CTF习题解答--查找正确的密码 [JS题型]

首先发题

小黑客们,坐好了,老司机要开车了!

这是密码错误的提示
CTF习题解答--查找正确的密码 [JS题型]_第1张图片
当我们输入正确的密码,会提示
CTF习题解答--查找正确的密码 [JS题型]_第2张图片

首页的代码

<html>
    <head>
        <title>Flag in your Handtitle>
        <style type="text/css">
            body {
                padding-left: 30%;
            }

            #flag {
                font-family: Garamond, serif;
                font-size: 36px;
            }

            #flagtitle {
                font-family: Garamond, serif;
                font-size: 24px;
            }

            .rightflag {
                color: green;
            }

            .wrongflag {
                color: red;
            }
        style>
        <script src="script-min.js">script>
        <script type="text/javascript">

            var ic = false;
            var fg = "";

            function getFlag() {
                var token = document.getElementById("secToken").value;
                ic = checkToken(token);             
                fg = bm(token); 
                showFlag()
            }
            function showFlag() {
                var t = document.getElementById("flagTitle");
                var f = document.getElementById("flag");
                t.innerText = !!ic ? "You got the flag below!!" : "Wronfg!";
                t.className = !!ic ? "rightflag" : "wrongflag";
                f.innerText = fg;
            }
        script>
    head>
    <body>
        <h1>Flag in your Handh1>
        <p>Type in some token to get the flag.p>
        <p>Tips: Flag is in your hand.p>
        <div>
            <p>
                <span>Token:span>
                <span><input type="text" id="secToken"/>span>
            p>
            <p>
                <input type="button" value="Get flag!" onclick="getFlag()" />
            p>
        div>
        <div>
            <p id="flagTitle">p>
            <p id="flag">p>
        div>
    body>
html>

JS代码

function hm(s) {
    return rh(rstr(str2rstr_utf8(s)));
}
function bm(s) {                                // 函 数 的 ==  入 口
    return rb(rstr(str2rstr_utf8(s)));
}
function rstr(s) {
    return binl2rstr(binl(rstr2binl(s), s.length * 8));
}
function checkToken(s) {
    var b = s === "FAKE-TOKEN"
    console.log(b)
    return b;
}
function rh(ip) {
    try {
        hc
    } catch (e) {
        hc = 0;
    }
    var ht = hc ? "0123456789ABCDEF" : "0123456789abcdef";
    var op = "";
    var x;
    for (var i = 0; i < ip.length; i++) {
        x = ip.charCodeAt(i);
        op += ht.charAt((x >>> 4) & 0x0F) + ht.charAt(x & 0x0F);
    }
    return op;
}
function rb(ip) {
    try {
        bp
    } catch (e) {
        bp = '';
    }
    var b = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
    var op = "";
    var len = ip.length;
    for (var i = 0; i < len; i += 3) {
        var t = (ip.charCodeAt(i) << 16) | (i + 1 < len ? ip.charCodeAt(i + 1) << 8 : 0) | (i + 2 < len ? ip.charCodeAt(i + 2) : 0);
        for (var j = 0; j < 4; j++) {
            if (i * 8 + j * 6 > ip.length * 8)
                op += bp;
            else
                op += b.charAt((t >>> 6 * (3 - j)) & 0x3F);
        }
    }
    return op;
}
function ck(s) {
    try {
        ic
    } catch (e) {
        return;
    }

