BCTF2014线上赛crypto400地铁难挤writeup

in writeups

地铁难挤: 400 描述

米特尼克需要用社工办法 拿到THU安全专家的磁盘镜像以了解更多信息，于是他收买了THU专家的博士生，来到BJ市需要与博士生当面联系。但是，来到BJ市之后遇到的第一个问题 就是交通。BJ市人满为患，上下地铁时人们也不先下后上，而是互相挤。左边的人想挤到右边下车，右边的人也想挤到左边上车。你作为米特尼克在BJ的一位小 伙伴，能否帮他和所有乘客设计一个尽量少移动次数的方案，使得需要上车的人都上车，需要下车的人都下车。218.2.197.242:6000 or 218.2.197.243:6000

提示

此题是PPC 1. 地铁和车都是背景描述而已，和题目没关系，本题的目标就是让左边的人 L 都到 右边去，右边的人 R 都到左边来 2. 人的移动规则和游戏规则需要大家遍历出来，每次输入一个数字(20以内) 都不知道PPC什么意思。。反正先连接上去看看 要算一个sha1，先写了个python跑了一次X是三位的，但是发现每次连接字符串都会变。。。只能取得了再跑，后来发现用python写的会超时。。。然后尝试用hashcat来破 命令如下command = ‘cudaHashcat64 –custom-charset1 ?l?u?d -m 100 -a 3 “+result+” “+base+”?1?1?1?1”‘ 连上之后是一个什么游戏。。。 玩了半天，再根据题意。规则是空格分隔左右两边，最后要让L都到右边，R都到左边。 一次输入一个数字，代表这个字符串的第几个位置（空格也算一个位置），让这个位置上的人往另外一侧移动 一次最多只能移动两个人 如LRLRLRLRL R 输入9,，就变成LRLRLRLR LR 如LRLRLRLRL R 输入8，就变成LRLRLRL LRR，即两个人都到另一侧，并且互换位置 若输入不合法则返回wrong answer游戏失败。 下面就是游戏的算法啦 xin5739写的

