84 lines
2.2 KiB
Markdown
84 lines
2.2 KiB
Markdown
---
|
|
id: 5900f3b61000cf542c50fec8
|
|
title: 'Problem 73: Counting fractions in a range'
|
|
challengeType: 5
|
|
forumTopicId: 302186
|
|
dashedName: 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.
|
|
|
|
```js
|
|
assert(typeof countingFractionsInARange(8) === 'number');
|
|
```
|
|
|
|
`countingFractionsInARange(8)` should return `3`.
|
|
|
|
```js
|
|
assert.strictEqual(countingFractionsInARange(8), 3);
|
|
```
|
|
|
|
`countingFractionsInARange(1000)` should return `50695`.
|
|
|
|
```js
|
|
assert.strictEqual(countingFractionsInARange(1000), 50695);
|
|
```
|
|
|
|
`countingFractionsInARange(6000)` should return `1823861`.
|
|
|
|
```js
|
|
assert.strictEqual(countingFractionsInARange(6000), 1823861);
|
|
```
|
|
|
|
`countingFractionsInARange(12000)` should return `7295372`.
|
|
|
|
```js
|
|
assert.strictEqual(countingFractionsInARange(12000), 7295372);
|
|
```
|
|
|
|
# --seed--
|
|
|
|
## --seed-contents--
|
|
|
|
```js
|
|
function countingFractionsInARange(limit) {
|
|
|
|
return true;
|
|
}
|
|
|
|
countingFractionsInARange(8);
|
|
```
|
|
|
|
# --solutions--
|
|
|
|
```js
|
|
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;
|
|
}
|
|
```
|