uidops

import math
import operator
import os
import secrets
import sys
import time


players = ('U', 'C', '0')
func = (min, max)

uidops

from ctypes import c_uint64, c_double

o = (c_uint64.from_buffer(c_double(1)).value >> 0x20) - 0xed81

def exp(x: float) -> float:
    x = int((0x171547 * x) + o) << 0x20
    return c_double.from_buffer(c_uint64(x)).value

def ln(x: float) -> float:
    x = c_uint64.from_buffer(c_double(x)).value >> 0x20

uidops

#include <time.h>
#include <stdio.h>
#include <stdlib.h>

#define clear_screen fputs("\x1b[2J\x1b[H", stdout);

int
main(void)
{
	const unsigned long long data[] = {

uidops

/*
* #59 XOR decryption - Project Euler [ https://projecteuler.net/problem=59 ]
* the numbers: https://projecteuler.net/project/resources/p059_cipher.txt
*/

#include <stdio.h>

int		 isvalid(const char);

int

uidops

/*
 * #89 Roman numerals - Project Euler [ https://projecteuler.net/problem=89 ]
 *
 *  The fastest way to solve this problem is replacing (but we will not replace).
 *  When we look at the numbers in the file, we found many numbers that can be written shorter.
 *  Those numbers are:
 *  	.   IIII = IV =   4
 *  	.  VIIII = IX =   9
 *  	.   XXXX = XL =  40
 *  	.  LXXXX = XC =  90

uidops

#include <err.h>
#include <errno.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MEMINFO "/proc/meminfo"


struct meminfo {

uidops

import sys

def bin_to_dec(b):
    number = 0
    l = len(b) - 1

    for char in b:
        number += int(char)*(2**l)
        l -= 1

uidops

/*
	CUBE (x86-64)
	Compile: cc -no-pie -nostartfiles cube.s -o cube
	C code:

	#include <stdio.h>
	#include <stdlib.h>

	long int
	cube(int x)

uidops

#!/usr/bin/python3
import subprocess
import random
import os

cows = [
    "ghost",
    "radio",
    "pod",
    "udder",

uidops

/*
	Fibonacci sequence in assembly language (x86)
	Compile: clang -m32 -no-pie -nostartfiles fib.s -o fib

	C code:

		#include <stdio.h>

		int
		main(void)