CSAW CTF 2013 / CTF Events

23/09/2013

[CSAW CTF 2013] keygenme (Re400)

keygenme – 400 Points Solved by 100 teams. nc 128.238.66.219 14549 keygenme32.elf File dự phòng: https://db.tt/LATd1AuP Nội dung chính Đây là note thì đúng hơn là một writeup Ở bài này chúng ta được...

CSAW CTF 2013 / CTF Events

23/09/2013

[CSAW CTF 2013] Impossible (Re500)

Ưu tiên viết bài này trước, vì đây là lần đầu tiên chạm tới ngưỡng 500 đầy danh giá, mà có lẽ sẽ mãi cho đến về sau cũng khó lặp lại được Impossible...

WarGame

17/09/2013

[WarGame] php? c?

http://wargame.kr:8080/prob/7/ :p Xem source, được gợi ý về file index.phps, mở nó ra coi:

<?php
 $pass="*** blind ***";
 if(isset($_POST['d1']) && isset($_POST['d2'])){
  $input1=(int)$_POST['d1'];
  $input2=(int)$_POST['d2'];
  if(!is_file("/tmp/p7")){exec("gcc -o /tmp/p7 ./p7.c");}
  $result=exec("/tmp/p7 ".$input1);
  if($result!=1 && $result==$input2){echo $pass;}else{echo "try again!";}
 }else{echo ":p";}
?>

<?php

$pass="*** blind ***";

if(isset($_POST['d1']) && isset($_POST['d2'])){

$input1=(int)$_POST['d1'];

$input2=(int)$_POST['d2'];

if(!is_file("/tmp/p7")){exec("gcc -o /tmp/p7 ./p7.c");}

$result=exec("/tmp/p7 ".$input1);

if($result!=1 && $result==$input2){echo $pass;}else{echo "try again!";}

}else{echo ":p";}

Mở tiếp file p7.c (http://wargame.kr:8080/prob/7/p7.c):

#include <stdio.h>
#include <stdlib.h>
void nono();
int main(int argc,char **argv){
	int i;
	if(argc!=2){nono();}
	i=atoi(argv[1]);
	if(i<0){nono();}
	i=i+5;
	if(i>4){nono();}
	if(i<5){printf("%d",i);}
	return 0;
}
void nono(){
	printf("%d",1);
	exit(1);
}

#include <stdio.h>

#include <stdlib.h>

void nono();

int main(int argc,char **argv){

int i;

if(argc!=2){nono();}

i=atoi(argv[1]);

if(i<0){nono();}

i=i+5;

if(i>4){nono();}

if(i<5){printf("%d",i);}

return 0;

}

void nono(){

printf("%d",1);

exit(1);

}

Chương trình này nhận 1 tham số, chính là cái đầu tiên mà...

CodeEngn

16/09/2013

[CodeEngn] Basic RCE – Level 20

http://codeengn.com/challenges/basic/20 This program needs a key file. What does the data in the file crackme3.key have to be to make it print the above message. Ex) 41424344454647 (This problem has multiple answers, so post...

Yêu CTF Blog

[Sharif University CTF Quals 2013] one Line encoder For Specialist pRogrammer (crypto200)

[Sharif University CTF Quals 2013] too hard? (crypto300)

[No cON Name Facebook CTF Quals 2013] Writeups

[CSAW CTF 2013] crackme (Re300)

[CSAW CTF 2013] onlythisprogram (Crypto300)

[CSAW CTF 2013] CSAWpad (Crypto100)

[CSAW CTF 2013] keygenme (Re400)

[CSAW CTF 2013] Impossible (Re500)

[WarGame] php? c?

[CodeEngn] Basic RCE – Level 20

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