| 'use strict' | |
| const alt_bit = function*() { | |
| let n = 0n | |
| let even = true | |
| while (true) { | |
| yield n | |
| n <<= 1n | |
| if (even) n |= 1n | |
| even = !even |
| #!/usr/bin/env python3 | |
| from typing import Final | |
| from string import digits, ascii_letters | |
| from sys import argv | |
| FULL_CHARSET: Final = digits + ascii_letters | |
| def to_radix(n: int, charset='0123456789', lil_endian=False): |
| from typing import Final, TypeVar | |
| T = TypeVar('T') | |
| def remove_dupes(inp: list[T], keep_last: bool = False) -> list[T]: | |
| ''' | |
| creates a list from input, | |
| such that all ocurrences (but 1) of each value are removed | |
| (order is preserved). |
| /// # Greatest Common Divisor / Highest Common Factor | |
| /// between `a` & `b`, | |
| /// using https://en.wikipedia.org/wiki/Euclidean_algorithm. | |
| /// | |
| /// It only supports unsigneds, | |
| /// because it's easier to avoid bugs, | |
| /// and allows wider input ranges. | |
| const fn gcd(mut a: u128, mut b: u128) -> u128 { | |
| // rustc doesn't have TCO, so recursion is not an option | |
| while b != 0 { |
This website seems to have a SQL-injection vulnerability, at the password reset and login sections.
I noticed this when using a password with random symbols.
I posted this gist to remind myself to tell them to fix it, and to notify everyone else about this
| from typing import Final | |
| BYTES_PER_PIXEL: Final = 3 # 24bit color, 8 per channel | |
| def decoded_video_size(*, w: Final[int], h: Final[int], s: Final[float], fps: Final[float]): | |
| pixels_per_frame: Final = w * h # px/px^2 = 1/px = px^-1 | |
| total_frames: Final = fps * s # px^2/s * s = px^2 | |
| total_pixels: Final = pixels_per_frame * total_frames # px^-1 * px^2 = px^2 / px^1 = px | |
| total_bytes: Final = total_pixels * BYTES_PER_PIXEL # px * B/px = B | |
| return total_bytes |
| from typing import Final, TypeVar | |
| _T = TypeVar('_T') | |
| # is there some way to avoid union type? | |
| def slice_by_search(inp: (list | tuple)[_T], start: _T, end: _T): | |
| '''LICENSE: Unlicense''' | |
| i: Final = 1 + inp.index(start) | |
| return inp[i: inp.index(end, i + 1)] |
| addEventListener('keydown', k => console.table(k.keyCode, k.code, k.key, String.fromCharCode(k.keyCode))) |
Use binary 111110000 as example...
Add up the powers of 2, then use bit shift/multiplication/addition identities to simplify the sum of all divisors. You'll get the original Perfect number
Divisors: 1, 10, 100, 1000, 10000, 11111, 111110, 1111100, 11111000
(WIP) I'm planning on improving this, then I'll make a YT video or a GHP-blog/post about it
divide each digit by 2 starting from the MSD using int div, if you have remainder it means you have to add 5 to the next digit AFTER dividing it by 2.
Examples: