2018-10-10 18:03:03 -04:00
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---
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id: 5900f3891000cf542c50fe9c
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2020-12-16 00:37:30 -07:00
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title: 问题29:不同的权力
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2018-10-10 18:03:03 -04:00
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challengeType: 5
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videoUrl: ''
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2021-01-13 03:31:00 +01:00
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dashedName: problem-29-distinct-powers
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2018-10-10 18:03:03 -04:00
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---
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2020-12-16 00:37:30 -07:00
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# --description--
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2020-02-18 01:40:55 +09:00
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考虑2≤a≤5和2≤b≤5的ab的所有整数组合:
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2020-12-16 00:37:30 -07:00
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2020-02-18 01:40:55 +09:00
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22 = 4、23 = 8、24 = 16、25 = 32
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2020-12-16 00:37:30 -07:00
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2020-02-18 01:40:55 +09:00
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32 = 9、33 = 27、34 = 81、35 = 243
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2020-12-16 00:37:30 -07:00
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2020-02-18 01:40:55 +09:00
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42 = 16、43 = 64、44 = 256、45 = 1024
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2020-12-16 00:37:30 -07:00
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2020-02-18 01:40:55 +09:00
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52 = 25、53 = 125、54 = 625、55 = 3125
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2020-12-16 00:37:30 -07:00
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2020-02-18 01:40:55 +09:00
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如果然后按数字顺序放置它们,并删除所有重复项,则会得到以下15个不同术语的序列:
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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4,8,9,16,25,27,32,64,81,125,243,256,625,1024,3125
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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对于2≤a≤n和2≤b≤n,由ab生成的序列中有多少个不同的项?
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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# --hints--
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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`distinctPowers(15)`应该返回177。
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2018-10-10 18:03:03 -04:00
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```js
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2020-12-16 00:37:30 -07:00
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assert.strictEqual(distinctPowers(15), 177);
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2018-10-10 18:03:03 -04:00
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```
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2020-12-16 00:37:30 -07:00
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`distinctPowers(20)`应该返回324。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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```js
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assert.strictEqual(distinctPowers(20), 324);
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```
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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`distinctPowers(25)`应该返回519。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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```js
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assert.strictEqual(distinctPowers(25), 519);
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```
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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`distinctPowers(30)`应该返回755。
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2018-10-10 18:03:03 -04:00
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```js
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2020-12-16 00:37:30 -07:00
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assert.strictEqual(distinctPowers(30), 755);
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2018-10-10 18:03:03 -04:00
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```
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2020-08-13 17:24:35 +02:00
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2021-01-13 03:31:00 +01:00
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# --seed--
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## --seed-contents--
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```js
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function distinctPowers(n) {
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return n;
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}
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distinctPowers(30);
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```
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2020-12-16 00:37:30 -07:00
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# --solutions--
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2021-01-13 03:31:00 +01:00
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```js
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const distinctPowers = (n) => {
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let list = [];
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for (let a=2; a<=n; a++) {
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for (let b=2; b<=n; b++) {
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let term = Math.pow(a, b);
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if (list.indexOf(term)===-1) list.push(term);
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}
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}
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return list.length;
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};
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```
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