diff --git a/content/posts/leetcode.md b/content/posts/leetcode.md
index 702e78f..1377c3d 100644
--- a/content/posts/leetcode.md
+++ b/content/posts/leetcode.md
@@ -5348,3 +5348,81 @@ func maximumHappinessSum(happiness []int, k int) int64 {
     return int64(result)
 }
 ```
+
+## day73 2024-05-10
+
+### 786. K-th Smallest Prime Fraction
+
+You are given a sorted integer array arr containing 1 and prime numbers, where all the integers of arr are unique. You are also given an integer k.
+
+For every i and j where 0 <= i < j < arr.length, we consider the fraction arr[i] / arr[j].
+
+Return the kth smallest fraction considered. Return your answer as an array of integers of size 2, where answer[0] == arr[i] and answer[1] == arr[j].
+
+![0510L1MMRZ5XYJzh](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0510L1MMRZ5XYJzh.png)
+
+### 题解
+
+保存每个分数的值, 和对应的分子分母, 对对应的分数值排序, 取第k小的即可
+
+### 代码
+
+```go
+
+func kthSmallestPrimeFraction(arr []int, k int) []int {
+    type point struct{
+        value float32
+        num []int
+    }
+    points := []point{}
+    length := len(arr)
+    i := 0
+    for index, value := range arr{
+        float32val := float32(value)
+        for i=index+1;i<length;i++{
+            points = append(points, point{float32val/float32(arr[i]),[]int{value, arr[i]}})
+        }
+    }
+    sort.Slice(points, func(i, j int)bool{return points[i].value < points[j].value})
+    return points[k-1].num
+}
+```
+
+### 总结
+
+看到了一种很有趣的解法, 使用二分法, 每次计算出中间值右侧的分数的个数, 根据与k的相对大小更新中间值. 代码如下, 这种方法比其他许多解法快的多得多, 在最快的平均时长为694ms的情况下竟然只需用时4ms. 这种解法关键在于其通过一开始取中间值为0.5的方式大大减少了需要遍历的分数的数目. 而且没有遍历的分数在后续因为中间值的调整也不会再遍历了, 也就是说每一次遍历的数目都是原本遍历的分数个数的一小部分, 如果数量特别多的情况下可以近似的认为分数的值分布在0-0.5和0.5-1之间的分数个数大致相同. 这样从期望角度讲每次都只需要遍历一般个数的分数, 这样遍历过程总体的时间复杂度为O(nlogn), 并且不需要排序, 最终可以直接返回结果, 因此能有很高的效率.
+
+```go
+func kthSmallestPrimeFraction(arr []int, k int) []int {
+    n := len(arr)
+    left := 0.0
+    right := 1.0
+    result := make([]int, 2)
+
+    for left < right {
+        mid := (left + right) / 2
+        count := 0
+        maxFraction := [2]int{0, 1}
+
+        for i, j := 0, 1; i < n-1; i++ {
+            for j < n && float64(arr[i])/float64(arr[j]) > mid {
+                j++
+            }
+            count += n - j
+            if j < n && float64(arr[i])/float64(arr[j]) > float64(maxFraction[0])/float64(maxFraction[1]) {
+                maxFraction = [2]int{arr[i], arr[j]}
+            }
+        }
+
+        if count == k {
+            return maxFraction[:]
+        } else if count < k {
+            left = mid
+        } else {
+            right = mid
+        }
+    }
+
+    return result
+}
+```