# Queries for Count of divisors of product of an Array in given range | Set 2 (MO’s Algorithm)

Given an array arr of size N and Q queries of the form [L, R], the task is to find the number of divisors of the product of this array in the given range.

Examples:

Input: arr[] = {4, 1, 9, 12, 5, 3}, Q = {{1, 3}, {3, 5}}
Output:
9
24

Input: arr[] = {5, 2, 3, 1, 4}, Q = {{2, 4}, {1, 5}}
Output:
4
16

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

The idea of MO’s algorithm is to pre-process all queries so that result of one query can be used in the next query.
Let a[0…n-1] be input array and q[0..m-1] be an array of queries.

1. Sort all queries in a way that queries with L values from 0 to √n – 1 are put together, then all queries from √n to 2×√n – 1, and so on. All queries within a block are sorted in increasing order of R values.
2. Process all queries one by one in a way that every query uses result computed in the previous query. Let ‘result’ be result of previous query
3. A number n can be represented as n = , where ai are prime factors and pi are integral power of them.
So, for this factorization we have formula to find total number of divisor of n and that is: 4. In Add function we increment counter array as i.e counter[a[i]]=counter[a[i]]+ pi. Let ‘prev’ stores the previous value of counter[a[i]]. Now as counter array changes, result changes as:

• result = result / (prev + 1) (Dividing by prev+1 neutralizes the effect of previous pi)
• result = (result × (counter[pi] + 1) (Now the previous result is neutralized so we multiply with the new count i.e counter[a[i]]+1)
5. In Remove function we decrement the counter array as counter[a[i]] = counter[a[i]] – pi. Now as counter array changes, result changes as:

• result = result / (prev + 1) (Dividing by prev+1 neutralizes the effect of previous pi)
• result = (result × (counter[pi] + 1) (Now the previous result is neutralized so we
multiply with the new count i.e counter[a[i]]+1)

