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leetcode/problem/count-distinct-numbers-on-board.html
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<p>You are given a positive integer <code>n</code>, that is initially placed on a board. Every day, for <code>10<sup>9</sup></code> days, you perform the following procedure:</p>
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<ul>
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<li>For each number <code>x</code> present on the board, find all numbers <code>1 <= i <= n</code> such that <code>x % i == 1</code>.</li>
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<li>Then, place those numbers on the board.</li>
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</ul>
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<p>Return<em> the number of <strong>distinct</strong> integers present on the board after</em> <code>10<sup>9</sup></code> <em>days have elapsed</em>.</p>
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<p><strong>Note:</strong></p>
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<ul>
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<li>Once a number is placed on the board, it will remain on it until the end.</li>
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<li><code>%</code> stands for the modulo operation. For example, <code>14 % 3</code> is <code>2</code>.</li>
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</ul>
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<p> </p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> n = 5
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<strong>Output:</strong> 4
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<strong>Explanation:</strong> Initially, 5 is present on the board.
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The next day, 2 and 4 will be added since 5 % 2 == 1 and 5 % 4 == 1.
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After that day, 3 will be added to the board because 4 % 3 == 1.
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At the end of a billion days, the distinct numbers on the board will be 2, 3, 4, and 5.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> n = 3
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<strong>Output:</strong> 2
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<strong>Explanation:</strong>
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Since 3 % 2 == 1, 2 will be added to the board.
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After a billion days, the only two distinct numbers on the board are 2 and 3.
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 <= n <= 100</code></li>
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</ul>
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