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//xin5739
#include <iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include "queue"
#define INF 100000000
using namespace std;
int sta\[1<<22\]\[22\];
int opp\[1<<22\]\[22\],ops\[1<<22\]\[22\],anss\[1<<22\]\[22\];
int len;
int judge(int s,int p)//判断是否达到目标,p是空格位置
{
     for (int i=0;i<len;i++)
     {
           if(i<p&&(s&(1<<i))==0) return 0;
           if(i>p&&(s&(1<<i))==1) return 0;
     }
     return 1;
}
int check(int s,int &p,int &now)//检查是否是合法移动,p是空格位置
{
       int x,y;
       x=p;y=now;
       if(x==y) return -1;
       int f0=0,f1=0;
       if(x>y)
       {
             x--;
             int t=x;x=y;y=t;
       }
       else x++;
       for (int i=x;i<=y;i++)
       {
            if(s&(1<<i)) f1=1;
            else f0=1;
       }
       if(f1==1&&f0==1)
       {
            if(y-x+1>2) return -1;
            if(p<x)//pxy->yxp
            {
                  if(s&(1<<y))
                  {
                      s|=(1<<p);
                      s^=(1<<y);
                  }
                  else if(s&(1<<p)) s^=(1<<p);
                  now=p;
                  p=y;
            }
            else //xyp->pyx
            {
                if(s&(1<<x))
                {
                    s|=(1<<p);
                    s^=(1<<x);
                }
                 else if(s&(1<<p)) s^=(1<<p);
                 now=p;
                p=x;
            }
            return s;
       }
       else
       {
           if(p<x) //pxxxx->xxxxp
           {
                 for (int j=p;j<y;j++)
                 {
                      if(s&(1<<(j+1)))
                      {
                            s|=(1<<j);
                      }
                      else if(s&(1<<j)) s^=(1<<j); 
                 }
                 if(s&(1<<y)) s^=(1<<y);
                 now=p;
                 p=y;
           }
           else //xxxxp->pxxxx
           {
              // if(p!=y+1) printf("1111213123");
                for (int j=p;j>x;j--)
                {
                    if(s&(1<<(j-1)))
                    {
                        s|=(1<<j);
                    }
                    else if(s&(1<<j)) s^=(1<<j);
                }
                if(s&(1<<x)) s^=(1<<x);
                now=p;
                p=x;
           }
           return s;
       }
       return -1;
}
struct node
{
    int s,p;
};
void bin2str(int s,int p)
{
    for (int i=0;i<len;i++)
    {
        if(i==p) printf(" ");
        else if(s&(1<<i)) printf("R");
        else printf("L");
    }
    printf("n");
}
void bfs()
{
    queue<node>q;
    node z;
    memset(sta,-1,sizeof(sta));
    int i;
    int s,p;
    for (i=0;i<len;i++)
    {
        s=0;p=i;
        int j;
        for(j=0;j<p;j++) s|=(1<<j);
        z.s=s;z.p=p;
        q.push(z);
    //  bin2str(s,p);
        sta\[s\]\[p\]=0;
    }
    while(!q.empty())
    {
        z=q.front();
        q.pop();
        for (i=0;i<len;i++)
        {
              int t=z.p,now=i;
              int x=check(z.s,t,now);
              if(sta\[x\]\[t\]==-1||sta\[z.s\]\[z.p\]+1<sta\[x\]\[t\])
              {
                    node a;
                    sta\[x\]\[t\]=sta\[z.s\]\[z.p\]+1;
                    a.s=x;a.p=t;
                    opp\[x\]\[t\]=z.p; ops\[x\]\[t\]=z.s;anss\[x\]\[t\]=now;
                    q.push(a);
//                  bin2str(x,t);
//                  x=check(x,t,now);
//                  printf("%dn",now);
//                  bin2str(x,t);
              }
        }
    }
}
void solve(int s,int p)
{
     if(sta\[s\]\[p\]==0) 
     {
    //    bin2str(s,p);
        // //if(judge(s,p))printf("11111n");
         return;
     }
//   bin2str(s,p);
     printf("%dn",anss\[s\]\[p\]+1);
     solve(ops\[s\]\[p\],opp\[s\]\[p\]);
}
int main() {
//  freopen("E:\\in.txt","r",stdin);
//  freopen("E:\\out.txt","w",stdout);
    char str\[22\];
    gets(str);
    //printf("%sn",str);
    len=strlen(str);
    int s=0,p;
    for (int i=0;i<len;i++)
    {
          if(str\[i\]==' ') p=i;
          else if(str\[i\]=='R') s|=(1<<i);
    }
//  printf("%d %dn",s,p);
//  int now=8;
//  s=check(s,p,now);
//  bin2str(s,p);
    bfs();
//  printf("%dn",sta\[s\]\[p\]);
    solve(s,p);
    return 0;
}

后来我写的

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//ROIS_yufan
//BCTF_crypto400
#include "stdio.h"
#include "string.h"
char czInput\[20\];
const int len = 16;
void PrintQueue()
{
    //printf("%sn", czInput);
}
void PrintRst(int rst)
{
    printf("%dn", rst + 1);
}
void ShiftSpace(int iWhere)
{
    int iSpace;
    for (iSpace = 0; iSpace < len; iSpace++)
    {
        if (czInput\[iSpace\] == ' ')
        {
            break;
        }
    }
    if (iSpace < iWhere)
    {
        while (iSpace != iWhere)
        {
            iSpace++;
            PrintRst(iSpace);
            czInput\[iSpace - 1\] = czInput\[iSpace\];
            czInput\[iSpace\] = ' ';
            PrintQueue();
        }
    }
    else if (iSpace > iWhere)
    {
        while (iSpace != iWhere)
        {
            iSpace--;
            PrintRst(iSpace);
            czInput\[iSpace + 1\] = czInput\[iSpace\];
            czInput\[iSpace\] = ' ';
            PrintQueue();
        }
    }
}
void RoolL(int curL)
{
    if (curL > 0)
    {
        ShiftSpace(curL + 1);
    }
    //最左的L特例
    else
    {
        ShiftSpace(curL + 2);
    }
}
void SwapToRight(int iSpace)
{
    //最左的L特例
    if (iSpace == 1)
    {
        iSpace++;
    }
    PrintRst(iSpace - 2);
    czInput\[iSpace\] = czInput\[iSpace - 2\];
    czInput\[iSpace - 2\] = ' ';
    PrintQueue();
}
void main()
{
    gets(czInput);
    int i;
    for (i = len - 1; i >= 0; i--)
    {
        if (czInput\[i\] != 'L')
        {
            for (int j = i - 1; j >= 0; j--)
            {
                if (czInput\[j\] == 'L')
                {
                    RoolL(j);
                    SwapToRight(j + 1);
                    i = len;
                    break;
                }
            }
        }
    }
    for (i = 0; i < len; i++)
    {
        if (czInput\[i\] == ' ')
        {
            PrintRst(i + 1);
            break;
        }
    }
}