    var a = [118, 104, 102, 120, 117, 108, 119, 124, 48, 100, 101, 102];
    var b = [115, 101, 99, 117, 114, 105, 116, 121, 45, 97, 98, 99]
    if (s.length == a.length) {
        for (i = 0; i < s.length; i++) {
            if (a[i] - s.charCodeAt(i) != 3){
                console.log(a[i] - s.charCodeAt(i))
                return ic = false;
            }
        }
        return ic = true;
    }
    return ic = false;
}
function str2rstr_utf8(input) {        //   字符串 第一层 处理 ::::字符串 str 转化 rstr_utf8
    var output = "";
    var i = -1;
    var x, y;                          // x: 本次字母        y: 下一次字母
    while (++i < input.length) {
        x = input.charCodeAt(i);                                    
        y = i + 1 < input.length ? input.charCodeAt(i + 1) : 0;     
        if (0xD800 <= x && x <= 0xDBFF && 0xDC00 <= y && y <= 0xDFFF) {  
         //  0b 1101 1000 0000 0000
         //  0b 1101 1011 1111 1111   >>>   11 1111 1111  = 2^10  
         //  0b 1101 1100 0000 0000
         //  0b 1101 1111 1111 1111   >>> 
            x = 0x10000 + ((x & 0x03FF) << 10) + (y & 0x03FF);
            // x 拼接 y 加上 0x10000
            i++;
        }
        if (x <= 0x7F)  // 0b 111 1111
            output += String.fromCharCode(x); 
        else if (x <= 0x7FF)                       // 无符号右移位运算符 (>>>)
            output += String.fromCharCode(0xC0 | ((x >>> 6) & 0x1F), 0x80 | (x & 0x3F));
        else if (x <= 0xFFFF)
            output += String.fromCharCode(0xE0 | ((x >>> 12) & 0x0F), 0x80 | ((x >>> 6) & 0x3F), 0x80 | (x & 0x3F));
        else if (x <= 0x1FFFFF)                   //  猜测 :应该运行这个东西
            output += String.fromCharCode(0xF0 | ((x >>> 18) & 0x07), 0x80 | ((x >>> 12) & 0x3F), 0x80 | ((x >>> 6) & 0x3F), 0x80 | (x & 0x3F));
    }                                               
    return output;
}
function rstr2binl(input) {
    var output = Array(input.length >> 2);   // 输出 字符串 是原来的 1/4
    for (var i = 0; i < output.length; i++)
        output[i] = 0;  // 初始化 = 0
    for (var i = 0; i < input.length * 8; i += 8)
        output[i >> 5] |= (input.charCodeAt(i / 8) & 0xFF) << (i % 32);
    return output;
}
function binl2rstr(i) {
    var o = "";
    for (var j = 0; j < i.length * 32; j += 8)
        o += String.fromCharCode((i[j >> 5] >>> (j % 32)) & 0xFF);
    return o;
}
function binl(x, len) {
    s = binl2rstr(x);
   // console.log(s);
    x[len >> 5] |= 0x80 << ((len) % 32);
    x[(((len + 64) >>> 9) << 4) + 14] = len;
    var a = 1732584193;
    var b = -271733879;
    var c = -1732584194;
    var d = 271733878;
    for (var i = 0; i < x.length; i += 16) {
        var olda = a;
        var oldb = b;
        var oldc = c;
        var oldd = d;
        a = ff(a, b, c, d, x[i + 0], 7, -680876936);
        d = ff(d, a, b, c, x[i + 1], 12, -389564586);
        c = ff(c, d, a, b, x[i + 2], 17, 606105819);
        b = ff(b, c, d, a, x[i + 3], 22, -1044525330);
        a = ff(a, b, c, d, x[i + 4], 7, -176418897);
        d = ff(d, a, b, c, x[i + 5], 12, 1200080426);
        c = ff(c, d, a, b, x[i + 6], 17, -1473231341);
        b = ff(b, c, d, a, x[i + 7], 22, -45705983);
        a = ff(a, b, c, d, x[i + 8], 7, 1770035416);
        d = ff(d, a, b, c, x[i + 9], 12, -1958414417);
        c = ff(c, d, a, b, x[i + 10], 17, -42063);
        b = ff(b, c, d, a, x[i + 11], 22, -1990404162);
        a = ff(a, b, c, d, x[i + 12], 7, 1804603682);
        d = ff(d, a, b, c, x[i + 13], 12, -40341101);
