* 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>
107 lines
1.9 KiB
Markdown
107 lines
1.9 KiB
Markdown
---
|
||
id: 5900f39c1000cf542c50feae
|
||
title: 问题47:不同的素数因素
|
||
challengeType: 5
|
||
videoUrl: ''
|
||
dashedName: problem-47-distinct-primes-factors
|
||
---
|
||
|
||
# --description--
|
||
|
||
前两个连续数字有两个不同的素数因子是:
|
||
|
||
14 = 2×7
|
||
|
||
15 = 3×5
|
||
|
||
前三个连续数字有三个不同的素因子:
|
||
|
||
644 =2²×7×23
|
||
|
||
645 = 3×5×43
|
||
|
||
646 = 2×17×19
|
||
|
||
找到前四个连续的整数,每个整数有四个不同的素数因子。这些数字中的第一个是什么?
|
||
|
||
# --hints--
|
||
|
||
`distinctPrimeFactors(2, 2)`应该返回14。
|
||
|
||
```js
|
||
assert.strictEqual(distinctPrimeFactors(2, 2), 14);
|
||
```
|
||
|
||
`distinctPrimeFactors(3, 3)`应该返回644。
|
||
|
||
```js
|
||
assert.strictEqual(distinctPrimeFactors(3, 3), 644);
|
||
```
|
||
|
||
`distinctPrimeFactors(4, 4)`应该返回134043。
|
||
|
||
```js
|
||
assert.strictEqual(distinctPrimeFactors(4, 4), 134043);
|
||
```
|
||
|
||
# --seed--
|
||
|
||
## --seed-contents--
|
||
|
||
```js
|
||
function distinctPrimeFactors(targetNumPrimes, targetConsecutive) {
|
||
|
||
return true;
|
||
}
|
||
|
||
distinctPrimeFactors(4, 4);
|
||
```
|
||
|
||
# --solutions--
|
||
|
||
```js
|
||
function distinctPrimeFactors(targetNumPrimes, targetConsecutive) {
|
||
function numberOfPrimeFactors(n) {
|
||
let factors = 0;
|
||
|
||
// Considering 2 as a special case
|
||
let firstFactor = true;
|
||
while (n % 2 == 0) {
|
||
n = n / 2;
|
||
if (firstFactor) {
|
||
factors++;
|
||
firstFactor = false;
|
||
}
|
||
}
|
||
// Adding other factors
|
||
for (let i = 3; i < Math.sqrt(n); i += 2) {
|
||
firstFactor = true;
|
||
while (n % i == 0) {
|
||
n = n / i;
|
||
if (firstFactor) {
|
||
factors++;
|
||
firstFactor = false;
|
||
}
|
||
}
|
||
}
|
||
|
||
if (n > 1) { factors++; }
|
||
|
||
return factors;
|
||
}
|
||
|
||
let number = 0;
|
||
let consecutive = 0;
|
||
|
||
while (consecutive < targetConsecutive) {
|
||
number++;
|
||
if (numberOfPrimeFactors(number) >= targetNumPrimes) {
|
||
consecutive++;
|
||
} else {
|
||
consecutive = 0;
|
||
}
|
||
}
|
||
return number - targetConsecutive + 1;
|
||
}
|
||
```
|