<p>You are given a sorted integer array <code>arr</code> containing <code>1</code> and <strong>prime</strong> numbers, where all the integers of <code>arr</code> are unique. You are also given an integer <code>k</code>.</p>

<p>For every <code>i</code> and <code>j</code> where <code>0 &lt;= i &lt; j &lt; arr.length</code>, we consider the fraction <code>arr[i] / arr[j]</code>.</p>

<p>Return <em>the</em> <code>k<sup>th</sup></code> <em>smallest fraction considered</em>. Return your answer as an array of integers of size <code>2</code>, where <code>answer[0] == arr[i]</code> and <code>answer[1] == arr[j]</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [1,2,3,5], k = 3
<strong>Output:</strong> [2,5]
<strong>Explanation:</strong> The fractions to be considered in sorted order are:
1/5, 1/3, 2/5, 1/2, 3/5, and 2/3.
The third fraction is 2/5.
</pre>

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

<pre>
<strong>Input:</strong> arr = [1,7], k = 1
<strong>Output:</strong> [1,7]
</pre>

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

<ul>
	<li><code>2 &lt;= arr.length &lt;= 1000</code></li>
	<li><code>1 &lt;= arr[i] &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>arr[0] == 1</code></li>
	<li><code>arr[i]</code> is a <strong>prime</strong> number for <code>i &gt; 0</code>.</li>
	<li>All the numbers of <code>arr</code> are <strong>unique</strong> and sorted in <strong>strictly increasing</strong> order.</li>
	<li><code>1 &lt;= k &lt;= arr.length * (arr.length - 1) / 2</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Can you solve the problem with better than <code>O(n<sup>2</sup>)</code> complexity?
// Tag: Heap, 编程语言

// Create Date: 2024/5/10
// Review Date: [2024/5/17, 2024/6/11, 2024/7/11, 2024/8/12] 

// #include <bits/stdc++.h>

using namespace std;

// 不需要visited变量。优化见法2
class Solution {
public:
    vector<int> kthSmallestPrimeFraction(vector<int>& arr, int k) {
        using ele_t = tuple<double, int, int>;  // (div, i, j)
        auto get_key = [](const int i, const int j) -> string
                       {
                            return to_string(i) + "," + to_string(j);
                       };

        priority_queue<ele_t, vector<ele_t>, greater<ele_t>> pq;

        const int n = arr.size();
        // ! 如果写成arr.front() / arr.back() * 1.0那么结果还是会略掉小数部分
        pq.push({arr.front() * 1.0 / arr.back(), 0, n - 1});
        unordered_set<string> visited;
        visited.insert(get_key(0, n - 1));

        int cnt = 1;
        while (cnt <= k)
        {
            auto tmp = pq.top();
            pq.pop();

            const int i = get<1>(tmp);
            const int j = get<2>(tmp);

            if (cnt == k)
            {
                return {arr[i], arr[j]};
            }

            const string key1 = get_key(i + 1, j);
            if (i + 1 < j && visited.find(key1) == visited.end())
            {
                visited.insert(key1);
                pq.push({arr[i + 1] * 1.0 / arr[j], i + 1, j});
            }

            const string key2 = get_key(i, j - 1);
            if (i < j - 1 && visited.find(key2) == visited.end())
            {
                visited.insert(key2);
                pq.push({arr[i] * 1.0 / arr[j - 1], i, j - 1});
            }
            
            ++cnt;
        }

        return {0, 0};
    }
};
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