125 lines
2.9 KiB
Markdown
125 lines
2.9 KiB
Markdown
---
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id: 5900f3cc1000cf542c50fede
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title: '问题 95:友好的数链'
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challengeType: 5
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forumTopicId: 302212
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dashedName: problem-95-amicable-chains
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---
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# --description--
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一个数的真因子是除自身以外的其他因子。 例如,28 的真因子是 1、2、4、7 和 14。 由于这些真因子之和等于 28,我们称 28 为完全数,又称完美数或完备数。
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有趣的是,220 的真因子之和为 284,而 284 的真因子之和为 220,形成了一条两个数构成的链。 因此,220 和 284 被称为友好数对。
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也许更长的链条鲜为人知。 例如,从 12496 开始,可以形成一条五个数字的数链:
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$$ 12496 → 14288 → 15472 → 14536 → 14264 \\,(→ 12496 → \cdots) $$
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由于该链返回其起始点,因此称为友好数链。
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找出最长友好数链中的最小数字,要求该链中的每一个数字均不能超过给定的 `limit`。
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# --hints--
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`amicableChains(300)` 应该返回一个数字。
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```js
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assert(typeof amicableChains(300) === 'number');
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```
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`amicableChains(300)` 应该返回 `220`。
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```js
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assert.strictEqual(amicableChains(300), 220);
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```
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`amicableChains(15000)` 应该返回 `220`。
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```js
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assert.strictEqual(amicableChains(15000), 220);
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```
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`amicableChains(100000)` 应该返回 `12496`。
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```js
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assert.strictEqual(amicableChains(100000), 12496);
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```
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`amicableChains(1000000)` 应该返回 `14316`。
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```js
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assert.strictEqual(amicableChains(1000000), 14316);
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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 amicableChains(limit) {
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return true;
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}
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amicableChains(300);
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```
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# --solutions--
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```js
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function amicableChains(limit) {
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function getSmallestMember(chain) {
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let smallest = chain[0];
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for (let i = 1; i < chain.length; i++) {
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if (smallest > chain[i]) {
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smallest = chain[i];
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}
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}
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return smallest;
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}
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function getFactorsSums(limit) {
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const factorsSums = new Array(limit + 1).fill(1);
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for (let i = 2; i <= limit / 2; i++) {
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for (let j = 2 * i; j <= limit; j += i) {
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factorsSums[j] += i;
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}
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}
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return factorsSums;
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}
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const factorsSums = getFactorsSums(limit);
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const checkedNumbers = new Set();
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let longestChain = 0;
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let smallestMember = 0;
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for (let i = 0; i <= limit; i++) {
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const curChain = [];
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let curNumber = i;
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while (!checkedNumbers.has(curNumber) && factorsSums[curNumber] <= limit) {
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curNumber = factorsSums[curNumber];
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const chainStart = curChain.indexOf(curNumber);
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if (chainStart === -1) {
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curChain.push(curNumber);
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continue;
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}
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const chainLength = curChain.length - chainStart;
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if (chainLength > longestChain) {
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longestChain = chainLength;
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smallestMember = getSmallestMember(curChain.slice(chainStart));
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}
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break;
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}
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for (let j = 0; j < curChain.length; j++) {
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checkedNumbers.add(curChain[j]);
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}
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}
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return smallestMember;
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}
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```
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