feat(curriculum): restore seed + solution to Chinese (#40683)
* 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>
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@ -3,11 +3,12 @@ id: 59713bd26bdeb8a594fb9413
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title: 计算硬币
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challengeType: 5
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videoUrl: ''
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dashedName: count-the-coins
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---
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# --description--
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<p> <a href='https://en.wikipedia.org/wiki/United_States' title='链接:https://en.wikipedia.org/wiki/United_States'>美国</a>货币有四种常见硬币: </p>季度(25美分)硬币(10美分)镍(5美分)和便士(1美分) <p>有六种方法可以换15美分: </p>一角钱和一角钱一角钱和5便士3镍2镍和5便士一镍和10便士15便士任务: <p>实现一个功能,以确定使用这些普通硬币改变一美元的方式有多少? (1美元= 100美分)。 </p>参考: <a href='http://mitpress.mit.edu/sicp/full-text/book/book-Z-H-11.html#%_sec_Temp_52' title='链接:http://mitpress.mit.edu/sicp/full-text/book/book-Z-H-11.html#%_sec_Temp_52'>麻省理工学院出版社的算法</a> 。
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<p> <a href='https://en.wikipedia.org/wiki/United_States' title='链接:https://en.wikipedia.org/wiki/United_States'>美国</a>货币有四种常见硬币: </p>季度(25美分)硬币(10美分)镍(5美分)和便士(1美分) <p>有六种方法可以换15美分: </p>一角钱和一角钱一角钱和5便士3镍2镍和5便士一镍和10便士15便士任务: <p>实现一个功能,以确定使用这些普通硬币改变一美元的方式有多少? (1美元= 100美分)。 </p>参考: <a href='http://mitpress.mit.edu/sicp/full-text/book/book-Z-H-11.html#%_sec_Temp_52' title='链接:http://mitpress.mit.edu/sicp/full-text/book/book-Z-H-11.html#%_sec_Temp_52'>麻省理工学院出版社的算法</a> 。
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# --hints--
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@ -23,5 +24,37 @@ assert(typeof countCoins === 'function');
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assert.equal(countCoins(), 242);
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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 countCoins() {
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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 countCoins() {
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let t = 100;
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const operands = [1, 5, 10, 25];
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const targetsLength = t + 1;
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const operandsLength = operands.length;
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t = [1];
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for (let a = 0; a < operandsLength; a++) {
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for (let b = 1; b < targetsLength; b++) {
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// initialise undefined target
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t[b] = t[b] ? t[b] : 0;
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// accumulate target + operand ways
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t[b] += (b < operands[a]) ? 0 : t[b - operands[a]];
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}
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}
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return t[targetsLength - 1];
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}
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```
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