“腾讯极客技术挑战赛 - 码上种树”随记
三月中旬看到了群友们发的腾讯极客技术挑战赛 - 码上种树，看了一眼感觉很有兴趣。同时发给了好友@yelexin，正值周五晚七点开始，程序员不用上班的时间，一起搞。
题目设置思路是系统下发参数，本地根据当前关卡脚本运算后提交答案，即为种下一棵树，且后端有限制种完一棵树才可继续下一棵。每秒种树两颗及以上可以触发加速BUFF，关键便是每秒提交答案越多越好。为了阻止你快速种树，出题方一是在种树 js 逻辑中放置了阻塞无用延时，二是通过增加计算函数复杂度，拖慢计算这个过程的时间。故通关关键也就是这两个，干掉一很容易，略去不提，有意思的主要是第二点的解决。
这里简单记录下各关卡当初解题思路。
关卡1 & 2：简单熟悉套路
//A274075A.js
window.A274075A=async function({a}){return new Promise(_=>setTimeout(__=>_(a[0]),2000))}
主要熟悉下题目套路，setTimeout 延时改0就可以了。
到达10万棵，换题。
关卡3：base64 & 代码混淆
//A5473788.js
eval(atob("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"))
将 base64 解码，并将混淆过的变量名恢复：
var arr = ['A5473788'];
(function (a, b) {
var func = function (_0x2858d9) {
while (--b) {
a['push'](a['shift']());
}
};
func(++b);
})(arr, 406);
var func2 = function (a, b) {
a = a - 0;
var func = arr[a];
return func;
};
window[func2('0')] = function (params) {
var a1 = 199999;
for (var i = 199999; i > 0; i--) {
var a2 = 0;
for (var j = 0; j < i; j++) {
a2 += params['a'][0];
}
a2 % params['a'][2] == params['a'][1] && i < a1 && (a1 = i);
}
return a1;
};
顺序看下来，发现前14行是没用的，像是出题者的小嘲讽，等效拿掉：
window["A5473788"] = function (params) {
var a1 = 199999;
for (var i = 199999; i > 0; i--) {
var a2 = 0;
for (var j = 0; j < i; j++) {
a2 += params['a'][0];
}
a2 % params['a'][2] == params['a'][1] && i < a1 && (a1 = i);
}
return a1;
};
一个传入参数样例：
{"c":"A5473788","a":[477,976,179600],"t":"0000071300100003A87626E66399860E"}
内层循环简化掉：
window["A5473788"] = function (params) {
var a1 = 199999;
for (var i = 199999; i > 0; i--) {
var a2 = 0;
a2 += params['a'][0] * i;
if(a2 % params['a'][2] === params['a'][1] && i < a1) a1 = i;
}
return a1;
};
这时候因为把短路操作符还原了，发现 a1 就是在不断被符合 a2 % params['a'][2] === params['a'][1] && i < a1 条件的 i 取代，而 199999 这个大数就是为了让你多运算的，这时候直接想到是不是可以把 for 循环的 i 倒序，返回第一个符合 if 条件的 i 呢？那就要看 ai 变量是不是有其他作用， 看到判断条件中有 i < a1，幸运的是在当前 i = 199999 : 1 逻辑下 i < a1 恒成立。
那么代码就变成：
window["A5473788"] = function (params) {
for (var i = 1; i <= 199999; i++) {
var a2 = params['a'][0] * i;
if(a2 % params['a'][2] === params['a'][1]) return i;
}
};
完成。
ps：这里在结束后想起来，并不严密，倒不是说代码简化，而是当时我没有论证符合条件的最小 i 究竟是离 1 近还是离 199999 近，只是凭一个入参运算结果以及出题者多半与我是对立的的猜测这样干了
关卡4：jsunfuck
//A593C8B8.js
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之前知道这个 JSFuck，但没有逆向还原经验。先放下 JSFuck 的科普链接：JSFuck，是利用 javascript 的语言特性实现将任意一段代码转化成只含有6种字符的等效代码。
这一关我卡了比较久，基本歇逼状态，因为我一直在想用 jsunfuck 或者其他规则一键来还原这些恶心代码。fuck！但是没有成功。同时 yelexin 采取的方式是根据 jsfuck 的正向规则人肉替换，事实证明这样是正确思路，这一关我在歇逼状态下被带飞了。
