henix

/**
 * Arbitrary-length tuples
 * =======================
 * 
 * Working with tuples of unknown length in TypeScript is a pain. Most library authors fall back on enumerating all possible
 * tuple lengths up to a threshold (see an example here https://github.com/pelotom/runtypes/blob/fe19290d375c8818d2c52243ddc2911c8369db37/src/types/tuple.ts )
 * 
 * In this gist, I'm attempting to leverage recursion to provide support for arbitrary length tuples. This has the potential
 * to make some kinds of declarative APIs nicer and enhance type inference in some cases.
 * This example shows how to take a variable-length tuple as an input, transform each of its types and use the resulting

henix

<script>
  if (typeof Promise == "undefined") {
    document.write('<script src="https://cdn.jsdelivr.net/npm/promise-polyfill@8.1.0/dist/polyfill.min.js"><\/script>');
  }
  if (window.URL && window.URL.prototype && ('href' in window.URL.prototype)) {
  } else {
    document.write('<script src="https://cdn.jsdelivr.net/gh/arv/DOM-URL-Polyfill@e54d67f/src/url.min.js"><\/script>');
  }
  if (typeof URLSearchParams == "undefined") {
    document.write('<script src="https://cdn.jsdelivr.net/npm/@ungap/url-search-params@0.1.2/min.js"><\/script>');

henix

#!/bin/sh
[ $# -ne 1 ] && echo "Usage: $0 name" && exit 1

p=$(printf "%d" "0x$(printf "%s" "$1" | md5sum | head -c 4)")
# pratical port range is 1024 ~ 65535 (for app that don't require root)
# 64499 is the first prime number that less than 65535 - 1024
port=$((p % 64499 + 1024))
echo "$port"

henix

有一存有整数的数组 A[n] (n <= 2^31 - 1, 但需要的话可以再扩大), 其中 0 <= i, A[i] <= n-1 且 i!=j 时 A[i]!=A[j] (即一个排列). 用你认为*最高效*的方法求 B=A^(-1) (即对范围内满足 A[B[i]]=i 的 B)

henix

rel() {
	[ ! -d 'target/universal/stage' ] && echo stage not exist! && return 1
	name=$(pwd | tr '/' '\n' | tail -n1)
	timestamp=$(date +%Y%m%d_%H%M%S)
	cp -r "target/universal/stage" "/apps/$name/stage-$timestamp"
	ln -sTv "stage-$timestamp" "/apps/$name/stage"
}

henix

/*************
* colors.js *
 *************
 *
 * You're almost at the exit. You just need to get past this
 * color lock.
 *
 * Changing your environment is no longer enough. You must
 * learn to change yourself. I've sent you a little something
 * that should help with that.

henix

int getint2(int *out) {
	register int s = 0;
	register int ch;
	ch = getchar();
	bool negative = false;
	while (ch < '0' || ch > '9') {
		if (ch == '-') {
			negative = true;
			ch = getchar();
			break;

henix

typedef int T;

inline bool less(T a, T b) {
	return a < b;
}

inline bool lessEq(T a, T b) {
	return a <= b;
}

henix

static inline void putllong(long long n)
{
	static char buf[20];
	register int pos;
	register long long x = n;
	if (x == 0) {
		putchar('0');
		return;
	}
	if (x == LLONG_MIN) { // x = -x do not work for the minimal value of long long, so process it first

henix

int getint(char end) {
	int s = 0;
	int ch;
	ch = getchar();
	while (ch != end && ch != EOF) {
		s = s * 10 + ch - '0';
		ch = getchar();
	}
	return s;
}