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gikf f559f18747 fix(curriculum): rework Project Euler 73 (#42300 )

* fix: rework challenge to use argument in function

* fix: add solution

* fix: use MathJax to improve math notation look

2021-06-03 14:00:28 +09:00

2.2 KiB

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id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
5900f3b61000cf542c50fec8	Problem 73: Counting fractions in a range	5	302186	problem-73-counting-fractions-in-a-range

--description--

Consider the fraction, \frac{n}{d}, where n and d are positive integers. If n < d and highest common factor, {HCF}(n, d) = 1, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we get:

\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \mathbf{\frac{3}{8}, \frac{2}{5}, \frac{3}{7}}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}

It can be seen that there are 3 fractions between \frac{1}{3} and \frac{1}{2}.

How many fractions lie between \frac{1}{3} and \frac{1}{2} in the sorted set of reduced proper fractions for d ≤ limit?

--hints--

countingFractionsInARange(8) should return a number.

assert(typeof countingFractionsInARange(8) === 'number');

countingFractionsInARange(8) should return 3.

assert.strictEqual(countingFractionsInARange(8), 3);

countingFractionsInARange(1000) should return 50695.

assert.strictEqual(countingFractionsInARange(1000), 50695);

countingFractionsInARange(6000) should return 1823861.

assert.strictEqual(countingFractionsInARange(6000), 1823861);

countingFractionsInARange(12000) should return 7295372.

assert.strictEqual(countingFractionsInARange(12000), 7295372);

--seed--

--seed-contents--

function countingFractionsInARange(limit) {

  return true;
}

countingFractionsInARange(8);

--solutions--

function countingFractionsInARange(limit) {
  let result = 0;
  const stack = [[3, 2]];
  while (stack.length > 0) {
    const [startDenominator, endDenominator] = stack.pop();
    const curDenominator = startDenominator + endDenominator;
    if (curDenominator <= limit) {
      result++;
      stack.push([startDenominator, curDenominator]);
      stack.push([curDenominator, endDenominator]);
    }
  }
  return result;
}

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