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---
id: 5900f3b61000cf542c50fec8
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title: 'Problema 73: contare frazioni in un range'
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challengeType: 5
forumTopicId: 302186
dashedName: problem-73-counting-fractions-in-a-range
---
# --description--
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Considera la frazione, $\frac{n}{d}$, dove `n` e `d` sono numeri interi positivi. Se `n` < `d` e il fattore commune più alto, ${HCF}(n, d) = 1$, viene chiamata frazione ridotta propria.
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Se elenchiamo la lista delle frazioni proprie ridotte per `d` ≤ 8 in ordine decrescente di dimensioni, otteniamo:
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$$\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \mathbf{\frac{3}{8}, \frac{2}{5}, \frac{3}{7}}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}$$
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Si vede che ci sono `3` frazioni tra $\frac{1}{3}$ e $\frac{1}{2}$.
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Quante frazioni ci sono tra $\frac{1}{3}$ e $\frac{1}{2}$ nel set ordinato di frazioni proprie semplificate per `d` ≤ `limit` ?
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# --hints--
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`countingFractionsInARange(8)` dovrebbe restituire un numero.
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```js
assert(typeof countingFractionsInARange(8) === 'number');
```
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`countingFractionsInARange(8)` dovrebbe restituire `3` .
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```js
assert.strictEqual(countingFractionsInARange(8), 3);
```
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`countingFractionsInARange(1000)` dovrebbe restituire `50695` .
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```js
assert.strictEqual(countingFractionsInARange(1000), 50695);
```
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`countingFractionsInARange(6000)` dovrebbe restituire `1823861` .
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```js
assert.strictEqual(countingFractionsInARange(6000), 1823861);
```
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`countingFractionsInARange(12000)` dovrebbe restituire `7295372` .
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```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;
}
```