Oliver Eyton-Williams ee1e8abd87
feat(curriculum): restore seed + solution to Chinese (#40683)
* feat(tools): add seed/solution restore script

* chore(curriculum): remove empty sections' markers

* chore(curriculum): add seed + solution to Chinese

* chore: remove old formatter

* fix: update getChallenges

parse translated challenges separately, without reference to the source

* chore(curriculum): add dashedName to English

* chore(curriculum): add dashedName to Chinese

* refactor: remove unused challenge property 'name'

* fix: relax dashedName requirement

* fix: stray tag

Remove stray `pre` tag from challenge file.

Signed-off-by: nhcarrigan <nhcarrigan@gmail.com>

Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
2021-01-12 19:31:00 -07:00

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id, title, challengeType, videoUrl, dashedName
id title challengeType videoUrl dashedName
5900f39a1000cf542c50fead 问题46哥德巴赫的另一个猜想 5 problem-46-goldbachs-other-conjecture

--description--

由Christian Goldbach提出每个奇数的复合数可以写成素数和两个平方的总和。 9 = 7 + 2×1 2 15 = 7 + 2×2 2 21 = 3 + 2×3 2 25 = 7 + 2×3 2 27 = 19 + 2×2 2 33 = 31 + 2×1 2转这个猜想是假的。什么是最小的奇数复合,不能写为素数和两倍平方的总和?

--hints--

goldbachsOtherConjecture()应返回5777。

assert.strictEqual(goldbachsOtherConjecture(), 5777);

--seed--

--seed-contents--

function goldbachsOtherConjecture() {

  return true;
}

goldbachsOtherConjecture();

--solutions--

function goldbachsOtherConjecture() {  function isPrime(num) {
    if (num < 2) {
      return false;
    } else if (num === 2) {
      return true;
    }
    const sqrtOfNum = Math.floor(num ** 0.5);
    for (let i = 2; i <= sqrtOfNum + 1; i++) {
      if (num % i === 0) {
        return false;
      }
    }
    return true;
  }

  function isSquare(num) {
    return Math.sqrt(num) % 1 === 0;
  }

  // construct a list of prime numbers
  const primes = [];
  for (let i = 2; primes.length < 1000; i++) {
    if (isPrime(i)) primes.push(i);
  }

  let num = 3;
  let answer;
  while (!answer) {
    num += 2;
    if (!isPrime(num)) {
      let found = false;
      for (let primeI = 0; primeI < primes.length && !found; primeI++) {
        const square = (num - primes[primeI]) / 2;
        if (isSquare(square)) {
          found = true;
          break;
        }
      }
      if (!found) answer = num;
    }
  }
  return answer;
}