* 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>
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id, title, challengeType, videoUrl, dashedName
id | title | challengeType | videoUrl | dashedName |
---|---|---|---|---|
5900f3711000cf542c50fe84 | 问题5:最小的倍数 | 5 | problem-5-smallest-multiple |
--description--
2520是可以除以1到10中的每个数字而没有任何余数的最小数字。从1到n
所有数字均可被整除的最小正数是多少?
--hints--
smallestMult(5)
应该返回60。
assert.strictEqual(smallestMult(5), 60);
smallestMult(7)
应该返回420。
assert.strictEqual(smallestMult(7), 420);
smallestMult(10)
应返回2520。
assert.strictEqual(smallestMult(10), 2520);
smallestMult(13)
应返回360360。
assert.strictEqual(smallestMult(13), 360360);
smallestMult(20)
应该返回232792560。
assert.strictEqual(smallestMult(20), 232792560);
--seed--
--seed-contents--
function smallestMult(n) {
return true;
}
smallestMult(20);
--solutions--
function smallestMult(n){
function gcd(a, b) {
return b === 0 ? a : gcd(b, a%b); // Euclidean algorithm
}
function lcm(a, b) {
return a * b / gcd(a, b);
}
var result = 1;
for(var i = 2; i <= n; i++) {
result = lcm(result, i);
}
return result;
}