Reverse｜Buuctf SimpleRev

ida64打开，查看伪代码

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 = 357761762382LL;// 数据转为16进制为‘0x534c43444e’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，故实际的字符顺序应为'0x4e44434c53'转为字符为'NDCLS'
  v7 = 0LL;
  v8 = 0;
  v9 = 512969957736LL;                      // 同理解出为'hadow'
  v10 = 0LL;
  v11 = 0;
  text = (char *)join(key3, &v9);      // key3='kills'  ;这个join函数，就是把key3和v9两个字符串相连即'killshadow'
  strcpy(key, key1);
  strcat(key, src);                       // key1 = 'ADSFK',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 )// 将key的字符转为小写的算法,即key = 'adsfkndcls'
      key[i] = key[v3 % v5] + 32;
    ++v3;
  }
  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;
      }
      else
      {
        str2[v2] = (v1 - 39 - key[v3++ % v5] + 97) % 26 + 97;
      }
      if ( !(v3 % v5) )
        putchar(32);
      ++v2;
    }
  }
  if ( !strcmp(text, str2) )
    puts("Congratulation!\n");
  else
    puts("Try again!\n");
  return __readfsqword(0x28u) ^ v12;
}

先是将两个字符串拼接成keys，然后转化为小写，然后将其进行了加密，再与text进行比较

脚本

text = 'killshadow'
key = 'adsfkndcls'

#暴力字典破解
v3=0
v5=len(key)
dict1="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"
flag = ""
for i in range(0,10):
	n=0
	for char in dict1:
		x = (ord(char) - 39 - ord(key[v3%v5])+97)%26+97
		if(chr(x)==text[i]):
			n = n+1
			if(n==1):
				print(char,end="")
	v3=v3+1

flag{KLDQCUDFZO}