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leetcode-problemset/leetcode/originData/count-good-numbers.json

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{
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"title": "Count Good Numbers",
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"content": "<p>A digit string is <strong>good</strong> if the digits <strong>(0-indexed)</strong> at <strong>even</strong> indices are <strong>even</strong> and the digits at <strong>odd</strong> indices are <strong>prime</strong> (<code>2</code>, <code>3</code>, <code>5</code>, or <code>7</code>).</p>\n\n<ul>\n\t<li>For example, <code>&quot;2582&quot;</code> is good because the digits (<code>2</code> and <code>8</code>) at even positions are even and the digits (<code>5</code> and <code>2</code>) at odd positions are prime. However, <code>&quot;3245&quot;</code> is <strong>not</strong> good because <code>3</code> is at an even index but is not even.</li>\n</ul>\n\n<p>Given an integer <code>n</code>, return <em>the <strong>total</strong> number of good digit strings of length </em><code>n</code>. Since the answer may be large, <strong>return it modulo </strong><code>10<sup>9</sup> + 7</code>.</p>\n\n<p>A <strong>digit string</strong> is a string consisting of digits <code>0</code> through <code>9</code> that may contain leading zeros.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 1\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> The good numbers of length 1 are &quot;0&quot;, &quot;2&quot;, &quot;4&quot;, &quot;6&quot;, &quot;8&quot;.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 4\n<strong>Output:</strong> 400\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 50\n<strong>Output:</strong> 564908303\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 10<sup>15</sup></code></li>\n</ul>\n",
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"code": "defmodule Solution do\n @spec count_good_numbers(n :: integer) :: integer\n def count_good_numbers(n) do\n \n end\nend",
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"hints": [
"Is there a formula we can use to find the count of all the good numbers?",
"Exponentiation can be done very fast if we looked at the binary bits of n."
],
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