* 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>
2.4 KiB
2.4 KiB
id, title, challengeType, videoUrl, dashedName
id | title | challengeType | videoUrl | dashedName |
---|---|---|---|---|
5900f3781000cf542c50fe8b | 问题12:高度可分的三角数 | 5 | problem-12-highly-divisible-triangular-number |
--description--
通过添加自然数生成三角数的序列。所以第7个三角形数字是1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.前十个术语是:
1,3,6,10,15,21,28,36,45,55 ......
让我们列出前七个三角形数字的因子:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
我们可以看到28是第一个超过五个除数的三角形数。超过n
除数的第一个三角形数的值是多少?
--hints--
divisibleTriangleNumber(5)
应该返回28。
assert.strictEqual(divisibleTriangleNumber(5), 28);
divisibleTriangleNumber(23)
应该返回630。
assert.strictEqual(divisibleTriangleNumber(23), 630);
divisibleTriangleNumber divisibleTriangleNumber(167)
应该返回1385280。
assert.strictEqual(divisibleTriangleNumber(167), 1385280);
divisibleTriangleNumber divisibleTriangleNumber(374)
应该返回17907120。
assert.strictEqual(divisibleTriangleNumber(374), 17907120);
divisibleTriangleNumber divisibleTriangleNumber(500)
应该返回76576500。
assert.strictEqual(divisibleTriangleNumber(500), 76576500);
--seed--
--seed-contents--
function divisibleTriangleNumber(n) {
return true;
}
divisibleTriangleNumber(500);
--solutions--
function divisibleTriangleNumber(n) {
if (n === 1) return 3;
let counter = 1;
let triangleNumber = counter++;
while (noOfFactors(triangleNumber) < n) {
triangleNumber += counter++;
}
return triangleNumber;
}
function noOfFactors(num) {
const primeFactors = getPrimeFactors(num);
let prod = 1;
for(let p in primeFactors) {
prod *= (primeFactors[p] + 1)
}
return prod;
}
function getPrimeFactors(num) {
let n = num;
let primes = {};
let p = 2;
let sqrt = Math.sqrt(num);
function checkAndUpdate(inc) {
if (n % p === 0) {
const curr = primes[p];
if (curr) {
primes[p]++
} else {
primes[p] = 1;
}
n /= p;
} else {
p += inc;
}
}
while(p === 2 && p <= n) {
checkAndUpdate(1);
}
while (p <= n && p <= sqrt) {
checkAndUpdate(2);
}
if(Object.keys(primes).length === 0) {
primes[num] = 1;
} else if(n !== 1) {
primes[n] = 1;
}
return primes;
}