Reversing-x64Elf-100
18最佳Writeup由 yuchouxuan 提供
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难度：1
方向：Reverse
题解数：15
解出人数：2460
题目来源:
题目描述: 暂无
note:
undefined8 FUN_004006fd(long param_1)
{
int local_2c;
char *local_28 [4];
local_28[0] = "Dufhbmf";
local_28[1] = "pG`imos";
local_28[2] = "ewUglpt";
local_2c = 0;
while( true ) {
if (0xb < local_2c) {
return 0;
}
if ((int)local_28[local_2c % 3][(local_2c / 3) * 2] - (int)*(char *)(param_1 + local_2c) != 1)
break;
local_2c = local_2c + 1;
}
return 1;
}
signed __int64 __fastcall sub_4006FD(__int64 a1)
{
signed int i; // [rsp+14h] [rbp-24h]
const char *v3; // [rsp+18h] [rbp-20h]
const char *v4; // [rsp+20h] [rbp-18h]
const char *v5; // [rsp+28h] [rbp-10h]
v3 = "Dufhbmf";
v4 = "pG`imos";
v5 = "ewUglpt";
for ( i = 0; i <= 11; ++i )
{
if ( (&v3)[i % 3][2 * (i / 3)] - *(char *)(i + a1) != 1 )
return 1LL;
}
return 0LL;
}
*(char *)(i + a1) =(&v3)[i % 3][2 * (i / 3)]-1
CcmbPZ[dd\ig
DepfU`blmfts____________CdoeT_aklesr____________67 100 111 101 84 95 97 107 108
101 115 114
wrong try!
#include<stdio.h>
int sub_4006FD(void)
{
signed int i; // [rsp+14h] [rbp-24h]
const char *v3; // [rsp+18h] [rbp-20h]
const char *v4; // [rsp+20h] [rbp-18h]
const char *v5; // [rsp+28h] [rbp-10h]
//int a1;
v3 = "Dufhbmf";
v4 = "pG`imos";
v5 = "ewUglpt";
for ( i = 0; i <= 11; ++i )
{
printf("%c",(&v3)[i % 3][2 * (i / 3)]);
// if ( (&v3)[i % 3][2 * (i / 3)] - *(char *)(i + a1) != 1 )
// return 1LL;
}
for ( i = 0; i <= 11; ++i )
{
printf("%s","-");
// if ( (&v3)[i % 3][2 * (i / 3)] - *(char *)(i + a1) != 1 )
// return 1LL;
}
for ( i = 0; i <= 11; ++i )
{
printf("%c",(&v3)[i % 3][2 * (i / 3)]-1);
// if ( (&v3)[i % 3][2 * (i / 3)] - *(char *)(i + a1) != 1 )
// return 1LL;
}
for ( i = 0; i <= 11; ++i )
{
printf("-");
// if ( (&v3)[i % 3][2 * (i / 3)] - *(char *)(i + a1) != 1 )
// return 1LL;
}
for ( i = 0; i <= 11; ++i )
{
printf("%d ",(&v3)[i % 3][2 * (i / 3)]-1);
// if ( (&v3)[i % 3][2 * (i / 3)] - *(char *)(i + a1) != 1 )
// return 1LL;
}
return 0LL;
}
int main(void){
sub_4006FD();
return 0;
}
DepfU`blmfts------------CdoeT_aklesr------------67 100 111 101 84 95 97 107 108
101 115 114
Process exited after 0.07565 seconds with return value 0
请按任意键继续. . .
┌──(kwkl㉿kwkl)-[~/HODL/adworld]
└─$ ./43ba593623a7434fa5437cdd5c7862a1.re
Enter the password: CcmbPZ[dd\ig
Incorrect password!
┌──(kwkl㉿kwkl)-[~/HODL/adworld]
└─$ ./43ba593623a7434fa5437cdd5c7862a1.re 1 ⨯
Enter the password: Code_Talkers
Nice!
