153 lines
3.8 KiB
Markdown
153 lines
3.8 KiB
Markdown
---
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id: 5900f3c81000cf542c50fedb
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title: '問題 92：平方數鏈'
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challengeType: 5
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forumTopicId: 302209
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dashedName: problem-92-square-digit-chains
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---
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# --description--
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將一個數字的每一位求平方再相加可以得到一個新的數字，不斷重複該過程，直到新的數字出現過爲止，可以得到一條數鏈。
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舉個例子：
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$$\begin{align} & 44 → 32 → 13 → 10 → \boldsymbol{1} → \boldsymbol{1}\\\\ & 85 → \boldsymbol{89} → 145 → 42 → 20 → 4 → 16 → 37 → 58 → \boldsymbol{89}\\\\ \end{align}$$
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可以發現，每條到達 1 或 89 的數鏈都會陷入循環。 最令人驚訝的是，從任意數字開始，數鏈最終都會到達 1 或 89。
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求出有多少個小於 `limit` 的數字最終會到達 89？
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# --hints--
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`squareDigitChains(100)` 應該返回一個數字。
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```js
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assert(typeof squareDigitChains(100) === 'number');
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```
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`squareDigitChains(100)` 應該返回 `80`。
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```js
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assert.strictEqual(squareDigitChains(100), 80);
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```
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`squareDigitChains(1000)` 應該返回 `857`。
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```js
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assert.strictEqual(squareDigitChains(1000), 857);
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```
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`squareDigitChains(100000)` 應該返回 `85623`。
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```js
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assert.strictEqual(squareDigitChains(100000), 85623);
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```
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`squareDigitChains(10000000)` 應該返回 `8581146`。
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```js
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assert.strictEqual(squareDigitChains(10000000), 8581146);
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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 squareDigitChains(limit) {
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return true;
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}
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squareDigitChains(100);
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```
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# --solutions--
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```js
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function squareDigitChains(limit) {
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// Based on https://www.xarg.org/puzzle/project-euler/problem-92/
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function getCombinations(neededDigits, curDigits) {
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if (neededDigits === curDigits.length) {
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return [curDigits];
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}
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const combinations = [];
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const lastDigit = curDigits.length !== 0 ? curDigits[0] : 9;
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for (let i = 0; i <= lastDigit; i++) {
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const results = getCombinations(neededDigits, [i].concat(curDigits));
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combinations.push(...results);
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}
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return combinations;
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}
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function getPossibleSums(limit) {
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const digitsCount = getDigits(limit).length - 1;
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const possibleSquaredSums = [false];
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for (let i = 1; i <= 81 * digitsCount; i++) {
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let curVal = i;
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while (curVal !== 1 && curVal !== 89) {
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curVal = addSquaredDigits(curVal);
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}
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possibleSquaredSums[i] = curVal === 89;
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}
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return possibleSquaredSums;
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}
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function addSquaredDigits(num) {
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const digits = getDigits(num);
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let result = 0;
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for (let i = 0; i < digits.length; i++) {
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result += digits[i] ** 2;
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}
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return result;
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}
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function getDigits(number) {
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const digits = [];
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while (number > 0) {
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digits.push(number % 10);
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number = Math.floor(number / 10);
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}
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return digits;
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}
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function getFactorials(number) {
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const factorials = [1];
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for (let i = 1; i < number; i++) {
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factorials[i] = factorials[i - 1] * i;
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}
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return factorials;
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}
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const neededDigits = getDigits(limit).length - 1;
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const combinations = getCombinations(neededDigits, []);
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const possibleSquaredDigitsSums = getPossibleSums(limit);
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const factorials = getFactorials(neededDigits + 1);
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let endingWith89 = 0;
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for (let i = 0; i < combinations.length; i++) {
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let counts = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0];
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let digits = combinations[i];
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let curSum = 0;
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for (let j = 0; j < digits.length; j++) {
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const curDigit = digits[j];
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curSum += curDigit ** 2;
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counts[curDigit]++;
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}
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if (possibleSquaredDigitsSums[curSum]) {
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let denominator = 1;
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for (let j = 0; j < counts.length; j++) {
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denominator = denominator * factorials[counts[j]];
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}
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endingWith89 += Math.floor(
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factorials[factorials.length - 1] / denominator
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);
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}
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}
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return endingWith89;
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}
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```
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