ee1e8abd87
* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
89 lines
1.3 KiB
Markdown
89 lines
1.3 KiB
Markdown
---
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id: 5900f3761000cf542c50fe88
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title: 问题9：特殊的毕达哥拉斯三重奏
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challengeType: 5
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videoUrl: ''
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dashedName: problem-9-special-pythagorean-triplet
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---
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# --description--
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毕达哥拉斯三元组是一组三个自然数， `a` < `b` < `c` ，其中，
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`a`
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<sup>2</sup>
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- `b`
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<sup>2</sup>
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= `c`
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<sup>2</sup>
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例如，3
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<sup>2</sup>
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- 4
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<sup>2</sup>
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= 9 + 16 = 25 = 5
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<sup>2</sup>
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。恰好存在一个毕达哥拉斯三元组，其中`a` + `b` + `c` = 1000.求产品`abc`使得`a` + `b` + `c` = `n` 。
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# --hints--
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`specialPythagoreanTriplet(1000)`应返回31875000。
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```js
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assert.strictEqual(specialPythagoreanTriplet(1000), 31875000);
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```
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`specialPythagoreanTriplet(24)`应该返回480。
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```js
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assert.strictEqual(specialPythagoreanTriplet(24), 480);
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```
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`specialPythagoreanTriplet(120)`应该返回49920。
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```js
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assert([49920, 55080, 60000].includes(specialPythagoreanTriplet(120)));
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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 specialPythagoreanTriplet(n) {
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let sumOfabc = n;
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return true;
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}
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specialPythagoreanTriplet(1000);
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```
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# --solutions--
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```js
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const specialPythagoreanTriplet = (n)=>{
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let sumOfabc = n;
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let a,b,c;
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for(a = 1; a<=sumOfabc/3; a++){
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for(b = a+1; b<=sumOfabc/2; b++){
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c = Math.sqrt(a*a+b*b);
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if((a+b+c) == sumOfabc){
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return a*b*c;
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}
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}
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}
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}
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```
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