然后写个脚本玩游戏就好啦。

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#!/usr/bin/env python  
#\-\*\- coding:utf-8 -*-
"""
BCTF_crypto300
ROIS_yufan
"""
import sys
import os
import base64,binascii,zlib
import hashlib
import re, socket
command = 'cudaHashcat64 --custom-charset1 ?l?u?d -m 100 -a 3 "+result+" "+base+"?1?1?1?1"'
def run(result, base):
    f = os.popen("oclHashcat-1.01cudaHashcat64.exe --custom-charset1 ?l?u?d -m 100 -a 3 "+result+" "+base+"?1?1?1?1");
    return f.read()
def getrst(str, result, base):
    partten = re.compile(result + r':('+base+'....'+')')
    return str\[str.find(result)+41:str.find(result)+61\]
def main():
    HOST = '218.2.197.242'    # The remote host
    PORT = 6000              # The same port as used by the server
    s = socket.socket(socket.AF\_INET, socket.SOCK\_STREAM)
    s.connect((HOST, PORT))
    print s.recv(1024)
    strrecv = s.recv(1024)
    print strrecv
\#    strrecv = 'SHA1("5kEQPr0sul1aWUBb" + X).hexdigest() == "952792a53660aac0cc4d9eb1277f9b00d0ebf48c", X is a string of alphanumeric'
    partten = re.compile(r'SHA1("(.*)" ')
    base = partten.findall(strrecv)\[0\]
    partten = re.compile(r'== "(.*)"')
    result = partten.findall(strrecv)\[0\]
    print base
    print result
\#    result = '848656a9e7f76052e8a73d42d17ec02400732e06'
\#    base = 'rVYQ76B4IJdCzy8m'
    rst = getrst(run(result, base), result, base)
    print rst
    s.sendall(rst\[-4:\]+'n')
    print s.recv(1024)
    while(1):
        strrecv = s.recv(1024)
        print repr(strrecv)
        pattern = re.compile(r'\[LR \]+n')
        if not pattern.search(strrecv):
            strrecv = s.recv(1024)
        strrecv = pattern.findall(strrecv)\[0\]
        print strrecv
        f = os.popen("c4.exe "+strrecv.replace(' ', '*'))
        for line in f:
            print repr(line)
            s.sendall(line)
            strrecv = s.recv(1024)
    #        pattern = re.compile(r'\[LR \]+n')
    #        strrecv = pattern.findall(strrecv)\[0\]
            print strrecv
    print s.recv(1024)
\#    while len(strrecv):
\#        input = raw_input()
\#        s.sendall(input+'n')
\#        strrecv = s.recv(1024)
\#        pattern = re.compile(r'\[LR \]+n')
\#        strrecv = pattern.findall(strrecv)\[0\]
\#        print strrecv
    return
if \_\_name\_\_ == '\_\_main\_\_':
    main()

Comment and share

BCTF2014线上赛crypto300比特币钱包writeup

in writeups

比特币钱包: 300 描述

来到中国后，米特尼克 身无分文了，怎样赚一大笔钱以备不时之需呢？比特币的爆发引起了他的注意。FBI 从丝绸之路缴获的大量比特币成了他的目标，他也想借此机会嘲笑一下 FBI。不费吹灰之力，他就搞定了这笔比特币巨款。你知道他是怎么搞定的吗？http://bctf.cn/files/downloads /robotum_9332cfdb5e503889e24e757d962a7454.html

