--- id: 5900f3ec1000cf542c50feff title: 'Problem 128: Hexagonal tile differences' challengeType: 1 forumTopicId: 301755 dashedName: problem-128-hexagonal-tile-differences --- # --description-- A hexagonal tile with number 1 is surrounded by a ring of six hexagonal tiles, starting at "12 o'clock" and numbering the tiles 2 to 7 in an anti-clockwise direction. New rings are added in the same fashion, with the next rings being numbered 8 to 19, 20 to 37, 38 to 61, and so on. The diagram below shows the first three rings. 前三圈排列好的六角砖,数字编号为 1 到 37,其中砖 8 和砖 17高亮 通过计算砖 $n$ 和它周围 6 块砖的数字差,我们定位 $PD(n)$ 为数字差中素数的个数。 例如,围绕砖 8 顺时针方向的差额分别为 12、29、11、6、1 和 13。 则 $PD(8) = 3$。 同理,围绕砖 17 的差额为 1、17、16、1、11 和 10,所以 $PD(17) = 2$。 可以发现 $PD(n)$ 的最大值是 $3$。 如果 $PD(n) = 3$ 的砖按升序排列,那么第 10 块砖将会是 271。 求序列中的第 2000 块砖。 # --hints-- `hexagonalTile(10)` should return `271`. ```js assert.strictEqual(hexagonalTile(10), 271); ``` `hexagonalTile(2000)` should return `14516824220`. ```js assert.strictEqual(hexagonalTile(2000), 14516824220); ``` # --seed-- ## --seed-contents-- ```js function hexagonalTile(tileIndex) { return true; } hexagonalTile(10); ``` # --solutions-- ```js class PrimeSeive { constructor(num) { const seive = Array(Math.floor((num - 1) / 2)).fill(true); const upper = Math.floor((num - 1) / 2); const sqrtUpper = Math.floor((Math.sqrt(num) - 1) / 2); for (let i = 0; i <= sqrtUpper; i++) { if (seive[i]) { // Mark value in seive array const prime = 2 * i + 3; // Mark all multiples of this number as false (not prime) const primeSqaredIndex = 2 * i ** 2 + 6 * i + 3; for (let j = primeSqaredIndex; j < upper; j += prime) { seive[j] = false; } } } this._seive = seive; } isPrime(num) { return num === 2 ? true : num % 2 === 0 ? false : this.isOddPrime(num); } isOddPrime(num) { return this._seive[(num - 3) / 2]; } }; function hexagonalTile(tileIndex) { const primeSeive = new PrimeSeive(tileIndex * 420); let count = 1; let n = 1; let number = 0; while (count < tileIndex) { if (primeSeive.isPrime(6*n - 1) && primeSeive.isPrime(6*n + 1) && primeSeive.isPrime(12*n + 5)) { number = 3*n*n - 3*n + 2; count++; if (count >= tileIndex) break; } if (primeSeive.isPrime(6*n + 5) && primeSeive.isPrime(6*n - 1) && primeSeive.isPrime(12*n - 7) && n != 1) { number = 3*n*n + 3*n + 1; count++; } n++; } return number; } ```