        c = ff(c, d, a, b, x[i + 14], 17, -1502002290);
        b = ff(b, c, d, a, x[i + 15], 22, 1236535329);
        ck(s);
        a = gg(a, b, c, d, x[i + 1], 5, -165796510);
        d = gg(d, a, b, c, x[i + 6], 9, -1069501632);
        c = gg(c, d, a, b, x[i + 11], 14, 643717713);
        b = gg(b, c, d, a, x[i + 0], 20, -373897302);
        a = gg(a, b, c, d, x[i + 5], 5, -701558691);
        d = gg(d, a, b, c, x[i + 10], 9, 38016083);
        c = gg(c, d, a, b, x[i + 15], 14, -660478335);
        b = gg(b, c, d, a, x[i + 4], 20, -405537848);
        a = gg(a, b, c, d, x[i + 9], 5, 568446438);
        d = gg(d, a, b, c, x[i + 14], 9, -1019803690);
        c = gg(c, d, a, b, x[i + 3], 14, -187363961);
        b = gg(b, c, d, a, x[i + 8], 20, 1163531501);
        a = gg(a, b, c, d, x[i + 13], 5, -1444681467);
        d = gg(d, a, b, c, x[i + 2], 9, -51403784);
        c = gg(c, d, a, b, x[i + 7], 14, 1735328473);
        b = gg(b, c, d, a, x[i + 12], 20, -1926607734);
        a = hh(a, b, c, d, x[i + 5], 4, -378558);
        d = hh(d, a, b, c, x[i + 8], 11, -2022574463);
        c = hh(c, d, a, b, x[i + 11], 16, 1839030562);
        b = hh(b, c, d, a, x[i + 14], 23, -35309556);
        a = hh(a, b, c, d, x[i + 1], 4, -1530992060);
        d = hh(d, a, b, c, x[i + 4], 11, 1272893353);
        c = hh(c, d, a, b, x[i + 7], 16, -155497632);
        b = hh(b, c, d, a, x[i + 10], 23, -1094730640);
        a = hh(a, b, c, d, x[i + 13], 4, 681279174);
        d = hh(d, a, b, c, x[i + 0], 11, -358537222);
        c = hh(c, d, a, b, x[i + 3], 16, -722521979);
        b = hh(b, c, d, a, x[i + 6], 23, 76029189);
        a = hh(a, b, c, d, x[i + 9], 4, -640364487);
        d = hh(d, a, b, c, x[i + 12], 11, -421815835);
        c = hh(c, d, a, b, x[i + 15], 16, 530742520);
        b = hh(b, c, d, a, x[i + 2], 23, -995338651);
        a = ii(a, b, c, d, x[i + 0], 6, -198630844);
        d = ii(d, a, b, c, x[i + 7], 10, 1126891415);
        c = ii(c, d, a, b, x[i + 14], 15, -1416354905);
        b = ii(b, c, d, a, x[i + 5], 21, -57434055);
        a = ii(a, b, c, d, x[i + 12], 6, 1700485571);
        d = ii(d, a, b, c, x[i + 3], 10, -1894986606);
        c = ii(c, d, a, b, x[i + 10], 15, -1051523);
        b = ii(b, c, d, a, x[i + 1], 21, -2054922799);
        a = ii(a, b, c, d, x[i + 8], 6, 1873313359);
        d = ii(d, a, b, c, x[i + 15], 10, -30611744);
        c = ii(c, d, a, b, x[i + 6], 15, -1560198380);
        b = ii(b, c, d, a, x[i + 13], 21, 1309151649);
        a = ii(a, b, c, d, x[i + 4], 6, -145523070);
        d = ii(d, a, b, c, x[i + 11], 10, -1120210379);
        c = ii(c, d, a, b, x[i + 2], 15, 718787259);
        b = ii(b, c, d, a, x[i + 9], 21, -343485551);
        a = sa(a, olda);
        b = sa(b, oldb);
        c = sa(c, oldc);
        d = sa(d, oldd);
    }
    return Array(a, b, c, d);
}
function cmn(q, a, b, x, s, t) {
    return sa(br(sa(sa(a, q), sa(x, t)), s), b);
}
function ff(a, b, c, d, x, s, t) {
    return cmn((b & c) | ((~b) & d), a, b, x, s, t);
}
function gg(a, b, c, d, x, s, t) {
    return cmn((b & d) | (c & (~d)), a, b, x, s, t);
}
function hh(a, b, c, d, x, s, t) {
    return cmn(b ^ c ^ d, a, b, x, s, t);
}
function ii(a, b, c, d, x, s, t) {
    return cmn(c ^ (b | (~d)), a, b, x, s, t);
}
function sa(x, y) {
    var lsw = (x & 0xFFFF) + (y & 0xFFFF);
    var msw = (x >> 16) + (y >> 16) + (lsw >> 16);
    return (msw << 16) | (lsw & 0xFFFF);
}
function br(n, c) {
    return (n << c) | (n >>> (32 - c));
}