提示

1. 机器人的眼睛是一个时钟，时钟是会走的阿，亲！2. 要做出此题，请先研究清楚比特币地址签名机制和 warpwallet 的用途3. 邮件发送格式说明，邮件内容那一行千万别写中文，verify 的部分不包含首尾的空格和回车(也就是 strip)，下面的 signature 是 base64 串，可以使用 bitcoin-qt 客户端进行 sign 和 verify message 来验证 http://bctf.cn/files/downloads/robotum_9332cfdb5e503889e24e757d962a7454.html内容如下

嘿！你们好！

我是 FBI 的比特币守护机器人，我的地址是 [email protected], 我正在守护着 FBI 的一个重要比特币地址 1Atk95NnaQDiegEkqjJvg6c2KkJbSr2BEL （听说里面有很多钱！）。

你们知道吗？最近我学会了使用 Warpwallet，它既简单，又强大，还易于使用。哈哈！你们肯定没法猜到我的密钥是什么，不要偷窥哦。

好了，如果我的主人想取回存储在我这里的机密信息的话，只需要向我发送邮件就可以了。当然，邮件要使用这个重要比特币地址来签名，另外，你还要告诉我你想要取什么东西，只有让我验证通过了我才能告诉你。

为了避免忘记，我再重复一遍邮件的格式。

—–BEGIN BITCOIN SIGNED MESSAGE—–
这里写邮件的内容啦！
—–BEGIN SIGNATURE—–
这里是签名串
—–END BITCOIN SIGNED MESSAGE—–

祝你们好运！

这道题完全是靠第一个提示做出来的。。。 在那个网页中发现机器人用的是Warpwallet，于是先google下 https://keybase.io/warp/warp_1.0.6_SHA256_e68d4587b0e2ec34a7b554fbd1ed2d0fedfaeacf3e47fbb6c5403e252348cbfc.html 这个是通过Warpwallet用Passphrase和salt生产比特币地址(公钥)和private key 同时还有相关算法的说明 推 测salt应该就是那个机器人的email，那passphrase呢。。。看第一个提示，猜测是个时间然后就用这个算法找碰撞。。。一开始只从时钟上的 时间试到12点，没跑出来，后面把所有时间都跑了。。结果出了结果20:20，然后就能跑出他的private key啦 算法的代码是github上来的

