2022-01-21 01:00:18 +05:30
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---
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id: 5900f3e61000cf542c50fef9
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title: '問題 122: 効率的な累乗法'
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challengeType: 5
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forumTopicId: 301749
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dashedName: problem-122-efficient-exponentiation
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---
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# --description--
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$n^{15}$ の最も単純な計算方法では、14 回の乗算が必要です。
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$$n × n × \ldots × n = n^{15}$$
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しかし、2 進法を使えば 6 回の乗算で計算できます。
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2022-04-02 14:16:30 +05:30
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$$\begin{align} & n × n = n^2\\\\
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& n^2 × n^2 = n^4\\\\ & n^4 × n^4 = n^8\\\\
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& n^8 × n^4 = n^{12}\\\\ & n^{12} × n^2 = n^{14}\\\\
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& n^{14} × n = n^{15} \end{align}$$
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2022-01-21 01:00:18 +05:30
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しかし、わずか 5 回の乗算で計算することも可能です。
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2022-04-02 14:16:30 +05:30
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$$\begin{align} & n × n = n^2\\\\
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& n^2 × n = n^3\\\\ & n^3 × n^3 = n^6\\\\
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& n^6 × n^6 = n^{12}\\\\ & n^{12} × n^3 = n^{15} \end{align}$$
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2022-01-21 01:00:18 +05:30
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ここで、$n^k$ を計算するための最小の乗算回数を $m(k)$ とします。例えば $m(15) = 5$ です。
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$1 ≤ a ≤ 200$ のとき、$\sum{m(k)}$ を求めなさい。
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# --hints--
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`efficientExponentation()` は `1582` を返す必要があります。
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```js
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assert.strictEqual(efficientExponentation(), 1582);
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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 efficientExponentation() {
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return true;
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}
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efficientExponentation();
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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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