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Sweet Like Apple Pie

The output is output = sin(flag % pi()).

We can get flag % pi() = arcsin(output) using a binary search. Note that there are 2 possible answers.

So, we want to calculate k such that flag - k * pi() = arcsin(output).

Due to the Decimal, these values are calculated in 300-digits precision. That means,

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Rubikrypto

This is only ElGamal cryptosystem on Rubik Cube.

The order of Rubik Cube Group is 43_252_003_274_489_856_000n and this is small enough to use Pohlig-Hellman.

So the algorithm is simple:

1. get g, h, c1, c2.
2. calculate y = log_g(c1) using Pohlig-Hellman.

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Modulus Amittendus

You can see that n, e, cf are exported in pubkey.json.

These values are a bit strange:

  def initialize
    # ...
    @cf = modinv(@q, @p)   # (1)

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TSG CTF 2020 writeups (日本語簡略版)

Beginner's Crypto

flag * 2^{10000} mod 10^{175} == val という形で、val が分かっている。

剰余の定義から flag * 2^{10000} - k * 10^{175} == val がわかる。

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Slowest Decryption

This is PPC problem.

For any i, v, x, we define cnt(i,v,x) as the number of P such that GCD(P)=v and P[i]=x. Then, the answer will be \sum_{i,v,x} cnt(i,v,x) * v * c[x] * i. Because cnt(i,v,x) = cnt(i',v,x), the answer can be written as \sum_{v,x} cnt(0,v,x) * v * c[x] * N*(N-1)/2.

To count cnt(0,v,x), we fix x. Then we check v from N-1 to 1 such that x % v = 0.

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Beginner's Misc

The problem is:

exploit = input('? ')
if eval(b64encode(exploit.encode('UTF-8'))) == math.pi:
  print(open('flag.txt').read())

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// main関数内以外は全てサンプルコードのままです

#include <iostream>
#include <vector>
#include <string>
#include <random>
#include <thread>
#include <chrono>
using namespace std;

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