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#!/usr/bin/env python
#
\# Author: [email protected]
#
\# Description: Implements the WarpWallet algorithm as descibed below:
#
\# s1 = scrypt(key=(passphrase||0x1), salt=(salt||0x1), N=2^18, r=8, p=1, dkLen=32)
\# s2 = pbkdf2(key=(passphrase||0x2), salt=(salt||0x2), c=2^16, dkLen=32, prf=HMAC_SHA256)
\# keypair = generate\_bitcoin\_keypair(s1 xor s2)
#
#modified by ROIS_yufan for BCTF2013 crypto300
import argparse
import binascii
import json
import scrypt
import sys
from passlib.utils import pbkdf2
from pycoin import ecdsa, encoding
from pycoin.ecdsa import secp256k1
class WarpWallet(object):
  def \_\_init\_\_(self, pbkdf2\_count, derived\_key\_len, scrypt\_power, scrypt_p,
               scrypt_r):
    self.dklen = derived\_key\_len
    self.pbkdf2\_count = pbkdf2\_count
    self.scrypt\_power = scrypt\_power
    self.scrypt\_r = scrypt\_r
    self.scrypt\_p = scrypt\_p
  def warp(self, passphrase, salt=""):
    """
    Return dictionary of WarpWallet public and private keys corresponding to
    the given passphrase and salt.
    """
    s1 = binascii.hexlify(self._scrypt(passphrase, salt))
    out = self._pbkdf2(passphrase, salt)
    s2 = binascii.hexlify(out)
    base = binascii.unhexlify(s1)
    s3 = binascii.hexlify(self._sxor(base,out))
    secret_exponent = int(s3, 16)
    public\_pair = ecdsa.public\_pair\_for\_secret\_exponent(secp256k1.generator\_secp256k1, secret_exponent)
    #private\_key = encoding.secret\_exponent\_to\_wif(secret_exponent, compressed=False)
    public\_key = encoding.public\_pair\_to\_bitcoin\_address(public\_pair, compressed=False)
    out = public_key
    return out
  def _scrypt(self, passphrase, salt=""):
    scrypt_key = passphrase + "x01"
    scrypt_salt = salt + "x01"
    out = scrypt.hash(scrypt\_key, scrypt\_salt, N=2**self.scrypt_power,
                      r=self.scrypt\_r, p=self.scrypt\_p, buflen=self.dklen)
    return out
  def _pbkdf2(self, passphrase, salt=""):
    hexlified_key = binascii.hexlify(passphrase) + "02"
    pbkdf2\_key = binascii.unhexlify(hexlified\_key)
    hexlified_salt = binascii.hexlify(salt) + "02"
    pbkdf2\_salt = binascii.unhexlify(hexlified\_salt)
    out = pbkdf2.pbkdf2(secret=pbkdf2\_key, salt=pbkdf2\_salt, keylen=self.dklen,
                        rounds=self.pbkdf2\_count, prf='hmac\_sha256')
    return out
  def _sxor (self, s1, s2):
    # Convert strings to a list of character pair tuples,
    # go through each tuple, converting them to ASCII code (ord),
    # perform exclusive or on the ASCII code,
    # then convert the result back to ASCII (chr),
    # merge the resulting array of characters as a string.
    return ''.join(chr(ord(a) ^ ord(b)) for a,b in zip(s1,s2))
def parse_args():
  parser = argparse.ArgumentParser()
  parser.add\_argument("-c", "--pbkdf2\_count",
                      help="iteration count",
                      type=int,
                      default=2**16)
  parser.add_argument("-d", "--dklen",
                      help="derived key length",
                      type=int,
                      default=32)
  parser.add\_argument("-n", "--scrypt\_power",
                      help="2^n passed as the 'N' param to scrypt",
                      type=int,
                      default=18)
  parser.add\_argument("-p", "--scrypt\_p",
                      help="'p' param to scrypt",
                      type=int,
                      default=1)
  parser.add\_argument("-r", "--scrypt\_r",
                      help="'r' param to scrypt",
                      type=int,
                      default=8)
  parser.add_argument("-P", "--passphrase" ,
                      help="passphrase",
                      type=str)
  parser.add_argument("-S", "--salt" ,
                      help="salt",
                      type=str)
  args = parser.parse_args()
  if not args.passphrase:
    print "Must provide passphrase (-P)"
    sys.exit(1)
  if not args.salt:
    print "Must provide salt (-S)"
    sys.exit(1)
  return args
if \_\_name\_\_ == "\_\_main\_\_":
    f = open("c300dict.txt", 'r')
    for line in f:
        strpass = line.strip()
        salt = '[email protected]'
        wallet = WarpWallet(2**16, 32, 18,1, 8)
        str = wallet.warp(strpass, salt)
        if str == '1Atk95NnaQDiegEkqjJvg6c2KkJbSr2BEL':
            print 'get!', strpass
            break
        print line

然后签名发邮件就ok啦，注意按题目要求的格式发。 http://p2pbucks.com/tools/brainwallet/index.html#sign

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BCTF2014线上赛crypto100混沌密码锁writeup

in writeups

BCTF结束了。。。有点可惜吧。。。那个窃密木马因为一个小错误到手的300分就这样没了。。。不然还能进前十的- -

题目 混沌密码锁: 100 描述

据 传说，米特尼克进任何门都是不需要钥匙的，无论是金锁银锁还是密码锁。使用伪造身份在BAT安全部门工作的时候，有一扇带着密码锁的大门吸引了他的注意。 门后面到底藏着什么呢？米特尼克决定一探究竟。 http://bctf.cn/files/downloads/passcode_396331980c645d184ff793fdcbcb739b.py 218.2.197.242:9991 218.2.197.243:9991 passcode_396331980c645d184ff793fdcbcb739b.py

