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    <p>If I understand you correctly, you have some sequence of numbers that can either be 1, 2, or 3, where the permutation of them doesn't matter (just the different combinations).</p> <p>That being the case:</p> <pre><code>std::vector&lt;int&gt; v{1, 2, 3}; std::sort(v.begin(), v.end()); </code></pre> <p>That will keep all of the different combinations properly aligned, and you could easily write a loop to test for equality.</p> <p>Alternatively, you could use a <code>std::array&lt;int, N&gt;</code> (where N is the number of possible values - in this case 3).</p> <pre><code>std::array&lt;int, 3&gt; a; </code></pre> <p>Where you would set <code>a[0]</code> equal to the number of <code>1</code>s you have, <code>a[1]</code> equal to the number of '2's, etc.</p> <pre><code>// if your string is 111 a[0] = 3; // if your string is 110 or 011 a[0] = 2; // if your string is 100 or 010 or 001 a[0] = 1; // if your string is 120 a[0] = 1; a[1] = 1; // if your string is 123 a[0] = 1; a[1] = 1; a[2] = 1; </code></pre> <p>If you are looking to store it in a single 32-bit integer:</p> <pre><code>unsigned long x = 1; // number of 1's in your string unsigned long y = 1; // number of 2's in your string unsigned long z = 1; // number of 3's in your string unsigned long result = x | y &lt;&lt; 8 | z &lt;&lt; 16; </code></pre> <p>To retrieve the number of each, you would do</p> <pre><code>unsigned long x = result &amp; 0x000000FF; unsigned long y = (result &gt;&gt; 8) &amp; 0x000000FF; unsigned long z = (result &gt;&gt; 16) &amp; 0x000000FF; </code></pre> <p>This is very similar to what happens in the <code>RBG</code> macros.</p>
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