Below is the implementation of the above approach

 `// C++ program to Count the divisors  ` `// of product of an Array in range  ` `// L to R for Q queries ` `#include ` `using` `namespace` `std; ` ` `  `#define MAX 1000000 ` `#define MOD 1000000007 ` `#define ll long long int ` ` `  `// Variable to represent block size. ` `// This is made global so compare() ` `// of sort can use it. ` `int` `block; ` ` `  `// Structure to represent a query range ` `struct` `Query { ` `    ``int` `L, R; ` `}; ` ` `  `// Store the prime factor of numbers ` `// till MAX ` `vector > store[MAX + 1]; ` ` `  `// Initialized to store the count  ` `// of prime fators ` `int` `counter[MAX + 1] = {}; ` ` `  `// Result is Initialized to 1 ` `int` `result = 1; ` ` `  `// Inverse array to store ` `// inverse of number from 1 to MAX ` `ll inverse[MAX + 1]; ` ` `  `// Function used to sort all queries so that  ` `// all queries of the same block are arranged  ` `// together and within a block, queries are ` `// sorted in increasing order of R values. ` `bool` `compare(Query x, Query y) ` `{ ` `    ``// Different blocks, sort by block. ` `    ``if` `(x.L / block != y.L / block) ` `        ``return` `x.L / block < y.L / block; ` ` `  `    ``// Same block, sort by R value ` `    ``return` `x.R < y.R; ` `} ` ` `  `// Function to calculate modular ` `// inverse and storing it in Inverse array ` `void` `modularInverse() ` `{ ` ` `  `    ``inverse = inverse = 1; ` `    ``for` `(``int` `i = 2; i <= MAX; i++) ` `        ``inverse[i] = inverse[MOD % i] ` `                    ``* (MOD - MOD / i)  ` `                      ``% MOD; ` `} ` ` `  `// Function to use Sieve to compute ` `// and store prime numbers ` `void` `sieve() ` `{ ` ` `  `    ``store.push_back({ 1, 0 }); ` `    ``for` `(``int` `i = 2; i <= MAX; i++) ` `    ``{ ` `        ``if` `(store[i].size() == 0) ` `        ``{ ` `            ``store[i].push_back({ i, 1 }); ` `             `  `            ``for` `(``int` `j = 2 * i; j <= MAX; j += i) ` `            ``{ ` `                ``int` `cnt = 0; ` `                ``int` `x = j; ` `                ``while` `(x % i == 0) ` `                    ``cnt++, x /= i; ` `                ``store[j].push_back({ i, cnt }); ` `            ``} ` `        ``} ` `    ``} ` `} ` ` `  `// Function to Add elements ` `// of current range ` `void` `add(``int` `currL, ``int` `a[]) ` `{ ` `    ``int` `value = a[currL]; ` `    ``for` `(``auto` `it = store[value].begin(); ` `         ``it != store[value].end(); it++) { ` `        ``// it->first is ai ` `        ``// it->second is its integral power ` `        ``int` `prev = counter[it->first]; ` `        ``counter[it->first] += it->second; ` `        ``result = (result * inverse[prev + 1]) ` `                 ``% MOD; ` `         `  `        ``result = (result * ` `                  ``(counter[it->first] + 1)) ` `                  ``% MOD; ` `    ``} ` `} ` ` `  `// Function to remove elements ` `// of previous range ` `void` `remove``(``int` `currR, ``int` `a[]) ` `{ ` `    ``int` `value = a[currR]; ` `    ``for` `(``auto` `it = store[value].begin();  ` `         ``it != store[value].end(); it++) { ` `        ``// it->first is ai ` `        ``// it->second is its integral power ` `        ``int` `prev = counter[it->first]; ` `        ``counter[it->first] -= it->second; ` `        ``result = (result * inverse[prev + 1]) ` `                  ``% MOD; ` `        ``result = (result * ` `                  ``(counter[it->first] + 1))  ` `                  ``% MOD; ` `    ``} ` `} ` ` `  `// Function to print the answer. ` `void` `queryResults(``int` `a[], ``int` `n, Query q[], ` `                  ``int` `m) ` `{ ` `    ``// Find block size ` `    ``block = (``int``)``sqrt``(n); ` ` `  `    ``// Sort all queries so that queries of  ` `    ``// same blocks are arranged together. ` `    ``sort(q, q + m, compare); ` ` `  `    ``// Initialize current L, current R and ` `    ``// current result ` `    ``int` `currL = 0, currR = 0; ` ` `  `    ``for` `(``int` `i = 0; i < m; i++) { ` `        ``// L and R values of current range ` `        ``int` `L = q[i].L, R = q[i].R; ` ` `  `        ``// Add Elements of current range ` `        ``while` `(currR <= R) { ` `            ``add(currR, a); ` `            ``currR++; ` `        ``} ` `        ``while` `(currL > L) { ` `            ``add(currL - 1, a); ` `            ``currL--; ` `        ``} ` ` `  `        ``// Remove element of previous range ` `        ``while` `(currR > R + 1) ` ` `  `        ``{ ` `            ``remove``(currR - 1, a); ` `            ``currR--; ` `        ``} ` `        ``while` `(currL < L) { ` `            ``remove``(currL, a); ` `            ``currL++; ` `        ``} ` ` `  `        ``cout << result << endl; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Precomputing the prime numbers  ` `    ``// using sieve ` `    ``sieve(); ` ` `  `    ``// Precomputing modular inverse of  ` `    ``// numbers from 1 to MAX ` `    ``modularInverse(); ` ` `  `    ``int` `a[] = { 5, 2, 3, 1, 4 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` `     `  `    ``Query q[] = { { 1, 3 }, { 0, 4 } }; ` `     `  `    ``int` `m = ``sizeof``(q) / ``sizeof``(q); ` `     `  `    ``// Answer the queries ` `    ``queryResults(a, n, q, m); ` `    ``return` `0; ` `} `

Output:

```4
16
```

Time Complexity: O(Q×sqrt(N)) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : Akanksha_Rai