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#!/usr/bin/env python2
#\-\*\- coding:utf-8 -*-
import base64,binascii,zlib
import os,random
base = \[str(x) for x in range(10)\] + \[ chr(x) for x in range(ord('A'),ord('A')+6)\]
def abc(str):
    return sha.new(str).hexdigest()
def bin2dec(string_num):
    return str(int(string_num, 2))
def hex2dec(string_num):
    return str(int(string_num.upper(), 16))
def dec2bin(string_num):
    num = int(string_num)
    mid = \[\]
    while True:
        if num == 0: break
        num,rem = divmod(num, 2)
        mid.append(base\[rem\])
    return ''.join(\[str(x) for x in mid\[::-1\]\])
def dec2hex(string_num):
    num = int(string_num)
    mid = \[\]
    while True:
        if num == 0: break
        num,rem = divmod(num, 16)
        mid.append(base\[rem\])
    return ''.join(\[str(x) for x in mid\[::-1\]\])
def hex2bin(string_num):
    return dec2bin(hex2dec(string_num.upper()))
def bin2hex(string_num):
    return dec2hex(bin2dec(string_num))
def reverse(string):
    return string\[::-1\]
def read_key():
    os.system('cat flag')
def gb2312(string):
    return string.decode('gb2312')
answer='78864179732635837913920409948348078659913609452869425042153399132863903834522365250250429645163517228356622776978637910679538418927909881502654275707069810737850807610916192563069593664094605159740448670132065615956224727012954218390602806577537456281222826375'
func_names = \['fun1', 'fun2', 'fun3', 'fun4', 'fun5', 'fun6', 'fun7', 'fun8', 'fun9'\]
f={}
f\['fun1'\]=reverse
f\['fun2'\]=base64.b64decode
f\['fun3'\]=zlib.decompress
f\['fun4'\]=dec2hex
f\['fun5'\]=binascii.unhexlify
f\['fun6'\]=gb2312
f\['fun7'\]=bin2dec
f\['fun8'\]=hex2bin
f\['fun9'\]=hex2dec
def check_equal(a, b):
    if a == b:
        return True
    try:
        if int(a) == int(b):
            return True
    except:
        return False
    return False
def main():
    print "Welcome to Secure Passcode System"
    print "First, please choose function combination:"
    print "2"
    for in1 in range(1,10):
        for in2 in range(1,10):
            for in3 in range(1,10):
                for in4 in range(1,10):
                    in1=str(in1)
                    in2=str(in2)
                    in3=str(in3)
                    in4=str(in4)
    f1='fun'+in1\[:1\]
    f2='fun'+in2\[:1\]
    f3='fun'+in3\[:1\]
    f4='fun'+in4\[:1\]
    if f1 not in func\_names or f2 not in func\_names or f3 not in func\_names or f4 not in func\_names:
        print 'invalid function combination'
        exit()
    try:
        answer_hash = f\['fun6'\](f\['fun2'\](f\[f1\](f\[f2\](f\[f3\](f\[f4\](answer))))))
    except:
        print "Wrong function combination, you bad guy!"
        exit()
    if len(answer_hash) == 0:
        print 'You must be doing some little dirty trick! Stop it!'
        exit()
    usercode = raw_input('Your passcode: ')
    try:
        user_hash = f\['fun6'\](f\['fun2'\](f\[f1\](f\[f2\](f\[f3\](f\[f4\](usercode))))))
        if user\_hash == answer\_hash:
            if check_equal(answer, usercode):
                print "This passcode has been locked, please use the new onen"
            else:
                print "Welcome back! The door always open for you, your majesty! "
                read_key()
        else:
            print "Sorry, bad passcode.n"
    except:
        print "Sorry, bad passcode. Please try again."
if \_\_name\_\_ == '\_\_main\_\_':
    main()

这个程序就是我们自选f1-f4四种函数answer=’78864179732635837913920409948348078659913609452869425042153399132863903834522365250250429645163517228356622776978637910679538418927909881502654275707069810737850807610916192563069593664094605159740448670132065615956224727012954218390602806577537456281222826375’ answer_hash = f[‘fun6’](f[‘fun2’](f[f1](f[f2](f[f3](f[f4](answer)))))) 然后将我们输入的usercode做同样变化 user_hash = f[‘fun6’](f[‘fun2’](f[f1](f[f2](f[f3](f[f4](usercode)))))) 相等即可通过（usercode不能与answer相同）。 如果输入的函数组合不符要求会”Wrong function combination, you bad guy!” 首先用暴力的方法求出函数的组合，只有唯一一种 f1=’fun3’ f2=’fun5’ f3=’fun1’ f4=’fun4’ 下面是关键

