JetBrains的推广活动,解谜后可以获取三个月的免费订阅。由于是推广活动,所以解密过程不是非常难。真正让人头痛的是那糟糕的网速,不管挂不挂代理页面的打开都非常的慢。
题目一
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48 61 76 65 20 79 6f 75 20 73 65 65 6e 20 74 68 65 20 73 6f 75 72 63 65 20 63 6f 64 65 20 6f 66 20 74 68 65 20 4a 65 74 42 72 61 69 6e 73 20 77 65 62 73 69 74 65 3f |
解题
很明显字符串的ASCII码,使用python很容易进行解码
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>>> s = "48 61 76 65 20 79 6f 75 20 73 65 65 6e 20 74 68 65 20 73 6f 75 72 63 65 20 63 6f 64 65 20 6f 66 20 74 68 65 20 4a 65 74 42 72 61 69 6e 73 20 77 65 62 73 69 74 65 3f" >>> ''.join(chr(int(e, 16)) for e in s.split(' ')) 'Have you seen the source code of the JetBrains website?' |
题目二
查看首页源代码找到解谜线索
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JetBrains has a lot of products, but there is one that looks like a joke on our Products page, you should start there... (hint: use Chrome Incognito mode) It’s dangerous to go alone take this key: Good luck! == Jrrg#oxfn$ |
根据提示,到产品页面。其中名为“JK”的产品介绍是“dare to lean more”,点击该产品继续进行挑战。
注:之前都不知道JetBrains居然已经有这多的产品了。
题目三
补完 https://jb.gg/###
后面确实的三个数字。数字为500到5000的质数个数。
解题
到网上找了个求质数的函数,跑了一下,很快得到结果574
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
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import math def isprime(n): if n < 2: return False for i in range(2, int(math.sqrt(n)) + 1): if n % i == 0: return False return True count = 0 for i in range(500, 5000): if isprime(i): count += 1 print(count) |
题目四
打开上面得到的链接,其中有张图片。图片中的字符为“MPS-31816”
解题
这个题目最坑的在于这个图片实在是太大了,约有15M,死活打不开。
- 查看图片属性,用编辑器直接打开图片,都没有获取到有效的信息。
- 直接访问
https://www.jetbrains.com/MPS-31816
显示没有这个网页。 - 于是直接启用Google搜索
MPS-31816
在JetBrains 网站找到对应页面。注:后续会知道图片上的图标是JebBrains网站的问题区的Logo。
题目五
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“The key is to think back to the beginning.” – The JetBrains Quest team Qlfh$#Li#|rx#duh#uhdglqj#wklv#|rx#pxvw#kdyh#zrunhg#rxw#krz#wr#ghfu|sw#lw1#Wklv#lv#rxu#lvvxh#wudfnhu#ghvljqhg#iru#djloh#whdpv1#Lw#lv#iuhh#iru#xs#wr#6#xvhuv#lq#Forxg#dqg#iru#43#xvhuv#lq#Vwdqgdorqh/#vr#li#|rx#zdqw#wr#jlyh#lw#d#jr#lq#|rxu#whdp#wkhq#zh#wrwdoo|#uhfrpphqg#lw1#|rx#kdyh#ilqlvkhg#wkh#iluvw#Txhvw/#qrz#lw“v#wlph#wr#uhghhp#|rxu#iluvw#sul}h1#Wkh#frgh#iru#wkh#iluvw#txhvw#lv#‟WkhGulyhWrGhyhors†1#Jr#wr#wkh#Txhvw#Sdjh#dqg#xvh#wkh#frgh#wr#fodlp#|rxu#sul}h1#kwwsv=22zzz1mhweudlqv1frp2surpr2txhvw2 |
解题
很明显要用到之前的 take this key: Good luck! == Jrrg#oxfn$
。我一开始将问题想的太复杂了,以为第一题里的内容是密钥,用xor进行解密。在网上找了个xor解密的函数,解出来一塌糊涂。
由回头仔细看了一下这所谓的密码,其实就是一个简单的字映射。
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>>> s = "Qlfh$#Li#|rx#duh#uhdglqj#wklv#|rx#pxvw#kdyh#zrunhg#rxw#krz#wr#ghfu|sw#lw1#Wklv#lv#rxu#lvvxh#wudfnhu#ghvljqhg#iru#djloh#whdpv1#Lw#lv#iuhh#iru#xs#wr#6#xvhuv#lq#Forxg#dq g#iru#43#xvhuv#lq#Vwdqgdorqh/#vr#li#|rx#zdqw#wr#jlyh#lw#d#jr#lq#|rxu#whdp#wkhq#zh#wrwdoo|#uhfrpphqg#lw1#|rx#kdyh#ilqlvkhg#wkh#iluvw#Txhvw/#qrz#lw“v#wlph#wr#uhghhp#|rxu#iluvw#s ul}h1#Wkh#frgh#iru#wkh#iluvw#txhvw#lv#‟WkhGulyhWrGhyhors†1#Jr#wr#wkh#Txhvw#Sdjh#dqg#xvh#wkh#frgh#wr#fodlp#|rxu#sul}h1#kwwsv=22zzz1mhweudlqv1frp2surpr2txhvw2" >>> >>> c = ord('J') - ord('G') >>> ''.join(chr(ord(e) - c) for e in s) 'Nice! If you are reading this you must have worked out how to decrypt it. This is our issue tracker designed for agile teams. It is free for up to 3 users in Cloud and for 10 users in Standalone, so if you want to give it a go in your team then we totally recommend it. you have finished the first Quest, now it’s time to redeem your first prize. Th e code for the first quest is “TheDriveToDevelop”. Go to the Quest Page and use the code to claim your prize. https://www.jetbrains.com/promo/quest/' |