ee1e8abd87
* 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>
90 lines
2.5 KiB
Markdown
90 lines
2.5 KiB
Markdown
---
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id: 5900f3a31000cf542c50feb6
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title: 问题55：Lychrel数字
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challengeType: 5
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videoUrl: ''
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dashedName: problem-55-lychrel-numbers
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---
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# --description--
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如果我们采取47，反向并添加，47 + 74 = 121，这是回文。并非所有数字都如此迅速地产生回文。例如，349 + 943 = 1292,1292 + 2921 = 4213 4213 + 3124 = 7337也就是说，349进行了三次迭代以到达回文。虽然还没有人证明这一点，但据认为有些数字，如196，从未产生回文。通过反向和添加过程从不形成回文的数字称为Lychrel数。由于这些数字的理论性质，并且出于这个问题的目的，我们将假设一个数字是Lychrel，直到证明不是这样。另外，对于每万个低于一万的数字，你将得到（i）在不到五十次迭代中成为回文，或者（ii）没有一个，具有所有存在的计算能力，到目前为止已经管理到将它映射到回文结构。事实上，10677是第一个在产生回文之前需要超过50次迭代的数字：4668731596684224866951378664（53次迭代，28位数）。令人惊讶的是，有一些回文数字本身就是Lychrel数字;第一个例子是4994.有多少Lychrel数字在`num`以下？注：2007年4月24日略微修改了措辞，以强调Lychrel数的理论性质。
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# --hints--
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`countLychrelNumbers(1000)`应该返回13。
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```js
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assert.strictEqual(countLychrelNumbers(1000), 13);
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```
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`countLychrelNumbers(5000)`应该返回76。
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```js
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assert.strictEqual(countLychrelNumbers(5000), 76);
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```
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`countLychrelNumbers(10000)`应该返回249。
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```js
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assert.strictEqual(countLychrelNumbers(10000), 249);
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```
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你的函数应该计算所有Lychrel数。
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```js
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assert.strictEqual(countLychrelNumbers(3243), 39);
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```
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您的函数应该通过所有测试用例。
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```js
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assert.strictEqual(countLychrelNumbers(7654), 140);
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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 countLychrelNumbers(num) {
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return true;
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}
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countLychrelNumbers(10000);
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```
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# --solutions--
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```js
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const countLychrelNumbers = (size) => {
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const numReverse = (num) => {
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return Number(num.toString().split('').reverse().join(''));
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};
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const isPalin = (num) => {
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if (numReverse(num) === num) {
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return true;
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}
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return false;
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};
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let total = 0;
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for (let i = 1; i < size; i++) {
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let loopCount = 1;
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let sum = i;
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while (loopCount < 50) {
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sum = sum + numReverse(sum);
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if (isPalin(sum)) {
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break;
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} else {
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loopCount++;
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}
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}
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if (loopCount === 50) {
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total++;
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}
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}
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return total;
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}
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```
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