经过f[f1](ff2))之后是一个BASE64的串，然后用fun2进行base64解密，再通过fun6生成hash f[f1](ff2)) = ‘ztLU2s/rxOPU2s/rztLKssO0tcTTw7nIuOi3rdLrv8+2qNK7teOyu7rD08O7ucrHsfDTw8HLv7TV4r7ku7C+wL3hy8DE4771tcPE2A==’ 解 码后ff1“>’fun2’ = ‘xcexd2xd4xdaxcfxebxc4xe3xd4xdaxcfxebxcexd2xcaxb2xc3xb4xb5xc4xd3xc3xb9xc8xb8xe8xb7xadxd2xebxbfxcfxb6xa8xd2xbbxb5xe3xb2xbbxbaxc3xd3xc3xbbxb9xcaxc7xb1xf0xd3xc3xc1xcbxbfxb4xd5xe2xbexe4xbbxb0xbexc0xbdxe1xcbxc0xc4xe3xbexf5xb5xc3xc4xd8’ 参考http://zh.wikipedia.org/zh-cn/Base64 Base64是一种基于64个可打印字符来表示二进制数据的表示方法。 转 换的时候，将三个byte的数据，先后放入一个24bit的缓冲区中，先来的byte占高位。数据不足3byte的话，于缓冲区中剩下的bit用0补足。 然后，每次取出6（因为2^6=64）个bit，按照其值选择 ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/中的字符作为编码 后的输出。不断进行，直到全部输入数据转换完成。 当原数据长度不是3的整数倍时, 如果最后剩下两个输入数据，在编码结果后加1个“=”；如果最后剩下一个输入数据，编码结果后加2个“=”；如果没有剩下任何数据，就什么都不要加，这样才可以保证资料还原的正确性。 回到题中，最后四位是‘2A==’，对应为110110 000000 000000 000000(不足的填0了) 因为两个’=’所以三个比特中的后两个是没有数据的，只有前8个bit对应数据x90 那如果是’2B==’呢，对应为110110 000001 000000 000000 后面的数据是无用的，有用的依然是前8bit 对应x90 so……‘ztLU2s/rxOPU2s /rztLKssO0tcTTw7nIuOi3rdLrv8+2qNK7teOyu7rD08O7ucrHsfDTw8HLv7TV4r7ku7C+wL3hy8DE4771tcPE2B==’ 通 过base64解码的结果和 ‘ztLU2s/rxOPU2s /rztLKssO0tcTTw7nIuOi3rdLrv8+2qNK7teOyu7rD08O7ucrHsfDTw8HLv7TV4r7ku7C+wL3hy8DE4771tcPE2A==’ 是 相同的 后面就简单啦，逆回去求usercode就好啦。

