buu-SimpleRev

64位

直接看main函数

输入d/D执行Decry()函数,输入q/Q退出
重点看Decry()函数

unsigned __int64 Decry()
{
  char v1; // [rsp+Fh] [rbp-51h]
  int v2; // [rsp+10h] [rbp-50h]
  int v3; // [rsp+14h] [rbp-4Ch]
  int i; // [rsp+18h] [rbp-48h]
  int v5; // [rsp+1Ch] [rbp-44h]
  char src[8]; // [rsp+20h] [rbp-40h]
  __int64 v7; // [rsp+28h] [rbp-38h]
  int v8; // [rsp+30h] [rbp-30h]
  __int64 v9; // [rsp+40h] [rbp-20h]
  __int64 v10; // [rsp+48h] [rbp-18h]
  int v11; // [rsp+50h] [rbp-10h]
  unsigned __int64 v12; // [rsp+58h] [rbp-8h]

  v12 = __readfsqword(0x28u);
  *(_QWORD *)src = 'SLCDN';
  v7 = 0LL;
  v8 = 0;
  v9 = 'wodah';
  v10 = '\0';
  v11 = 0;
  text = (char *)join(key3, &v9);
  //key3='kills'
  //text='killshadow'
  strcpy(key, key1);
  //key1='ADSFK'
  strcat(key, src);
  //key='ADSFKNDCLS'
  v2 = 0;
  v3 = 0;
  getchar();
  v5 = strlen(key);
  //v5=10
  for ( i = 0; i < v5; ++i )
  {
    if ( key[v3 % v5] > 64 && key[v3 % v5] <= 90 )
      key[i] = key[v3 % v5] + 32;
    ++v3;
  }
  //遍历key，大写转小写
  //key='adsfkndcls'
  printf("Please input your flag:", src);
  while ( 1 )
  {
    v1 = getchar();
    if ( v1 == 10 )
      break;
    if ( v1 == 32 )
    {
      ++v2;
    }
    else
    {
      if ( v1 <= 96 || v1 > 122 )
      {
        if ( v1 > 64 && v1 <= 90 )
          str2[v2] = (v1 - 39 - key[v3++ % v5] + 97) % 26 + 97;
      }
      //除去了'[\]^_'五个字符
      else
      {
        str2[v2] = (v1 - 39 - key[v3++ % v5] + 97) % 26 + 97;
      }
      if ( !(v3 % v5) )
        putchar(32);
      ++v2;
    }
  }
  if ( !strcmp(text, str2) )
  //比较str2和text
    puts("Congratulation!\n");
  else
    puts("Try again!\n");
  return __readfsqword(0x28u) ^ v12;
}

查看v9的地址发现上大下小
数据在内存中是小端顺序，高位在高地址处，低位在低地址处
(暂定此种方式判断）

事实上正确的scr和v9是转换之后字符串的倒序，即’NDCLS’和’hadow’
text则为’killshadow’
key为’ADSFKNDCLS’

加密函数重点就这一句
str2[v2] = (v1 - 39 - key[v3++ % v5] + 97) % 26 + 97;
逆推一下
(str2[v2]-97)+n26+key[v3++ % v5]-58 = v1
用循环的n26来代替%26
v1=(ord(text[j])-97)+26*i+ord(key[v3%v5])-58

str2 = 'killshadow'
key = 'adsfkndcls'
v3=0
v5=len(key)
flag=[0,0,0,0,0,0,0,0,0,0]
for i in range(0,4):
    for j in range(0,10):
        v1=(ord(str2[j])-97)+26*i+ord(key[v3%v5])-58
        if(v1>65 and v1<=90):
            flag[j]=chr(v1)
        v3=v3+1
for i in flag:
    print(i,end="")

结果即是flag