Given an array of integers, evaluate queries of the form LCM(l, r). There might be many queries, hence evaluate the queries efficiently.

LCM (l, r) denotes the LCM of array elements that lie between the index l and r (inclusive of both indices) Mathematically, LCM(l, r) = LCM(arr[l], arr[l+1] , ......... , arr[r-1], arr[r])

Examples:

Inputs : Array = {5, 7, 5, 2, 10, 12 ,11, 17, 14, 1, 44} Queries: LCM(2, 5), LCM(5, 10), LCM(0, 10) Outputs: 60 15708 78540 Explanation : In the first query LCM(5, 2, 10, 12) = 60, similarly in other queries.

A naive solution would be to traverse the array for every query and calculate the answer by using,

LCM(a, b) = (a*b) / GCD(a,b)

However as the number of queries can be large, this solution would be impractical.

An efficient solution would be to use segment tree. Recall that in this case, where no update is required, we can build the tree once and can use that repeatedly to answer the queries. Each node in the tree should store the LCM value for that particular segment and we can use the same formula as above to combine the segments. Hence we can answer each query efficiently!

Below is a C++ solution for the same.

`// LCM of given range queries using Segment Tree ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `#define MAX 1000 ` ` ` `// allocate space for tree ` `int` `tree[4*MAX]; ` ` ` `// declaring the array globally ` `int` `arr[MAX]; ` ` ` `// Function to return gcd of a and b ` `int` `gcd(` `int` `a, ` `int` `b) ` `{ ` ` ` `if` `(a == 0) ` ` ` `return` `b; ` ` ` `return` `gcd(b%a, a); ` `} ` ` ` `//utility function to find lcm ` `int` `lcm(` `int` `a, ` `int` `b) ` `{ ` ` ` `return` `a*b/gcd(a,b); ` `} ` ` ` `// Function to build the segment tree ` `// Node starts beginning index of current subtree. ` `// start and end are indexes in arr[] which is global ` `void` `build(` `int` `node, ` `int` `start, ` `int` `end) ` `{ ` ` ` `// If there is only one element in current subarray ` ` ` `if` `(start==end) ` ` ` `{ ` ` ` `tree[node] = arr[start]; ` ` ` `return` `; ` ` ` `} ` ` ` ` ` `int` `mid = (start+end)/2; ` ` ` ` ` `// build left and right segments ` ` ` `build(2*node, start, mid); ` ` ` `build(2*node+1, mid+1, end); ` ` ` ` ` `// build the parent ` ` ` `int` `left_lcm = tree[2*node]; ` ` ` `int` `right_lcm = tree[2*node+1]; ` ` ` ` ` `tree[node] = lcm(left_lcm, right_lcm); ` `} ` ` ` `// Function to make queries for array range )l, r). ` `// Node is index of root of current segment in segment ` `// tree (Note that indexes in segment tree begin with 1 ` `// for simplicity). ` `// start and end are indexes of subarray covered by root ` `// of current segment. ` `int` `query(` `int` `node, ` `int` `start, ` `int` `end, ` `int` `l, ` `int` `r) ` `{ ` ` ` `// Completely outside the segment, returning ` ` ` `// 1 will not affect the lcm; ` ` ` `if` `(end<l || start>r) ` ` ` `return` `1; ` ` ` ` ` `// completely inside the segment ` ` ` `if` `(l<=start && r>=end) ` ` ` `return` `tree[node]; ` ` ` ` ` `// partially inside ` ` ` `int` `mid = (start+end)/2; ` ` ` `int` `left_lcm = query(2*node, start, mid, l, r); ` ` ` `int` `right_lcm = query(2*node+1, mid+1, end, l, r); ` ` ` `return` `lcm(left_lcm, right_lcm); ` `} ` ` ` `//driver function to check the above program ` `int` `main() ` `{ ` ` ` `//initialize the array ` ` ` `arr[0] = 5; ` ` ` `arr[1] = 7; ` ` ` `arr[2] = 5; ` ` ` `arr[3] = 2; ` ` ` `arr[4] = 10; ` ` ` `arr[5] = 12; ` ` ` `arr[6] = 11; ` ` ` `arr[7] = 17; ` ` ` `arr[8] = 14; ` ` ` `arr[9] = 1; ` ` ` `arr[10] = 44; ` ` ` ` ` `// build the segment tree ` ` ` `build(1, 0, 10); ` ` ` ` ` `// Now we can answer each query efficiently ` ` ` ` ` `// Print LCM of (2, 5) ` ` ` `cout << query(1, 0, 10, 2, 5) << endl; ` ` ` ` ` `// Print LCM of (5, 10) ` ` ` `cout << query(1, 0, 10, 5, 10) << endl; ` ` ` ` ` `// Print LCM of (0, 10) ` ` ` `cout << query(1, 0, 10, 0, 10) << endl; ` ` ` ` ` `return` `0; ` `} ` |

Output:

60 15708 78540

This article is contributed by **Ashutosh Kumar**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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