# 數組元素的頻率范圍查詢 > 原文： [https://www.geeksforgeeks.org/range-queries-for-frequencies-of-array-elements/](https://www.geeksforgeeks.org/range-queries-for-frequencies-of-array-elements/) 給定`n`個非負整數的數組。 任務是在`array[]`的任意范圍內查找特定元素的頻率。 范圍以數組中的位置（不是基于 0 的索引）的形式給出。 可以有多個給定類型的查詢。 **示例**： ``` Input : arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11}; left = 2, right = 8, element = 8 left = 2, right = 5, element = 6 Output : 3 1 The element 8 appears 3 times in arr[left-1..right-1] The element 6 appears 1 time in arr[left-1..right-1] ``` **樸素的方法**：是從左到右遍歷并在找到元素時更新計數變量。 以下是樸素方法的代碼：- ## C++ ```cpp // C++ program to find total count of an element // in a range #include<bits/stdc++.h> using namespace std; // Returns count of element in arr[left-1..right-1] int findFrequency(int arr[], int n, int left, ?????????????????????????int right, int element) { ????int count = 0; ????for (int i=left-1; i<=right; ++i) ????????if (arr[i] == element) ????????????++count; ????return count; } // Driver Code int main() { ????int arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11}; ????int n = sizeof(arr) / sizeof(arr[0]); ????// Print frequency of 2 from position 1 to 6 ????cout << "Frequency of 2 from 1 to 6 = " ?????????<< findFrequency(arr, n, 1, 6, 2) << endl; ????// Print frequency of 8 from position 4 to 9 ????cout << "Frequency of 8 from 4 to 9 = " ?????????<< findFrequency(arr, n, 4, 9, 8); ????return 0; } ```