拯救大兵艾伦

题目

众所周知，艾伦(eren)身怀坐标之力（flag）可以变身成为巨人(kyojin)于是，他又双叒叕被抓走啦233三爷当然是选择原谅，啊呸，救回他不过救回来的是巨人状态的艾伦，我们需要把他复原回去

解题报告

打开提供的链接，查看源代码发现如下Javascript

function check() {    
  let p = document.getElementById("eren").value;    
  let flag = document.getElementById("flag").value;    
  let kyojin = "~dyl[fj!XHAu)(Hj~$HT}i";    
  let f = (x, y) => (x == 1) ? 1 : f(y % x, y) * (y - parseInt(y / x)) % y;     let fail = flag.length != kyojin.length;    
  Array.from(flag).forEach((x, i) => fail |= String.fromCharCode(f(x.charCodeAt(0), p)) != kyojin[i]);    
  fail ? alert("不正常") : alert("正常"); 
}

可以用脚本直接暴力破解

let kyojin = "~dyl[fj!XHAu)(Hj~$HT}i"; 
let result = "" 
let f = (x, y) => (x == 1) ? 1 : f(y % x, y) * (y - parseInt(y / x)) % y; 
for (let i = 0; i < kyojin.length; ++i) {  
  for (let j = 0x20; j <= 0x7E; ++j) {    
    if (String.fromCharCode(f(j, 193)) == kyojin[i]) { 
      // console.log(String.fromCharCode(j));
      result += String.fromCharCode(j);
      break;    
    }  
  } 
} 
console.log(result);

或者阅读js代码，发现f函数是求模p乘法逆元

跟异或一样，把“整容后”再整一次容就可以变回“整容前”

然后问题是连p也未知

不过p一定是素数，候选项不会太多，果断爆破

// copy from 地下室.html
let f = (x, y) => (x == 1) ? 1 : f(y % x, y) * (y - parseInt(y / x)) % y;
let kyojin = "~dyl[fj!XHAu)(Hj~$HT}i";

// get prime
const MAXP = 1000000;
let prime = new Array(MAXP).fill(0);
for (let i = 2; i <= MAXP; ++i) {
    if (!prime[i])
        prime[++prime[0]] = i;
    for (let j = 1; j <= prime[0] && prime[j] * i <= MAXP; ++j) {
        prime[prime[j]*i] = 1;
        if (i % prime[j] == 0)
            break;
    }
}

//try prime
for (let i = 1; i <= prime[0]; ++i) {
    let eren = prime[i];
    let flag = [];
    Array.from(kyojin).forEach(x => {
        if (x.charCodeAt(0) >= eren)
            return;
        let y = f(x.charCodeAt(0), eren);
        if (y < 33 || y > 126)
            return;
        flag.push(y);
    });
    if (flag.length == kyojin.length) {
        console.log(eren);
        console.log(flag.map(x => String.fromCharCode(x)).join(""));
    }
}

Flag

HSCTF{zuD~b!qo~zH;~li}