# 计算将`L-R`范围内的所有数字相除的元素 > 原文： [https://www.geeksforgeeks.org/count-elements-which-divide-all-numbers-in-range-l-r/](https://www.geeksforgeeks.org/count-elements-which-divide-all-numbers-in-range-l-r/) 给定`N`个数字和`Q`个查询，每个查询都由`L`和`R`组成。任务是编写一个打印数字计数的程序，该程序将给定范围`L-R`中的所有数字相除。 **示例**： ``` Input : a = {3, 4, 2, 2, 4, 6} Q = 2 L = 1 R = 4 L = 2 R = 6 Output : 0 2 Explanation : The range 1-4 has {3, 4, 2, 2} which does not have any number that divides all the numbers in this range. The range 2-6 has {4, 2, 2, 4, 6} which has 2 numbers {2, 2} which divides all numbers in the given range. Input: a = {1, 2, 3, 5} Q = 2 L = 1 R = 4 L = 2 R = 4 Output: 1 0 ``` **朴素的方法**：针对每个查询从范围`L-R`进行迭代，并检查索引`i`处的给定元素是否将范围中的所有数字相除。 我们对所有元素进行计数，将所有数字相除。 在最坏情况下，每个查询的复杂度将为`O(n^2)`。 **以下是朴素方法的实现**： ## C++ ```cpp // CPP program to Count elements which // divides all numbers in range L-R #include <bits/stdc++.h> using namespace std; // function to count element // Time complexity O(n^2) worst case int answerQuery(int a[], int n,                  int l, int r) {     // answer for query     int count = 0;     // 0 based index     l = l - 1;     // iterate for all elements     for (int i = l; i < r; i++)      {         int element = a[i];         int divisors = 0;         // check if the element divides         // all numbers in range         for (int j = l; j < r; j++)          {             // no of elements             if (a[j] % a[i] == 0)                 divisors++;             else                 break;         }         // if all elements are divisible by a[i]         if (divisors == (r - l))             count++;     }     // answer for every query     return count; } // Driver Code int main() {     int a[] = { 1, 2, 3, 5 };     int n = sizeof(a) / sizeof(a[0]);     int l = 1, r = 4;     cout << answerQuery(a, n, l, r) << endl;     l = 2, r = 4;         cout << answerQuery(a, n, l, r) << endl;     return 0; } ```