| n_cases = input() | |
| for case in xrange(1, n_cases + 1): | |
| bot = {'O': [1, 0], 'B': [1, 0]} | |
| buttons = raw_input().split()[1:] | |
| buttons = zip(buttons[::2], map(int, buttons[1::2])) | |
| time = 0 | |
| 0x10c Radio Communications | |
| For each solar system, there are 256 channels. | |
| Each channel holds 128 words of data, and gets reset to zeros 8 times per second. | |
| A radio can broadcast 1-128 words once per second to any channel, and receive | |
| 1-128 words eight times per second from any channel. | |
| Broadcasting XORs the sent data with the current contents of the channel. |
| //===-- DCPU16InstrFormats.td - DCPU16 Instruction Formats -*- tablegen -*-===// | |
| // | |
| // The LLVM Compiler Infrastructure | |
| // | |
| // This file is distributed under the University of Illinois Open Source | |
| // License. See LICENSE.TXT for details. | |
| // | |
| //===----------------------------------------------------------------------===// | |
| //===----------------------------------------------------------------------===// |
This makes instructions that address the stack twice meaningful instead of confusing and useless.
ADD PEEK, POP should pop the top of the stack and add it to the second entry, but currently the PEEK is evaluated before the POP occurs.
| lines = [x.strip() for x in open("seq.txt")] | |
| while '0000' in lines: | |
| # strip the time codes | |
| i = lines.index('0000') | |
| lines = lines[:i] + lines[i+4:] | |
| subst = {'0590': " ", '22f2': "e", '1225': "a", '77ee': "t", | |
| '9036': "h", '0817': "n", '5819': "d", '29d3': "o", | |
| '4079': "g", 'f6de': "y", '906b': "m", '5bcd': "l", |
| /***************************************************************************** | |
| * | |
| * Gradient: an Artificial Life Experiment | |
| * Copyright (C) 2011 Maxime Chevalier-Boisvert | |
| * | |
| * This file is part of Gradient. | |
| * | |
| * Gradient is free software: you can redistribute it and/or modify | |
| * it under the terms of the GNU General Public License as published by | |
| * the Free Software Foundation, either version 3 of the License, or |
| def enc(id1, id2, id3): | |
| ''' encrypt the first three bytes of the ID | |
| * the 4th byte of the ID is id1 ^ id2 ^ id3 | |
| * the top three bits of id1 are lost, meaning | |
| there are 2^21 ~= 2M unique clicker IDs | |
| source: reversed iclicker1 firmware ''' | |
| a = (id1 & 0x1F) << 3 | |
| a |= (id2 & 0x80) >> 5 | |
| a |= maskFrom3rd(a) |
| SRCS=iGtalk.c friends.c | |
| BIN=iGtalk | |
| CC=gcc | |
| CFLAGS=-g3 -DDEBUG=1 -I include | |
| LDLIBS=-L lib -lgnutls -lgsasl -lrecv_xml_nonblock | |
| OBJS=$(SRCS:.c=.o) | |
| DEP=makefile.dep |
I reverse engineered the binary to find the encoding function, converted it to Python, and verified correctness. I then wasted a bunch of time making an iterative bruteforcer, until I realized that the encoding function is basically equivalent to:
bits('RTFM') + tab[buf[0]] + '00' + tab[buf[1]] + '00' + ... + tab[buf[n]]
where tab is a mapping from ascii characters to variable-length bitstrings, with the important property that they start and end with 1s, and contain no 00s. (match /^1.*1$/ but not /00/)
After that, decoding is simple.
| SET A, plain | |
| SET B, 8 | |
| SET C, key | |
| SET X, iv | |
| JSR crypt_hummingbird2_enc | |
| SET A, plain | |
| SET B, 8 | |
| SET C, key | |
| SET X, iv |