false => ![] true => !![] undefined => [][[]] NaN => +[![]] 0 => +[] 1 => +!+[] 2 => !+[]+!+[] 10 => +[[+!+[]]+[+[]]] Array => [] Number => +[] String => []+[] Boolean => ![] Function => []["filter"] run => []["filter"]["constructor"]( CODE )() eval => []["filter"]["constructor"]("return eval")()( CODE ) window => []["filter"]["constructor"]("return this")()
yelexin 人肉还原结果：
window.A593C8B8 = async param =>
(($, param, m, e, d) => {
let a = (function*() {
while (true) {
let f1 = (param, m) => param + m
let f2 = (param, m) => param - m
let f3 = (param, m) => param * m
yield [f1,f2,f3][++m % 3]['bind'](0, e, d);
}
})();
let b = function(a, b, c) {
d = a;
e = b['next']()['value']();
m == param['a']['length'] && c(-e);
};
return new Promise(m => param.a.forEach(e => $.setTimeout(d => b(e, a, m), e)));
})(window, param, 0, 0, 0);
于是敏感点来了，又见 setTimeout 为了通关，开始尝试，0、e/100、最后 yelexin 找到 e/99 可以使用。遂通关。
ps：yelexin 反应过来这背后是睡眠排序。唉，哭笑不得。
关卡5：wasm
这一关考点是WebAssembly(wasm)，运算过程是汇编码，所以为了简化，需要读汇编码。之前知道 ffmpeg 有人做成了 wasm 所以对 wasm 有耳闻，但并无实操。这次被迫补习。
但已经深夜，就先睡了。第二天起床，yelexin 告诉我他没去研究汇编码，而是写了个暴力脚本跑了一晚过了，漂亮！他发现结果与入参差距不大，直接在入参附近挂机暴力破解。于是我一遍挂着暴力脚本，一遍继续看汇编码，最后手动反编译：
function myfunc(a1, a2) {
let maxd, mind
let a1_back = a1
for (let i = 1; a2 - i > 0; i++) {
maxd = 0
mind = 10
let temp = a1_back
while (a1_back > 0) {
let d = a1_back % 10
a1_back = parseInt(a1_back / 10)
maxd = Math.max(d, maxd)
mind = Math.min(d, mind)
}
a1_back = temp + maxd * mind
}
return a1_back
}
先想到的是把循环 /10 操作换成等价的取最大最小字符：
function myfunc(a1,a2){
console.time();
let a1_back = a1
for(let i = 1; a2-i>0; i++){
let temp = a1_back
const arr = a1_back.toString().split("")
a1_back = temp + Math.max(...arr)*Math.min(...arr)
}
console.timeEnd();
return a1_back
}
但不行，每棵树还在 3s 左右。
纳闷一会儿，突然开窍，找 0，某一次最小值出现 0， 那么 a1_back 便不会再变：
function myfunc(a1,a2){
let a1_back = a1
for(let i = 1; a2-i>0; i++){
let temp = a1_back
const arr = a1_back.toString().split("")
if(arr.includes("0")) break
a1_back = temp + Math.max(...arr)*Math.min(...arr)
}
return a1_back
}
遂完成。
关卡6 & 7：栈式虚拟机
(待更...
到这一关，了解了下这个栈式虚拟机特点后决定先不继续了，因为这周末之后有两家面试，感觉精力不足以让我兼得。不得已把这个放一放，没再继续。推荐继续看 yelexin 的腾讯极客技术挑战赛-植树节 | 端点星。
写在最后
最后停留在了 100w 棵🌲的等级。
总的来说，整个过程自己都处在一个不停歇的探索状态，这是一个高度快乐的过程。讲实话日常工作中是很少有这种体验的。
期间 yelexin 暴力挂机一晚上解决掉了一道题是给我触动很大的一点，他观察到了输入数据与输出数据相差不大这个特点，并决定晚上挂机，我拍手叫绝。反过来想一下，我当时根据前面题目的套路，一来思想已经逐渐局限，二来因为汇编代码在眼前不愿舍弃。导致没有快速意识到两个不利点：
- 这是不限手段的比赛，善于使用任何手段是优势，而且目的清晰且唯一：让系统认可你的答案 & 尽可能得抢先
- 时间点正好到深夜，自己思维其实已经逐渐不清醒，这时候分析汇编代码效率很可能会很低。而挂机暴力在夜晚又正合适
我发现，也许这恰恰又与我最近在准备面试有关系，因为不可否认面试中“循规蹈矩”内容其实占了不小一部分。好久没有“解放思想🚀”了。
整场比赛中我与好友有交流，这在一定程度上叫做违规，即使比赛并为写明。但我依然享受这个过程，有独立破关也有与好友协作，有思维上的碰撞💥，这是高等动物特有的快乐。
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