Duelist's Crackme #1Duelist's Crackme #1
Posted at 2011. 1. 17. 17:42 | Posted in Computer/Reverse EngineeringDuelist's Crackme #1 Soluction
#include <stdio.h> #include <stdlib.h> void main(void) { int iCount = 0; int cipher[] = { 0x7B, 0x61, 0x65, 0x78, 0x64, 0x6D, 0x26, 0x6B, 0x7A, 0x69, 0x6B, 0x63, 0x65, 0x6D, 0x26, 0x3C, 0x26, 0x66, 0x6D, 0x7F, 0x6A, 0x61, 0x6D, 0x7B, 0x26, 0x6A, 0x71, 0x26, 0x6C, 0x7D, 0x6D, 0x64, 0x61, 0x7B, 0x7C }; int *plainText = malloc(sizeof(cipher)); for(iCount = 0; iCount < (sizeof(cipher) / sizeof(int)); iCount++) { plainText[iCount] = cipher[iCount] ^ 0x43 ^ 0x1E ^ 0x55; printf("%c", plainText[iCount]); } printf("\n"); }
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