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matinkaboli / generator_promise.js
Created September 2, 2017 19:59
use Generator and Promise to impelement async functions
function* getResponse() {
yield new Promise(resolve => {
request('example.com', res => {
resolve();
});
});
// you can have more functions. It's just a example. just do it
// yield new Promise ....
}
#include <stdio.h>
int checkNumbers(int n) {
if (n % 3 == 0 || n % 5 == 0) {
return 1;
}
return 0;
}
int main() {
#include <stdio.h>
int is_even(int n) {
if (n % 2 == 0) {
return 1;
}
return 0;
}
int main() {
let primes = [];
let inProgress = 600851475143;
let primeFactors = [];
function isPrime(n) {
for (let i = 2; i < n; i++) {
if (n % i === 0) {
return false;
}
}
let arr = [];
for (let i = 100; i < 1000; i++) {
for (let j = 100; j < 1000; j++) {
let answer = i * j;
let aStr = answer.toString().split('');
if (aStr[0] === aStr[aStr.length - 1]) {
if (aStr[1] === aStr[aStr.length - 2]) {
if (aStr[2] === aStr[aStr.length - 3]) {
arr.push(answer);
// This is slow
for (let i = 1;; i++) {
let count = 0;
for (let j = 1; j < 20; j++) {
if (i % j === 0) {
count += 1;
}
}
if (count === 19) {
console.log(i);
let arrSq = [];
let arrSum = [];
for (let i = 1; i <= 100; i++) {
arrSq.push(i * i);
}
for (let i = 0; i <= 100; i++) {
arrSum.push(i);
}
arrSum = arrSum.reduce((a, b) => a + b);
i = 3
n = s = 1
while (i <= 10001):
if (s == 1):
x = 6 * n -1
s = 0
else:
x = 6 * n +1
s = 1
n = n + 1
strings = [
"73167176531330624919225119674426574742355349194934",
"96983520312774506326239578318016984801869478851843",
"85861560789112949495459501737958331952853208805511",
"12540698747158523863050715693290963295227443043557",
"66896648950445244523161731856403098711121722383113",
"62229893423380308135336276614282806444486645238749",
"30358907296290491560440772390713810515859307960866",
"70172427121883998797908792274921901699720888093776",
"65727333001053367881220235421809751254540594752243",
#include <stdio.h>
int pythagorean(int num) {
for (int i = 1; i < num; i++) {
for (int j = 1; j < num - 1; j++) {
const int k = num - i - j;
if ((i * i) + (j * j) == (k * k)) {
return i * j * k;
}
}