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>
97 lines
2.5 KiB
Markdown
97 lines
2.5 KiB
Markdown
---
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id: 594810f028c0303b75339ad6
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title: 勾选村号码表示
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challengeType: 5
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videoUrl: ''
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dashedName: zeckendorf-number-representation
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---
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# --description--
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<p>正如数字可以用位置表示法表示为十（十进制）或二（二进制）的幂的倍数之和;所有正整数都可以表示为Fibonacci系列的不同成员的一次或零次的总和。 </p><p>回想一下，第一六个不同斐波那契数是： <code>1, 2, 3, 5, 8, 13</code> 。十进制数11可以用位置表示法写成<code>0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1</code>或<code>010100</code> ，其中列表示乘以序列的特定成员。前导零被丢弃，因此十进制11变为<code>10100</code> 。 </p><p> 10100不是从斐波那契数字中得到11的唯一方法，但是<code>0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1</code>或010011也代表十进制11。对于真正的Zeckendorf数字还有一个额外的限制，即“不能使用两个连续的Fibonacci数”，这导致了前一个独特的解决方案。 </p><p>任务：编写一个函数，按顺序生成并返回前N个Zeckendorf数的数组。 </p>
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# --hints--
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zeckendorf必须是功能
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```js
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assert.equal(typeof zeckendorf, 'function');
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```
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你的`zeckendorf`函数应该返回正确的答案
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```js
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assert.deepEqual(answer, solution20);
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```
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# --seed--
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## --after-user-code--
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```js
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const range = (m, n) => (
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Array.from({
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length: Math.floor(n - m) + 1
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}, (_, i) => m + i)
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);
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const solution20 = [
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'1', '10', '100', '101', '1000', '1001', '1010', '10000', '10001',
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'10010', '10100', '10101', '100000', '100001', '100010', '100100', '100101',
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'101000', '101001', '101010'
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];
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const answer = range(1, 20).map(zeckendorf);
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```
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## --seed-contents--
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```js
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function zeckendorf(n) {
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}
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```
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# --solutions--
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```js
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// zeckendorf :: Int -> String
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function zeckendorf(n) {
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const f = (m, x) => (m < x ? [m, 0] : [m - x, 1]);
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return (n === 0 ? ([0]) :
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mapAccumL(f, n, reverse(
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tail(fibUntil(n))
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))[1]).join('');
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}
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// fibUntil :: Int -> [Int]
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let fibUntil = n => {
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const xs = [];
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until(
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([a]) => a > n,
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([a, b]) => (xs.push(a), [b, a + b]), [1, 1]
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);
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return xs;
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};
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let mapAccumL = (f, acc, xs) => (
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xs.reduce((a, x) => {
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const pair = f(a[0], x);
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return [pair[0], a[1].concat(pair[1])];
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}, [acc, []])
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);
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let until = (p, f, x) => {
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let v = x;
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while (!p(v)) v = f(v);
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return v;
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};
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const tail = xs => (
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xs.length ? xs.slice(1) : undefined
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);
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const reverse = xs => xs.slice(0).reverse();
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```
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