# Count elements which divide all numbers in range L-R

Given N numbers and Q queries, each query consists of L and R. Task is to write a program which prints the count of numbers which divides all numbers in the given range L-R.

Examples :

```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
```

Naive approach : Iterate from range L-R for every query and check if the given element at index-i divide all the numbers in the range. We keep a count for of all the elements which divides all the numbers. The complexity of every query at worst case will be O(n2).

Below is the implementation of Naive Approach :

## C++

 `// CPP program to Count elements which ` `// divides all numbers in range L-R ` `#include ` `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); ` ` `  `    ``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; ` `} `

## Java

 `// Java program to Count elements which ` `// divides all numbers in range L-R ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// function to count element ` `// Time complexity O(n^2) worst case ` `static` `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 ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `a[] = { ``1``, ``2``, ``3``, ``5` `}; ` `    ``int` `n = a.length; ` `     `  `    ``int` `l = ``1``, r = ``4``; ` `    ``System.out.println( answerQuery(a, n, l, r)); ` `     `  `    ``l = ``2``; r = ``4``;  ` `    ``System.out.println( answerQuery(a, n, l, r)); ` `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

## Python3

 `# Python 3 program to Count elements which ` `# divides all numbers in range L-R ` ` `  `# function to count element ` `# Time complexity O(n^2) worst case ` `def` `answerQuery(a, n, l, r): ` `     `  `    ``# answer for query ` `    ``count ``=` `0` ` `  `    ``# 0 based index ` `    ``l ``=` `l ``-` `1` ` `  `    ``# iterate for all elements ` `    ``for` `i ``in` `range``(l, r, ``1``): ` `        ``element ``=` `a[i] ` `        ``divisors ``=` `0` ` `  `        ``# check if the element divides ` `        ``# all numbers in range ` `        ``for` `j ``in` `range``(l, r, ``1``): ` `             `  `            ``# no of elements ` `            ``if` `(a[j] ``%` `a[i] ``=``=` `0``): ` `                ``divisors ``+``=` `1` `            ``else``: ` `                ``break` `         `  `        ``# if all elements are divisible ` `        ``# by a[i] ` `        ``if` `(divisors ``=``=` `(r ``-` `l)): ` `            ``count ``+``=` `1` ` `  `    ``# answer for every query ` `    ``return` `count ` ` `  `# Driver Code ` `if` `__name__ ``=``=``'__main__'``: ` `    ``a ``=` `[``1``, ``2``, ``3``, ``5``] ` `    ``n ``=` `len``(a) ` ` `  `    ``l ``=` `1` `    ``r ``=` `4` `    ``print``(answerQuery(a, n, l, r)) ` ` `  `    ``l ``=` `2` `    ``r ``=` `4` `    ``print``(answerQuery(a, n, l, r)) ` ` `  `# This code is contributed by ` `# Shashank_Sharma `

## C#

 `// C# program to Count elements which ` `// divides all numbers in range L-R ` `using System; ` ` `  `class` `GFG  ` `{ ` ` `  `// function to count element ` `// Time complexity O(n^2) worst case ` `static` `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 ` `public` `static` `void` `Main ()  ` `{ ` `    ``int` `[]a = { ``1``, ``2``, ``3``, ``5` `}; ` `    ``int` `n = a.Length; ` `     `  `    ``int` `l = ``1``, r = ``4``; ` `    ``Console.WriteLine(answerQuery(a, n, l, r)); ` `     `  `    ``l = ``2``; r = ``4``;  ` `    ``Console.WriteLine(answerQuery(a, n, l, r)); ` `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

## PHP

 `

Output:

```1
0
```

Efficient approach : Use Segment Trees to solve this problem. If an element divides all the numbers in a given range, then the element is the minimum number in that range and it is the gcd of all elements in the given range L-R. So the count of the number of minimums in range L-R, given that minimum is equal to the gcd of that range will be our answer to every query. The problem boils down to finding the GCD, MINIMUM and countMINIMUM for every range using Segment trees. On every node of the tree, three values are stored.

On querying for a given range, if the gcd and minimum of the given range are equal, countMINIMUM is returned as the answer. If they are unequal, 0 is returned as the answer.

Below is the implementation of efficient approach :

