# 給定范圍內具有相等元素的索引數 > 原文： [https://www.geeksforgeeks.org/number-indexes-equal-elements-given-range/](https://www.geeksforgeeks.org/number-indexes-equal-elements-given-range/) 給定`N`個數字和`Q`個查詢，每個查詢都由`L`和`R`組成，任務是找到這樣的整數`i`（`L <= i < R`）的數量，以使`A[i] = A[i+1]`。 考慮基于 0 的索引。 **示例**： ``` Input : A = [1, 2, 2, 2, 3, 3, 4, 4, 4] Q = 2 L = 1 R = 8 L = 0 R = 4 Output : 5 2 Explanation: We have 5 index i which has Ai=Ai+1 in range [1, 8). We have 2 indexes i which have Ai=Ai+1 in range [0, 4). Input :A = [3, 3, 4, 4] Q = 2 L = 0 R = 3 L = 2 R = 3 Output : 2 1 ``` **樸素的方法**是從`L`遍歷到`R`（不包括`R`）并分別針對每個查詢，計算滿足條件`A[i] = A[i+1]`的索引`i`的數量。 以下是樸素方法的實現： ## C++ ```cpp // CPP program to count the number of indexes // in range L R such that Ai = Ai+1 #include <bits/stdc++.h> using namespace std; // function that answers every query in O(r-l) int answer_query(int a[], int n, int l, int r) { ????// traverse from l to r and count ????// the required indexes ????int count = 0; ????for (int i = l; i < r; i++) ????????if (a[i] == a[i + 1]) ????????????count += 1; ????return count; } // Driver Code int main() { ????int a[] = { 1, 2, 2, 2, 3, 3, 4, 4, 4 }; ????int n = sizeof(a) / sizeof(a[0]); ????// 1-st query ????int L, R; ????L = 1; ????R = 8; ????cout << answer_query(a, n, L, R) << endl; ????// 2nd query ????L = 0; ????R = 4; ????cout << answer_query(a, n, L, R) << endl; ????return 0; } ```