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
5900f39c1000cf542c50feae 问题47不同的素数因素 5 problem-47-distinct-primes-factors

--description--

前两个连续数字有两个不同的素数因子是:

14 = 2×7

15 = 3×5

前三个连续数字有三个不同的素因子:

644 =2²×7×23

645 = 3×5×43

646 = 2×17×19

找到前四个连续的整数,每个整数有四个不同的素数因子。这些数字中的第一个是什么?

--hints--

distinctPrimeFactors(2, 2)应该返回14。

assert.strictEqual(distinctPrimeFactors(2, 2), 14);

distinctPrimeFactors(3, 3)应该返回644。

assert.strictEqual(distinctPrimeFactors(3, 3), 644);

distinctPrimeFactors(4, 4)应该返回134043。

assert.strictEqual(distinctPrimeFactors(4, 4), 134043);

--seed--

--seed-contents--

function distinctPrimeFactors(targetNumPrimes, targetConsecutive) {

  return true;
}

distinctPrimeFactors(4, 4);

--solutions--

function distinctPrimeFactors(targetNumPrimes, targetConsecutive) {
  function numberOfPrimeFactors(n) {
    let factors = 0;

    //  Considering 2 as a special case
    let firstFactor = true;
    while (n % 2 == 0) {
      n = n / 2;
      if (firstFactor) {
        factors++;
        firstFactor = false;
      }
    }
    // Adding other factors
    for (let i = 3; i < Math.sqrt(n); i += 2) {
      firstFactor = true;
      while (n % i == 0) {
        n = n / i;
        if (firstFactor) {
          factors++;
          firstFactor = false;
        }
      }
    }

    if (n > 1) { factors++; }

    return factors;
  }

  let number = 0;
  let consecutive = 0;

  while (consecutive < targetConsecutive) {
    number++;
    if (numberOfPrimeFactors(number) >= targetNumPrimes) {
      consecutive++;
    } else {
      consecutive = 0;
    }
  }
  return number - targetConsecutive + 1;
}