 `// CPP program to Count elements ` `// which divides all numbers in ` `// range L-R efficient approach ` `#include ` `using` `namespace` `std; ` ` `  `#define N 100005 ` ` `  `// predefines the tree with nodes ` `// storing gcd, min and count ` `struct` `node  ` `{ ` `    ``int` `gcd; ` `    ``int` `min; ` `    ``int` `cnt; ` `} tree[5 * N]; ` ` `  `// function to construct the tree ` `void` `buildtree(``int` `low, ``int` `high,  ` `               ``int` `pos, ``int` `a[]) ` `{ ` `    ``// base condition ` `    ``if` `(low == high)  ` `    ``{ ` `        ``// initially always gcd and min ` `        ``// are same at leaf node ` `        ``tree[pos].min =  tree[pos].gcd = a[low]; ` `        ``tree[pos].cnt = 1; ` `         `  `        ``return``; ` `    ``} ` ` `  `    ``int` `mid = (low + high) >> 1; ` `     `  `    ``// left-subtree ` `    ``buildtree(low, mid, 2 * pos + 1, a); ` ` `  `    ``// right-subtree ` `    ``buildtree(mid + 1, high, 2 * pos + 2, a); ` ` `  `    ``// finds gcd of left and rigth subtree ` `    ``tree[pos].gcd = __gcd(tree[2 * pos + 1].gcd,  ` `                      ``tree[2 * pos + 2].gcd); ` ` `  `    ``// left subtree has the minimum element ` `    ``if` `(tree[2 * pos + 1].min < tree[2 * pos + 2].min)  ` `    ``{ ` `        ``tree[pos].min = tree[2 * pos + 1].min; ` `        ``tree[pos].cnt = tree[2 * pos + 1].cnt; ` `    ``} ` `     `  `    ``// right subtree has the minimum element ` `    ``else`  `    ``if` `(tree[2 * pos + 1].min > tree[2 * pos + 2].min)  ` `    ``{ ` `        ``tree[pos].min = tree[2 * pos + 2].min; ` `        ``tree[pos].cnt = tree[2 * pos + 2].cnt; ` `    ``} ` `     `  `    ``// both subtree has the same minimum element ` `    ``else`  `    ``{ ` `        ``tree[pos].min = tree[2 * pos + 1].min; ` `        ``tree[pos].cnt = tree[2 * pos + 1].cnt +  ` `                        ``tree[2 * pos + 2].cnt; ` `    ``} ` `} ` ` `  `// function that answers every query ` `node query(``int` `s, ``int` `e, ``int` `low, ``int` `high, ``int` `pos) ` `{ ` `    ``node dummy; ` `     `  `    ``// out of range ` `    ``if` `(e < low or s > high)  ` `    ``{ ` `        ``dummy.gcd = dummy.min = dummy.cnt = 0; ` `        ``return` `dummy; ` `    ``} ` `     `  `    ``// in range ` `    ``if` `(s >= low and e <= high)  ` `    ``{ ` `        ``node dummy; ` `        ``dummy.gcd = tree[pos].gcd; ` `        ``dummy.min = tree[pos].min; ` `        ``if` `(dummy.gcd != dummy.min) ` `            ``dummy.cnt = 0; ` `        ``else` `            ``dummy.cnt = tree[pos].cnt; ` `         `  `        ``return` `dummy; ` `    ``} ` ` `  `    ``int` `mid = (s + e) >> 1; ` `     `  `    ``// left-subtree ` `    ``node ans1 = query(s, mid, low,  ` `                ``high, 2 * pos + 1); ` `     `  `    ``// right-subtree ` `    ``node ans2 = query(mid + 1, e, low,  ` `                   ``high, 2 * pos + 2); ` ` `  `    ``node ans; ` `     `  `    ``// when both left subtree and ` `    ``// right subtree is in range ` `    ``if` `(ans1.gcd and ans2.gcd)  ` `    ``{ ` `        ``// merge two trees ` `        ``ans.gcd = __gcd(ans1.gcd, ans2.gcd); ` `        ``ans.min = min(ans1.min, ans2.min); ` ` `  `        ``// when gcd is not equal to min ` `        ``if` `(ans.gcd != ans.min)          ` `            ``ans.cnt = 0;         ` `        ``else`  `        ``{ ` `            ``// add count when min is ` `            ``// same of both subtree ` `            ``if` `(ans1.min == ans2.min)              ` `                ``ans.cnt = ans2.cnt + ans1.cnt;             ` `             `  `            ``// store the minimal's count ` `            ``else`  `            ``if` `(ans1.min < ans2.min)              ` `                ``ans.cnt = ans1.cnt;             ` `            ``else`              `                ``ans.cnt = ans2.cnt;             ` `        ``} ` `         `  `        ``return` `ans; ` `    ``} ` `     `  `    ``// only left subtree is in range ` `    ``else` `if` `(ans1.gcd) ` `        ``return` `ans1; ` `     `  `    ``// only right subtree is in range ` `    ``else` `if` `(ans2.gcd) ` `        ``return` `ans2; ` `} ` ` `  `// function to answer query in range l-r ` `int` `answerQuery(``int` `a[], ``int` `n, ``int` `l, ``int` `r) ` `{ ` `    ``// calls the function which returns ` `    ``// a node this function returns the ` `    ``// count which will be the answer ` `    ``return` `query(0, n - 1, l - 1, r - 1, 0).cnt; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `a[] = { 3, 4, 2, 2, 4, 6 }; ` ` `  `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` `    ``buildtree(0, n - 1, 0, a); ` `    ``int` `l = 1, r = 4; ` ` `  `    ``// answers 1-st query ` `    ``cout << answerQuery(a, n, l, r) << endl; ` ` `  `    ``l = 2, r = 6; ` `    ``// answers 2nd query ` `    ``cout << answerQuery(a, n, l, r) << endl; ` `    ``return` `0; ` `} `

Output:

```0
2
```

Time Complexity: Time Complexity for tree construction is O(n logn) since tree construction takes O(n) and finding out gcd takes O(log n). The time taken for every query in worst case will be O(log n * log n) since the inbuilt function __gcd takes O(log n)

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