Duelist's Crackme #1Duelist's Crackme #1

Posted at 2011. 1. 17. 17:42 | Posted in Computer/Reverse Engineering


Duelist'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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