2018-10-10 22:03:03 +00:00
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---
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id: 5900f3811000cf542c50fe94
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2020-12-16 07:37:30 +00:00
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title: 问题21:友好的数字
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2018-10-10 22:03:03 +00:00
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challengeType: 5
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videoUrl: ''
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2021-01-13 02:31:00 +00:00
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dashedName: problem-21-amicable-numbers
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2018-10-10 22:03:03 +00:00
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---
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2020-12-16 07:37:30 +00:00
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# --description--
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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设d( `n` )定义为`n`的适当除数之`和` (小于`n的`数均匀分成`n` )。如果d( `a` )= `b`并且d( `b` )= `a` ,其中`a` ≠ `b` ,则`a`和`b`是友好对,并且`a`和`b`中的每`一个`被称为友好数字。例如,220的适当除数是1,2,4,5,10,11,20,22,44,55和110;因此d(220)= 284. 284的适当除数是1,2,4,71和142;所以d(284)= 220.评估`n`下所有友好数字的总和。
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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# --hints--
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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`sumAmicableNum(1000)`应返回504。
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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```js
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assert.strictEqual(sumAmicableNum(1000), 504);
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2018-10-10 22:03:03 +00:00
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```
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2020-12-16 07:37:30 +00:00
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`sumAmicableNum(2000)`应该返回2898。
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2018-10-10 22:03:03 +00:00
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```js
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2020-12-16 07:37:30 +00:00
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assert.strictEqual(sumAmicableNum(2000), 2898);
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2018-10-10 22:03:03 +00:00
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```
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2020-12-16 07:37:30 +00:00
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`sumAmicableNum(5000)`应该返回8442。
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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```js
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assert.strictEqual(sumAmicableNum(5000), 8442);
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```
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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`sumAmicableNum(10000)`应返回31626。
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2018-10-10 22:03:03 +00:00
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```js
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2020-12-16 07:37:30 +00:00
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assert.strictEqual(sumAmicableNum(10000), 31626);
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2018-10-10 22:03:03 +00:00
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```
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2020-08-13 15:24:35 +00:00
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2021-01-13 02:31:00 +00:00
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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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2020-12-16 07:37:30 +00:00
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# --solutions--
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2021-01-13 02:31:00 +00:00
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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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