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85 lines
1.8 KiB
Markdown
85 lines
1.8 KiB
Markdown
---
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id: 5900f3811000cf542c50fe94
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title: 'Problem 21: Amicable numbers'
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challengeType: 1
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forumTopicId: 301851
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dashedName: problem-21-amicable-numbers
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---
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# --description--
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Let d(`n`) be defined as the sum of proper divisors of `n` (numbers less than `n` which divide evenly into `n`).
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If d(`a`) = `b` and d(`b`) = `a`, where `a` ≠ `b`, then `a` and `b` are an amicable pair and each of `a` and `b` are called amicable numbers.
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Zum Beispiel sind die richtigen Teiler von 220 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 und 110; daher ist d(220) = 284. Die richtigen Teiler von 284 sind 1, 2, 4, 71 und 142; also ist d(284) = 220.
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Berechne die Summe aller gütigen Zahlen unter `n`.
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# --hints--
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`sumAmicableNum(1000)` sollte eine Zahl zurückgeben.
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```js
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assert(typeof sumAmicableNum(1000) === 'number');
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```
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`sumAmicableNum(1000)` sollte 504 zurückgeben.
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```js
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assert.strictEqual(sumAmicableNum(1000), 504);
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```
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`sumAmicableNum(2000)` sollte 2898 zurückgeben.
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```js
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assert.strictEqual(sumAmicableNum(2000), 2898);
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```
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`sumAmicableNum(5000)` sollte 8442 zurückgeben.
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```js
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assert.strictEqual(sumAmicableNum(5000), 8442);
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```
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`sumAmicableNum(10000)` sollte 31626 zurückgeben.
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```js
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assert.strictEqual(sumAmicableNum(10000), 31626);
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```
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# --seed--
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## --seed-contents--
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```js
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function sumAmicableNum(n) {
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return n;
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}
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sumAmicableNum(10000);
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```
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# --solutions--
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```js
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const sumAmicableNum = (n) => {
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const fsum = (n) => {
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let sum = 1;
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for (let i = 2; i <= Math.floor(Math.sqrt(n)); i++)
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if (Math.floor(n % i) === 0)
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sum += i + Math.floor(n / i);
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return sum;
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};
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let d = [];
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let amicableSum = 0;
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for (let i=2; i<n; i++) d[i] = fsum(i);
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for (let i=2; i<n; i++) {
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let dsum = d[i];
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if (d[dsum]===i && i!==dsum) amicableSum += i+dsum;
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}
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return amicableSum/2;
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};
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```
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