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133 lines
2.6 KiB
Markdown
133 lines
2.6 KiB
Markdown
---
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id: 5900f38f1000cf542c50fea2
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title: 'Problem 35: Kreisförmige Primzahlen'
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challengeType: 1
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forumTopicId: 302009
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dashedName: problem-35-circular-primes
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---
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# --description--
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Die Zahl 197 wird als kreisförmige Primzahl bezeichnet, weil alle Drehungen der Ziffern: 197, 971, und 719, selbst Primzahlen sind.
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Es gibt dreizehn solcher Primzahlen unter 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79 und 97.
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Wie viele kreisförmige Primzahlen gibt es unter `n`, wobei 100 ≤ `n` ≤ 1000000?
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**Hinweis:**
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Kreisförmige Primzahlen können eine individuelle Drehung von `n` überschreiten.
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# --hints--
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`circularPrimes(100)` sollte eine Zahl zurückgeben.
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```js
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assert(typeof circularPrimes(100) === 'number');
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```
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`circularPrimes(100)` sollte 13 ergeben.
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```js
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assert(circularPrimes(100) == 13);
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```
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`circularPrimes(100000)` sollte 43 zurückgeben.
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```js
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assert(circularPrimes(100000) == 43);
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```
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`circularPrimes(250000)` sollte 45 zurückgeben.
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```js
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assert(circularPrimes(250000) == 45);
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```
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`circularPrimes(500000)` sollte 49 zurückgeben.
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```js
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assert(circularPrimes(500000) == 49);
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```
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`circularPrimes(750000)` sollte 49 zurückgeben.
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```js
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assert(circularPrimes(750000) == 49);
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```
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`circularPrimes(1000000)` sollte 55 zurückgeben.
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```js
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assert(circularPrimes(1000000) == 55);
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```
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# --seed--
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## --seed-contents--
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```js
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function circularPrimes(n) {
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return n;
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}
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circularPrimes(1000000);
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```
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# --solutions--
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```js
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function rotate(n) {
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if (n.length == 1) return n;
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return n.slice(1) + n[0];
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}
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function circularPrimes(n) {
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// Nearest n < 10^k
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const bound = 10 ** Math.ceil(Math.log10(n));
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const primes = [0, 0, 2];
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let count = 0;
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// Making primes array
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for (let i = 4; i <= bound; i += 2) {
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primes.push(i - 1);
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primes.push(0);
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}
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// Getting upperbound
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const upperBound = Math.ceil(Math.sqrt(bound));
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// Setting other non-prime numbers to 0
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for (let i = 3; i < upperBound; i += 2) {
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if (primes[i]) {
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for (let j = i * i; j < bound; j += i) {
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primes[j] = 0;
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}
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}
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}
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// Iterating through the array
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for (let i = 2; i < n; i++) {
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if (primes[i]) {
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let curr = String(primes[i]);
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let tmp = 1; // tmp variable to hold the no of rotations
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for (let x = rotate(curr); x != curr; x = rotate(x)) {
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if (x > n && primes[x]) {
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continue;
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}
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else if (!primes[x]) {
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// If the rotated value is 0 then it isn't a circular prime, break the loop
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tmp = 0;
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break;
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}
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tmp++;
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primes[x] = 0;
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}
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count += tmp;
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}
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}
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return count;
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}
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```
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