┌──(kwkl㉿kwkl)-[~/HODL/adworld]
└─$
Final try!
#include<stdio.h>
int FUN_004006fd( void)
{
int local_2c,n,i;
char *local_28 [4];
int param_1[11]={99};
local_28[0] = "Dufhbmf";
local_28[1] = "pG`imos";
local_28[2] = "ewUglpt";
local_2c = 0;
for ( local_2c = 0; local_2c <= 11; ++local_2c ) {
for (n=0;n<222;n++){
if (local_28[local_2c % 3][(local_2c / 3) * 2] - (n + local_2c) == 1)
param_1[local_2c]=n;
}
local_2c = local_2c + 1;
}
for ( i = 0; i <= 11; ++i ){
printf("%c",param_1[i]);
}
printf("\n------------------------\n");
for ( i = 0; i <= 11; ++i ){
printf("%c",local_28[i % 3][(i / 3) * 2]-1);
}
return 1;
}
int sub_4006FD(void)
{
int i,n; // [rsp+14h] [rbp-24h]
const char *v3; // [rsp+18h] [rbp-20h]
const char *v4; // [rsp+20h] [rbp-18h]
const char *v5; // [rsp+28h] [rbp-10h]
//int a1;
v3 = "Dufhbmf";
v4 = "pG`imos";
v5 = "ewUglpt";
int a1[11]={1};
// for ( i = 0; i <= 11; ++i )
// {
// printf("%c",(&v3)[i % 3][2 * (i / 3)]);
if ( (&v3)[i % 3][2 * (i / 3)] - *(char *)(i + a1) != 1 )
return 1LL;
// }
//
//
//
// printf("\n------------------------\n");
//
//
// for ( i = 0; i <= 11; ++i )
// {
// printf("%c",(&v3)[i % 3][2 * (i / 3)]-1);
//
// }
printf("\n------------------------\n");
for ( i = 0; i <= 11; ++i )
{
// printf("%c",(&v3)[i % 3][2 * (i / 3)]-1);
for ( n = 0; n < 222; n++)
{
if ((&v3)[i % 3][2 * (i / 3)]- n == 1)
a1[i] = n;
}
}
for ( i = 0; i <= 11; ++i ){
printf("%c",a1[i]);
}
printf("\n------------------------\n");
return 0LL;
}
int main(void){
sub_4006FD();
FUN_004006fd();
return 0;
}
Useful code:
int FUN_004006fd( void)
{
int local_2c,n,i;
char *local_28 [4];
int param_1[11]={99};
local_28[0] = "Dufhbmf";
local_28[1] = "pG`imos";
local_28[2] = "ewUglpt";
local_2c = 0;
printf("\n------------------------\n");
for ( i = 0; i <= 11; ++i ){
printf("%c",local_28[i % 3][(i / 3) * 2]-1);
}
return 1;
}
int local_2c,n,i;
char *local_28 [4];
int param_1[11]={99};
local_28[0] = "Dufhbmf";
local_28[1] = "pG`imos";
local_28[2] = "ewUglpt";
local_2c = 0;
printf("\n------------------------\n");
for ( i = 0; i <= 11; ++i ){
printf("%c",local_28[i % 3][(i / 3) * 2]-1);
}
https://www.freebuf.com/articles/network/327979.html算法分析丨Reversing-x64Elf-100逆向题
https://blog.csdn.net/czw2479/article/details/124996658
res = ""
word = ["Dufhbmf", "pG`imos", "ewUglpt"]
for i in range(12):
res += chr(ord(word[i % 3][2 * (i // 3)]) - 1)
print(res)
"""output
Code_Talkers
"""
https://www.freebuf.com/articles/network/327979.html算法分析丨Reversing-x64Elf-100逆向题
res = “”
word = [“Dufhbmf”, “pG`imos”, “ewUglpt”]
for i in range(12):
res += chr(ord(word[i % 3][2 * (i // 3)]) - 1)
print(res)
“”“output
Code_Talkers
“””
