[pico CTF] Client-side-again

1. 개요

난독화 해석 문제

https://play.picoctf.org/practice?category=1&page=1&search=Client-side-again 

 

2. 분석

https://beautifier.io/ 에서 코드 예쁘게 만들긴 했지만, 그래도 분석이 필요합니다.

var _0x4b5b=function(_0x2d8f05,_0x4b81bb){_0x2d8f05=_0x2d8f05-0x0;var _0x4d74cb=_0x5a46[_0x2d8f05];return _0x4d74cb;};

function verify() {
    checkpass = document[_0x4b5b('0x0')]('pass')[_0x4b5b('0x1')];
    const split = 0x4;
    if (checkpass[_0x4b5b('0x2')](0x0, split * 0x2) == _0x4b5b('0x3')) {
        if (checkpass[_0x4b5b('0x2')](0x7, 0x9) == '{n') {
            if (checkpass[_0x4b5b('0x2')](split * 0x2, split * 0x2 * 0x2) == _0x4b5b('0x4')) {
                if (checkpass[_0x4b5b('0x2')](0x3, 0x6) == 'oCT') {
                    if (checkpass[_0x4b5b('0x2')](split * 0x3 * 0x2, split * 0x4 * 0x2) == _0x4b5b('0x5')) {
                        if (checkpass['substring'](0x6, 0xb) == 'F{not') {
                            if (checkpass[_0x4b5b('0x2')](split * 0x2 * 0x2, split * 0x3 * 0x2) == _0x4b5b('0x6')) {
                                if (checkpass[_0x4b5b('0x2')](0xc, 0x10) == _0x4b5b('0x7')) {
                                    alert(_0x4b5b('0x8'));
                                }
                            }
                        }
                    }
                }
            }
        }
    } else {
        alert(_0x4b5b('0x9'));
    }
}

 

 

3. exploit

0 ~ 6 - picoCTF

6 ~ 11 - F{not

12 ~ 16 - this

16 ~ 24 - _again_e

24 ~ 32 - f49bf}

picoCTF{not_this_again_ef49bf}