题目
题目为simple_js
首先，进入网站
提示输入密码
随便输入一个，提示输入错误，之后查看源码，得到js源代码，是一个解码函数
<html>
<head>
<title>JS</title>
<script type="text/javascript">
function dechiffre(pass_enc){
var pass = "70,65,85,88,32,80,65,83,83,87,79,82,68,32,72,65,72,65";
var tab = pass_enc.split(',');
var tab2 = pass.split(',');var i,j,k,l=0,m,n,o,p = "";i = 0;j = tab.length;
k = j + (l) + (n=0);
n = tab2.length;
for(i = (o=0); i < (k = j = n); i++ ){o = tab[i-l];p += String.fromCharCode((o = tab2[i]));
if(i == 5)break;}
for(i = (o=0); i < (k = j = n); i++ ){
o = tab[i-l];
if(i > 5 && i < k-1)
p += String.fromCharCode((o = tab2[i]));
}
p += String.fromCharCode(tab2[17]);
pass = p;return pass;
}
String["fromCharCode"](dechiffre("\x35\x35\x2c\x35\x36\x2c\x35\x34\x2c\x37\x39\x2c\x31\x31\x35\x2c\x36\x39\x2c\x31\x31\x34\x2c\x31\x31\x36\x2c\x31\x30\x37\x2c\x34\x39\x2c\x35\x30"));
h = window.prompt('Enter password');
alert( dechiffre(h) );
</script>
</head>
</html>
分析源码，发现随便输入什么都会输出
pass=”70,65,85,88,32,80,65,83,83,87,79,82,68,32,72,65,72,65″对应的字符
用python处理一下，得到该字符串为: FAUX PASSWORD HAHA,正好是提示错误的字符串
看到有有一串16进制的字符
\x35\x35\x2c\x35\x36\x2c\x35\x34\x2c\x37\x39\x2c\x31\x31\x35\x2c\x36\x39\x2c\x31\x31\x34\x2c\x31\x31\x36\x2c\x31\x30\x37\x2c\x34\x39\x2c\x35\x30
用python跑一下（转为10进制），得
55,56,54,79,115,69,114,116,107,49,50
再转换为ASCII码相对应的，得
FLAG
结合题目开始所说的flag格式，flag为Cyberpeace{786OsErtk12}
原创文章，作者：陌涛，如若转载，请注明出处：https://imotao.com/4087.html
