evandrix

Level 1. Make a nasa.gov URL display a text of My Choice 
Level 2. debfKNH1AvtBo deGH9Aq./kiSY denjFRfA8kzL2 
Level 3. Recognize 
Level 4. IP address is 72.14.221.104. What contains artvЂ â € ™ That Points to IT? 
Level 5. Password is the owlvЂ ™ s Name 
Level 6. LetvЂ ™ s See you do some Easy SQL ninjitsu 
Level 7. snifflog.txt â € "ngrep format 
Level 8. Password is on a Picture: not available from Anywhere 
Level 9. Elementary Maths 
Level 10. A poem 

evandrix

packets = []
f = open('packets', 'rb')
data = f.readlines()
f.close()

for line in data:
	line = line.strip()
	packets.append({'index': line.split('\t')[0], 'data': line.split('\t')[1]})
packets = sorted(packets, key=lambda packet: packet['index'])

evandrix

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <cstddef>
#include <cstdio>
#include <cstdlib>

evandrix

/*
First off, logAdd() takes two arguments of type double in log space, logX and logY.
It returns a double value which closely approximates log(X + Y). Consequently, the inputs and output are in log space, and we need not convert in and out of log space to complete this operation.
*/
public static double logAdd(double logX, double logY) {
       // 1. make X the max
       if (logY > logX) {
           double temp = logX;
           logX = logY;
           logY = temp;

evandrix

#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <fcntl.h>
#include <sys/types.h>
#include <sys/event.h>
#include <sys/time.h>
#include <errno.h>
#include <string.h>
#include <inttypes.h>

evandrix

#!/usr/bin/env python

"""
=====================================
PEP 20 (The Zen of Python) by example
=====================================

Usage: %prog

:Author: Hunter Blanks, hblanks@artifex.org / hblanks@monetate.com

evandrix

//
//  Created by Michael Petrov on 12-02-23.
//  Copyright (c) 2012 TenthBit Inc. All rights reserved.
//  http://michaelpetrov.com ([email protected])
//
//
//  This solution performs a timing attack on the fork system call. By monitoring the process closely
//  it is possible to discover where the fork likely happened. With some basic heuristics, it's possible
//  to infer where the wrong character is. With very minor brute force searching it becomes very easy
//  to find the password one letter at a time.

evandrix

Stripe CTF - Work Notes
mpetrov ([email protected])

These notes are very rough. They should give a general idea of how each level was solved.

---- LEVEL 01 (login: e9gx26YEb2) -----
Solution: modifying PATH env variable
Password: kxlVXUvzv

date.c

evandrix

#!/bin/sh

if [ ! -t 0 ]; then
    echo >&2 'STDIN is not a terminal'
    exit 1
fi

clear

cd "$(mktemp -d)"

evandrix

#include <iostream>
#include <algorithm>
#include <vector>
#include <stdio.h>
#include <string.h>
using namespace std;
#define REP(i, a, b)    for(int i=int(a); i<int(b); i++)
typedef long long       ll;

const int MAXS=10001, MAXQ=100001, MAXM=100001;