92 lines
		
	
	
		
			1.9 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
		
		
			
		
	
	
			92 lines
		
	
	
		
			1.9 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
|   | --- | ||
|  | id: 5900f3931000cf542c50fea6 | ||
|  | title: '問題 39: 整数辺を持つ直角三角形' | ||
|  | challengeType: 5 | ||
|  | forumTopicId: 302054 | ||
|  | dashedName: problem-39-integer-right-triangles | ||
|  | --- | ||
|  | 
 | ||
|  | # --description--
 | ||
|  | 
 | ||
|  | 長さが整数である辺 {a,b,c} を持つ直角三角形の周長を `p` とします。p = 120 のとき、ちょうど 3 つの解があります。 | ||
|  | 
 | ||
|  | {20,48,52}, {24,45,51}, {30,40,50} | ||
|  | 
 | ||
|  | `p` ≤ `n` のとき、解の個数が最大になる `p` を求めなさい。 | ||
|  | 
 | ||
|  | # --hints--
 | ||
|  | 
 | ||
|  | `intRightTriangles(500)` は数値を返す必要があります。 | ||
|  | 
 | ||
|  | ```js | ||
|  | assert(typeof intRightTriangles(500) === 'number'); | ||
|  | ``` | ||
|  | 
 | ||
|  | `intRightTriangles(500)` は 420 を返す必要があります。 | ||
|  | 
 | ||
|  | ```js | ||
|  | assert(intRightTriangles(500) == 420); | ||
|  | ``` | ||
|  | 
 | ||
|  | `intRightTriangles(800)` は 720 を返す必要があります。 | ||
|  | 
 | ||
|  | ```js | ||
|  | assert(intRightTriangles(800) == 720); | ||
|  | ``` | ||
|  | 
 | ||
|  | `intRightTriangles(900)` は 840 を返す必要があります。 | ||
|  | 
 | ||
|  | ```js | ||
|  | assert(intRightTriangles(900) == 840); | ||
|  | ``` | ||
|  | 
 | ||
|  | `intRightTriangles(1000)` は 840 を返す必要があります。 | ||
|  | 
 | ||
|  | ```js | ||
|  | assert(intRightTriangles(1000) == 840); | ||
|  | ``` | ||
|  | 
 | ||
|  | # --seed--
 | ||
|  | 
 | ||
|  | ## --seed-contents--
 | ||
|  | 
 | ||
|  | ```js | ||
|  | function intRightTriangles(n) { | ||
|  | 
 | ||
|  |   return n; | ||
|  | } | ||
|  | 
 | ||
|  | intRightTriangles(500); | ||
|  | ``` | ||
|  | 
 | ||
|  | # --solutions--
 | ||
|  | 
 | ||
|  | ```js | ||
|  | // Original idea for this solution came from | ||
|  | // https://www.xarg.org/puzzle/project-euler/problem-39/ | ||
|  | 
 | ||
|  | function intRightTriangles(n) { | ||
|  |   // store the number of triangles with a given perimeter | ||
|  |   let triangles = {}; | ||
|  |   // a is the shortest side | ||
|  |   for (let a = 3; a < n / 3; a++) | ||
|  |   // o is the opposite side and is at least as long as a | ||
|  |     for (let o = a; o < n / 2; o++) { | ||
|  |       let h = Math.sqrt(a * a + o * o); // hypotenuse | ||
|  |       let p = a + o + h;  // perimeter | ||
|  |       if ((h % 1) === 0 && p <= n) { | ||
|  |         triangles[p] = (triangles[p] || 0) + 1; | ||
|  |       } | ||
|  |     } | ||
|  | 
 | ||
|  |   let max = 0, maxp = null; | ||
|  |   for (let p in triangles) { | ||
|  |     if (max < triangles[p]) { | ||
|  |       max = triangles[p]; | ||
|  |       maxp = parseInt(p); | ||
|  |     } | ||
|  |   } | ||
|  |   return maxp; | ||
|  | } | ||
|  | ``` |