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>
82 lines
1.6 KiB
Markdown
82 lines
1.6 KiB
Markdown
---
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id: 5900f3931000cf542c50fea6
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title: 问题39：整数直角三角形
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challengeType: 5
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videoUrl: ''
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dashedName: problem-39-integer-right-triangles
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---
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# --description--
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如果p是具有整数长度边的直角三角形的周长{a，b，c}，则对于p = 120，恰好有三个解。{20,48,52}，{24,45,51}，{ 30,40,50}对于p≤n的值，最大化解的数量是多少？
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# --hints--
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`intRightTriangles(500)`应该返回420。
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```js
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assert(intRightTriangles(500) == 420);
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```
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`intRightTriangles(800)`应该返回420。
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```js
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assert(intRightTriangles(800) == 720);
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```
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`intRightTriangles(900)`应该返回840。
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```js
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assert(intRightTriangles(900) == 840);
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```
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`intRightTriangles(1000)`应该返回840。
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```js
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assert(intRightTriangles(1000) == 840);
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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 intRightTriangles(n) {
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return n;
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}
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intRightTriangles(500);
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```
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# --solutions--
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```js
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// Original idea for this solution came from
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// https://www.xarg.org/puzzle/project-euler/problem-39/
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function intRightTriangles(n) {
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// store the number of triangles with a given perimeter
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let triangles = {};
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// a is the shortest side
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for (let a = 3; a < n / 3; a++)
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// o is the opposite side and is at least as long as a
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for (let o = a; o < n / 2; o++) {
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let h = Math.sqrt(a * a + o * o); // hypotenuse
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let p = a + o + h; // perimeter
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if ((h % 1) === 0 && p <= n) {
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triangles[p] = (triangles[p] || 0) + 1;
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}
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}
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let max = 0, maxp = null;
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for (let p in triangles) {
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if (max < triangles[p]) {
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max = triangles[p];
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maxp = parseInt(p);
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}
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}
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return maxp;
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}
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```
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