ADCTF2014 [10] xor
xor+shiftの暗号かぁー
712249146f241d31651a504a1a7372384d173f7f790c2b115f47
Source Code:
[c]
#include <stdio.h>
#include <string.h>
int main() {
char flag = "ADCTF_XXXXXXXXXXXXXXXXXXXX";
int len = strlen(flag);
for (int i = 0; i < len; i++) {
if (i > 0) flag[i] ^= flag[i-1];
flag[i] ^= flag[i] >> 4;
flag[i] ^= flag[i] >> 3;
flag[i] ^= flag[i] >> 2;
flag[i] ^= flag[i] >> 1;
printf("%02x", (unsigned char)flag[i]);
}
return 0;
}
[/c]
昔、某k○nctfの暗号問題を解くときに書いたプログラムをそのまま流用するか
[c]
include <stdio.h>
include <string.h>
unsigned long decXORShift_R(unsigned long,int);
int main(void){ int i; char flag = {0x71,0x22,0x49,0x14,0x6f,0x24,0x1d,0x31,0x65,0x1a,0x50,0x4a,0x1a,0x73,0x72,0x38,0x4d,0x17,0x3f,0x7f,0x79,0x0c,0x2b,0x11,0x5f,0x47,0x00}; int len = strlen(flag);
for(i=len-1; i>=0; i--){ flag[i]=decXORShift_R(flag[i],1); flag[i]=decXORShift_R(flag[i],2); flag[i]=decXORShift_R(flag[i],3); flag[i]=decXORShift_R(flag[i],4); if(i>0) flag[i] ^= flag[i-1]; } printf("%s",flag);
return 0; }
unsigned long decXORShift_R(unsigned long x,int t){ unsigned long y=x, z=0; for(unsigned long mask=((1U<<t)-1)<<(sizeof(long)*8-t);mask;mask>>=t){ z |= (y&mask)>>t; y = zx; } return y; }
[/c]
元のプログラムはand演算を加味したものも解けるようにしてあったのだけど、今回は必要ないから取り除きました
FLAG： ADCTF_51mpl3_X0R_R3v3r51n6