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#!/usr/bin/env python2
#\-\*\- coding:utf-8 -*-
#BCTF_Crypto100
#ROIS_yufan
import base64,binascii,zlib
import os,random
base = \[str(x) for x in range(10)\] + \[ chr(x) for x in range(ord('A'),ord('A')+6)\]
def abc(str):
    return sha.new(str).hexdigest()
def bin2dec(string_num):
    return str(int(string_num, 2))
def hex2dec(string_num):
    return str(int(string_num.upper(), 16))
def dec2bin(string_num):
    num = int(string_num)
    mid = \[\]
    while True:
        if num == 0: break
        num,rem = divmod(num, 2)
        mid.append(base\[rem\])
    return ''.join(\[str(x) for x in mid\[::-1\]\])
def dec2hex(string_num):
    num = int(string_num)
    mid = \[\]
    while True:
        if num == 0: break
        num,rem = divmod(num, 16)
        mid.append(base\[rem\])
    return ''.join(\[str(x) for x in mid\[::-1\]\])
def hex2bin(string_num):
    return dec2bin(hex2dec(string_num.upper()))
def bin2hex(string_num):
    return dec2hex(bin2dec(string_num))
def reverse(string):
    return string\[::-1\]
def read_key():
    os.system('cat flag')
def gb2312(string):
    return string.decode('gb2312')
answer='78864179732635837913920409948348078659913609452869425042153399132863903834522365250250429645163517228356622776978637910679538418927909881502654275707069810737850807610916192563069593664094605159740448670132065615956224727012954218390602806577537456281222826375'
func_names = \['fun1', 'fun2', 'fun3', 'fun4', 'fun5', 'fun6', 'fun7', 'fun8', 'fun9'\]
f={}
f\['fun1'\]=reverse
f\['fun2'\]=base64.b64decode
f\['fun3'\]=zlib.decompress
f\['fun4'\]=dec2hex
f\['fun5'\]=binascii.unhexlify
f\['fun6'\]=gb2312
f\['fun7'\]=bin2dec
f\['fun8'\]=hex2bin
f\['fun9'\]=hex2dec
def check_equal(a, b):
    if a == b:
        return True
    try:
        if int(a) == int(b):
            return True
    except:
        return False
    return False
def main():
    print "Welcome to Secure Passcode System"
    print "First, please choose function combination:"
\#                    in1=raw_input('f1: ')
    f1='fun'+'3'
\#                    in2=raw_input('f2: ')
    f2='fun'+'5'
\#                    in3=raw_input('f3: ')
    f3='fun'+'1'
\#                    in4=raw_input('f4: ')
    f4='fun'+'4'
    print f1, f2, f3, f4
    if f1 not in func\_names or f2 not in func\_names or f3 not in func\_names or f4 not in func\_names:
        print 'invalid function combination'
        exit()
    try:
        answer_hash = f\['fun6'\](f\['fun2'\](f\[f1\](f\[f2\](f\[f3\](f\[f4\](answer))))))
\#        print f\[f4\](answer)
\#        print f\[f3\](f\[f4\](answer))
\#        print repr(f\[f2\](f\[f3\](f\[f4\](answer))))
        print 'original base64', f\[f1\](f\[f2\](f\[f3\](f\[f4\](answer))))
        print 'original decoded base64', repr(f\['fun2'\](f\[f1\](f\[f2\](f\[f3\](f\[f4\](answer))))))
        print 'my base64 decoded', repr(base64.b64decode('ztLU2s/rxOPU2s/rztLKssO0tcTTw7nIuOi3rdLrv8+2qNK7teOyu7rD08O7ucrHsfDTw8HLv7TV4r7ku7C+wL3hy8DE4771tcPE2'+'B'+'=='))
    except:
        print "Wrong function combination, you bad guy!"
        exit()
    if len(answer_hash) == 0:
        print 'You must be doing some little dirty trick! Stop it!'
        exit()
    for c in range(ord('A'), ord('Q')):
        usercode = str (hex2dec(reverse(binascii.hexlify(zlib.compress('ztLU2s/rxOPU2s/rztLKssO0tcTTw7nIuOi3rdLrv8+2qNK7teOyu7rD08O7ucrHsfDTw8HLv7TV4r7ku7C+wL3hy8DE4771tcPE2'+chr(c)+'==')))))
        print usercode
        try:
\#            print f\[f4\](usercode)
\#            print f\[f3\](f\[f4\](usercode))
\#            print repr(f\[f2\](f\[f3\](f\[f4\](usercode))))
\#            print f\[f1\](f\[f2\](f\[f3\](f\[f4\](usercode))))
\#            print repr(f\['fun2'\](f\[f1\](f\[f2\](f\[f3\](f\[f4\](usercode))))))
            user_hash = f\['fun6'\](f\['fun2'\](f\[f1\](f\[f2\](f\[f3\](f\[f4\](usercode))))))
            if user\_hash == answer\_hash:
                if check_equal(answer, usercode):
                    print "This passcode has been locked, please use the new onen"
                else:
                    print "Welcome back! The door always open for you, your majesty! "
            else:
                print "Sorry, bad passcode.n"
        except:
            print "Sorry, bad passcode. Please try again."
if \_\_name\_\_ == '\_\_main\_\_':
    main()

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Author's picture

Eadom

NO PWN NO FUN


@Alibaba


Hangzhou