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>
61 lines
1.4 KiB
Markdown
61 lines
1.4 KiB
Markdown
---
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id: 5900f3871000cf542c50fe9a
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title: 问题27：二次素数
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challengeType: 5
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videoUrl: ''
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dashedName: problem-27-quadratic-primes
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---
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# --description--
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欧拉发现了显着的二次公式：$ n ^ 2 + n + 41 $事实证明，公式将为连续的整数值$ 0 \\ le n \\ le 39 $产生40个素数。但是，当$ n = 40时，40 ^ 2 + 40 + 41 = 40（40 + 1）+ 41 $可被41整除，当然$ n = 41时，41 ^ 2 + 41 + 41 $显然可以被整除41.发现了令人难以置信的公式$ n ^ 2 - 79n + 1601 $，它为连续值$ 0 \\ le n \\ le 79 $产生80个素数。系数-79和1601的乘积是-126479。考虑形式的二次方：
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$ n ^ 2 + an + b $，其中$ | a | <range $和$ | b | \\ le $ $其中$ | n | $是$ n $的模数/绝对值，例如$ | 11 | = 11 $和$ | -4 | = 4 $
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找到系数的乘积，$ a $和$ b $，用于生成连续值$ n $的最大素数数的二次表达式，从$ n = 0 $开始。
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# --hints--
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`quadraticPrimes(200)`应返回-4925。
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```js
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assert(quadraticPrimes(200) == -4925);
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```
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`quadraticPrimes(500)`应返回-18901。
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```js
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assert(quadraticPrimes(500) == -18901);
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```
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`quadraticPrimes(800)`应返回-43835。
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```js
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assert(quadraticPrimes(800) == -43835);
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```
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`quadraticPrimes(1000)`应返回-59231。
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```js
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assert(quadraticPrimes(1000) == -59231);
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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 quadraticPrimes(range) {
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return range;
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}
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quadraticPrimes(1000);
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```
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# --solutions--
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```js
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// solution required
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```
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