<p>Given an integer <code>n</code>, return <em>a list of all <strong>simplified</strong> fractions between </em><code>0</code><em> and </em><code>1</code><em> (exclusive) such that the denominator is less-than-or-equal-to </em><code>n</code>. You can return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> [&quot;1/2&quot;]
<strong>Explanation:</strong> &quot;1/2&quot; is the only unique fraction with a denominator less-than-or-equal-to 2.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> [&quot;1/2&quot;,&quot;1/3&quot;,&quot;2/3&quot;]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 4
<strong>Output:</strong> [&quot;1/2&quot;,&quot;1/3&quot;,&quot;1/4&quot;,&quot;2/3&quot;,&quot;3/4&quot;]
<strong>Explanation:</strong> &quot;2/4&quot; is not a simplified fraction because it can be simplified to &quot;1/2&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 100</code></li>
</ul>

class Solution {
    int gcd(int x, int y)
    {
        int p, q, t;

        p = std::abs(x);
        q = std::abs(y);
        if (p > q)
            std::swap(p, q);
        if (q <= 1)
            return q == 0 ? p : q;

        do {
            t = q;
            q = p % q;
            p = t;
        } while (q >= 2);

        return q > 0 ? q : p;
    }

public:
    vector<string> simplifiedFractions(int n)
    {
        vector<string>* v;
        int p, q;

        v = new vector<string>();
        for (q = 2; q <= n; q++)
            for (p = 1; p < q; p++)
                if (gcd(p, q) == 1)
                    v->push_back(std::to_string(p) + "/" +
                                 std::to_string(q));

        return *v;
    }
};
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