117 lines
2.4 KiB
Markdown
117 lines
2.4 KiB
Markdown
---
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id: 5949b579404977fbaefcd736
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title: 9 billion names of God the integer
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challengeType: 5
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forumTopicId: 302219
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dashedName: 9-billion-names-of-god-the-integer
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---
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# --description--
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This task is a variation of the [short story by Arthur C. Clarke](https://en.wikipedia.org/wiki/The Nine Billion Names of God#Plot_summary "wp: The Nine Billion Names of God#Plot_summary").
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(Solvers should be aware of the consequences of completing this task.)
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In detail, to specify what is meant by a "name":
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<ul>
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<li>The integer 1 has 1 name "1".</li>
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<li>The integer 2 has 2 names "1+1" and "2".</li>
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<li>The integer 3 has 3 names "1+1+1", "2+1", and "3".</li>
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<li>The integer 4 has 5 names "1+1+1+1", "2+1+1", "2+2", "3+1", "4".</li>
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<li>The integer 5 has 7 names "1+1+1+1+1", "2+1+1+1", "2+2+1", "3+1+1", "3+2", "4+1", "5".</li>
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</ul>
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This can be visualized in the following form:
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<pre> 1
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1 1
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1 1 1
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1 2 1 1
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1 2 2 1 1
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1 3 3 2 1 1
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</pre>
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Where row $n$ corresponds to integer $n$, and each column $C$ in row $m$ from left to right corresponds to the number of names beginning with $C$.
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Optionally note that the sum of the $n$-th row $P(n)$ is the integer partition function.
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# --instructions--
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Implement a function that returns the sum of the $n$-th row.
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# --hints--
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`numberOfNames` should be function.
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```js
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assert(typeof numberOfNames === 'function');
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```
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`numberOfNames(5)` should equal 7.
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```js
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assert.equal(numberOfNames(5), 7);
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```
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`numberOfNames(12)` should equal 77.
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```js
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assert.equal(numberOfNames(12), 77);
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```
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`numberOfNames(18)` should equal 385.
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```js
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assert.equal(numberOfNames(18), 385);
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```
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`numberOfNames(23)` should equal 1255.
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```js
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assert.equal(numberOfNames(23), 1255);
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```
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`numberOfNames(42)` should equal 53174.
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```js
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assert.equal(numberOfNames(42), 53174);
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```
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`numberOfNames(123)` should equal 2552338241.
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```js
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assert.equal(numberOfNames(123), 2552338241);
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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 numberOfNames(num) {
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return true;
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}
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```
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# --solutions--
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```js
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function numberOfNames(num) {
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const cache = [
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[1]
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];
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for (let l = cache.length; l < num + 1; l++) {
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let Aa;
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let Mi;
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const r = [0];
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for (let x = 1; x < l + 1; x++) {
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r.push(r[r.length - 1] + (Aa = cache[l - x < 0 ? cache.length - (l - x) : l - x])[(Mi = Math.min(x, l - x)) < 0 ? Aa.length - Mi : Mi]);
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}
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cache.push(r);
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}
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return cache[num][cache[num].length - 1];
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}
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```
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