1.8 KiB
1.8 KiB
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f3931000cf542c50fea6 | Problem 39: Integer right triangles | 5 | 302054 | problem-39-integer-right-triangles |
--description--
If p
is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.
{20,48,52}, {24,45,51}, {30,40,50}
For which value of p
≤ n
, is the number of solutions maximized?
--hints--
intRightTriangles(500)
should return a number.
assert(typeof intRightTriangles(500) === 'number');
intRightTriangles(500)
should return 420.
assert(intRightTriangles(500) == 420);
intRightTriangles(800)
should return 720.
assert(intRightTriangles(800) == 720);
intRightTriangles(900)
should return 840.
assert(intRightTriangles(900) == 840);
intRightTriangles(1000)
should return 840.
assert(intRightTriangles(1000) == 840);
--seed--
--seed-contents--
function intRightTriangles(n) {
return n;
}
intRightTriangles(500);
--solutions--
// 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;
}