# 给定范围内具有相等元素的索引数 > 原文： [https://www.geeksforgeeks.org/number-indexes-equal-elements-given-range/](https://www.geeksforgeeks.org/number-indexes-equal-elements-given-range/) 给定`N`个数字和`Q`个查询，每个查询都由`L`和`R`组成，任务是找到这样的整数`i`（`L <= i < R`）的数量，以使`A[i] = A[i+1]`。 考虑基于 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 ``` **朴素的方法**是从`L`遍历到`R`（不包括`R`）并分别针对每个查询，计算满足条件`A[i] = A[i+1]`的索引`i`的数量。 以下是朴素方法的实现： ## 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; } ```