119 lines
2.2 KiB
Markdown
119 lines
2.2 KiB
Markdown
---
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id: 5900f38f1000cf542c50fea2
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title: 问题35：循环素数
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challengeType: 5
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videoUrl: ''
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dashedName: problem-35-circular-primes
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---
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# --description--
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这个数字197被称为循环素数，因为数字的所有旋转：197,971和719本身都是素数。在100：2,3,5,7,11,13,17,31,37,71,73,79和97之下有十三个这样的素数。在n下面有多少个圆形素数，而100 <= n < = 1000000？
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# --hints--
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`circularPrimes(100)`应该返回13。
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```js
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assert(circularPrimes(100) == 13);
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```
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`circularPrimes(100000)`应该返回43。
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```js
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assert(circularPrimes(100000) == 43);
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```
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`circularPrimes(250000)`应该返回45。
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```js
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assert(circularPrimes(250000) == 45);
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```
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`circularPrimes(500000)`应该返回49。
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```js
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assert(circularPrimes(500000) == 49);
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```
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`circularPrimes(750000)`应该返回49。
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```js
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assert(circularPrimes(750000) == 49);
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```
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`circularPrimes(1000000)`应该返回55。
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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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