正式 开始了

当看到这个题,我们第一反应可能是看js代码,然后 找字符串,然后试一试。就像我第一次找了“FAKE-TOKEN”这个字符串,其实结果也是很正常的,一定是错误的。
接下来,就开始看html的代码。发现了按钮执行的函数,然后 按照一步一步的调用,逐行逐句的进行代码的阅读,一开始阅读代码还不是特别的费力,比如大家看到我在代码中的注释,基本上把这个加密算法的一部分的原理过程给破译了,但有用么??这才是关键,破译解密算法是没有用的。应该怎么做?,我们可以继续看看下面这个函数:

function checkToken(s) {
    return s === "FAKE-TOKEN";
}

这个函数就是我第一次找到字符串的,误以为密码的。其实 输入“FAKE-TOKEN”返回值确实是True;那么问题到底出在哪哪里??在一头雾水的时候,看html文件吧;

            function getFlag() {
                var token = document.getElementById("secToken").value;  
                ic = checkToken(token);  //  1)
                fg = bm(token);          //  2)
                showFlag()               //  3)
            }

            function showFlag() {
                var t = document.getElementById("flagTitle");
                var f = document.getElementById("flag");
                t.innerText = !!ic ? "You got the flag below!!" : "Wronfg!";
                t.className = !!ic ? "rightflag" : "wrongflag";
                f.innerText = fg;
            }

可以看得出来,ic的值确实是True,而执行到showFlag() 这个函数的时候,发现,ic=False,那么就一定是因为执行fg = bm(token); 的时候,导致ic的值发生了变化,那么,我们就去js文件找一找是否存在导致ic变化的东西。从这里就可以得出了一个结论,其实对于加密算法,是没有用的,是完全不会影响到你的解题。
我改变ic值的代码拷贝出来,我们来细细的分析一下;

function ck(s) {
    try {
        ic
    } catch (e) {
        return;
    }
    var a = [118, 104, 102, 120, 117, 108, 119, 124, 48, 100, 101, 102];
    if (s.length == a.length) {
        for (i = 0; i < s.length; i++) {
            if (a[i] - s.charCodeAt(i) != 3)
                return ic = false;
        }
        return ic = true;
    }
    return ic = false;
}

是不是有一点: 山重水复疑无路,柳暗花明又一村;的感觉。简化一下,只看核心

function ck(s) {
    var a = [118, 104, 102, 120, 117, 108, 119, 124, 48, 100, 101, 102];

    if (s.length == a.length) {
        for (i = 0; i < s.length; i++) {
            if (a[i] - s.charCodeAt(i) != 3)
                return ic = false;
        }
        return ic = true;
    }

    return ic = false;
}

为了True,所以 必须 s.length = a.length; 所以我们知道了密码的长度 就是数组的长度;接下来 我们必须执行return ic = true; 才行;
重任 就落在了for循环了:

        for (i = 0; i < s.length; i++) {
            if (a[i] - s.charCodeAt(i) != 3)
                return ic = false;
        }

必须让if都执行失败;那么就是

if (a[i] - s.charCodeAt(i) = 3)              

我们通过换算,得到

s.charCodeAt(i) = a[i] - 3

终于得出了答案了:我们发现了;
s就是最后要得出的密码;

解决脚本

var bb = [115, 101, 99, 117, 114, 105, 116, 121, 45, 97, 98, 99]
var s = ""
function qwert() {
    for (var i = 0; i < bb.length; i++) 
        s += String.fromCharCode(bb[i])
}


consolo.log(s)

喜欢CTF的黑客 大家一起交流哇!

总结:

这些题 确实有意思。大家还是提升自己阅读代码能力,,200行的代码量筛选也是挺麻烦的。
我js是前端中学的,迁移到信安也是第一次,不过,看代码,还是需要忽略一些 东西的,太看重语言问题,那就没戏了。

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