2018-10-10 18:03:03 -04:00
|
|
|
|
---
|
|
|
|
|
id: 5900f3811000cf542c50fe94
|
2020-12-16 00:37:30 -07:00
|
|
|
|
title: 问题21:友好的数字
|
2018-10-10 18:03:03 -04:00
|
|
|
|
challengeType: 5
|
|
|
|
|
videoUrl: ''
|
|
|
|
|
---
|
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
# --description--
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
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。
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
# --hints--
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
`sumAmicableNum(1000)`应返回504。
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
```js
|
|
|
|
|
assert.strictEqual(sumAmicableNum(1000), 504);
|
2018-10-10 18:03:03 -04:00
|
|
|
|
```
|
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
`sumAmicableNum(2000)`应该返回2898。
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
|
|
|
|
```js
|
2020-12-16 00:37:30 -07:00
|
|
|
|
assert.strictEqual(sumAmicableNum(2000), 2898);
|
2018-10-10 18:03:03 -04:00
|
|
|
|
```
|
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
`sumAmicableNum(5000)`应该返回8442。
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
```js
|
|
|
|
|
assert.strictEqual(sumAmicableNum(5000), 8442);
|
|
|
|
|
```
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
`sumAmicableNum(10000)`应返回31626。
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
|
|
|
|
```js
|
2020-12-16 00:37:30 -07:00
|
|
|
|
assert.strictEqual(sumAmicableNum(10000), 31626);
|
2018-10-10 18:03:03 -04:00
|
|
|
|
```
|
2020-08-13 17:24:35 +02:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
# --solutions--
|
|
|
|
|
|