Given Q queries, of type: **L R**, for each query you must print the maximum number of divisors that a number **x (L <= x <= R)** has.

Examples:

L = 1 R = 10: 1 has 1 divisor. 2 has 2 divisors. 3 has 2 divisors. 4 has 3 divisors. 5 has 2 divisors. 6 has 4 divisors. 7 has 2 divisors. 8 has 4 divisors. 9 has 3 divisors. 10 has 4 divisors. So the answer for above query is 4, as it is the maximum number of divisors a number has in [1, 10].

**Pre-requisites : **Eratosthenes Sieve, Segment Tree

Below are steps to solve the problem.

- Firstly, let’s see how many number of divisors does a number
**n = p**(where p_{1}^{k1}* p_{2}^{k2}* … * p_{n}^{kn}_{1}, p_{2}, …, p_{n}are prime numbers) has; the answer is**(k**. How? For each prime number in the prime factorization, we can have its_{1}+ 1)*(k_{2}+ 1)*…*(k_{n}+ 1)**k**possible powers in a divisor (0, 1, 2,…, k_{i}+ 1_{i}). - Now let’s see how can we find the prime factorization of a number, we firstly build an array,
**smallest_prime[]**, which stores the smallest prime divisor of**i**at**i**index, we divide a number by its smallest prime divisor to obtain a new number (we also have the smallest prime divisor of this new number stored), we keep doing it until the smallest prime of the number changes, when the smallest prime factor of the new number is different from the previous number’s, we have k^{th}_{i}for the i^{th}prime number in the prime factorization of the given number. - Finally, we obtain the number of divisors for all the numbers and store these in a segment tree that maintains the maximum numbers in the segments. We respond to each query by querying the segment tree.

`// A C++ implementation of the above idea to process ` `// queries of finding a number with maximum divisors. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `#define maxn 1000005 ` `#define INF 99999999 ` ` ` `int` `smallest_prime[maxn]; ` `int` `divisors[maxn]; ` `int` `segmentTree[4 * maxn]; ` ` ` `// Finds smallest prime factor of all numbers in ` `// range[1, maxn) and stores them in smallest_prime[], ` `// smallest_prime[i] should contain the smallest prime ` `// that divides i ` `void` `findSmallestPrimeFactors() ` `{ ` ` ` `// Initialize the smallest_prime factors of all ` ` ` `// to infinity ` ` ` `for` `(` `int` `i = 0 ; i < maxn ; i ++ ) ` ` ` `smallest_prime[i] = INF; ` ` ` ` ` `// to be built like eratosthenes sieve ` ` ` `for` `(` `long` `long` `i = 2; i < maxn; i++) ` ` ` `{ ` ` ` `if` `(smallest_prime[i] == INF) ` ` ` `{ ` ` ` `// prime number will have its smallest_prime ` ` ` `// equal to itself ` ` ` `smallest_prime[i] = i; ` ` ` `for` `(` `long` `long` `j = i * i; j < maxn; j += i) ` ` ` ` ` `// if 'i' is the first prime number reaching 'j' ` ` ` `if` `(smallest_prime[j] > i) ` ` ` `smallest_prime[j] = i; ` ` ` `} ` ` ` `} ` `} ` ` ` `// number of divisors of n = (p1 ^ k1) * (p2 ^ k2) ... (pn ^ kn) ` `// are equal to (k1+1) * (k2+1) ... (kn+1) ` `// this function finds the number of divisors of all numbers ` `// in range [1, maxn) and stores it in divisors[] ` `// divisors[i] stores the number of divisors i has ` `void` `buildDivisorsArray() ` `{ ` ` ` `for` `(` `int` `i = 1; i < maxn; i++) ` ` ` `{ ` ` ` `divisors[i] = 1; ` ` ` `int` `n = i, p = smallest_prime[i], k = 0; ` ` ` ` ` `// we can obtain the prime factorization of the number n ` ` ` `// n = (p1 ^ k1) * (p2 ^ k2) ... (pn ^ kn) using the ` ` ` `// smallest_prime[] array, we keep dividing n by its ` ` ` `// smallest_prime until it becomes 1, whilst we check ` ` ` `// if we have need to set k zero ` ` ` `while` `(n > 1) ` ` ` `{ ` ` ` `n = n / p; ` ` ` `k ++; ` ` ` ` ` `if` `(smallest_prime[n] != p) ` ` ` `{ ` ` ` `//use p^k, initialize k to 0 ` ` ` `divisors[i] = divisors[i] * (k + 1); ` ` ` `k = 0; ` ` ` `} ` ` ` ` ` `p = smallest_prime[n]; ` ` ` `} ` ` ` `} ` `} ` ` ` `// builds segment tree for divisors[] array ` `void` `buildSegtmentTree(` `int` `node, ` `int` `a, ` `int` `b) ` `{ ` ` ` `// leaf node ` ` ` `if` `(a == b) ` ` ` `{ ` ` ` `segmentTree[node] = divisors[a]; ` ` ` `return` `; ` ` ` `} ` ` ` ` ` `//build left and right subtree ` ` ` `buildSegtmentTree(2 * node, a, (a + b) / 2); ` ` ` `buildSegtmentTree(2 * node + 1, ((a + b) / 2) + 1, b); ` ` ` ` ` `//combine the information from left ` ` ` `//and right subtree at current node ` ` ` `segmentTree[node] = max(segmentTree[2 * node], ` ` ` `segmentTree[2 *node + 1]); ` `} ` ` ` `//returns the maximum number of divisors in [l, r] ` `int` `query(` `int` `node, ` `int` `a, ` `int` `b, ` `int` `l, ` `int` `r) ` `{ ` ` ` `// If current node's range is disjoint with query range ` ` ` `if` `(l > b || a > r) ` ` ` `return` `-1; ` ` ` ` ` `// If the current node stores information for the range ` ` ` `// that is completely inside the query range ` ` ` `if` `(a >= l && b <= r) ` ` ` `return` `segmentTree[node]; ` ` ` ` ` `// Returns maximum number of divisors from left ` ` ` `// or right subtree ` ` ` `return` `max(query(2 * node, a, (a + b) / 2, l, r), ` ` ` `query(2 * node + 1, ((a + b) / 2) + 1, b,l,r)); ` `} ` ` ` `// driver code ` `int` `main() ` `{ ` ` ` `// First find smallest prime divisors for all ` ` ` `// the numbers ` ` ` `findSmallestPrimeFactors(); ` ` ` ` ` `// Then build the divisors[] array to store ` ` ` `// the number of divisors ` ` ` `buildDivisorsArray(); ` ` ` ` ` `// Build segment tree for the divisors[] array ` ` ` `buildSegtmentTree(1, 1, maxn - 1); ` ` ` ` ` `cout << ` `"Maximum divisors that a number has "` ` ` `<< ` `" in [1, 100] are "` ` ` `<< query(1, 1, maxn - 1, 1, 100) << endl; ` ` ` ` ` ` ` `cout << ` `"Maximum divisors that a number has"` ` ` `<< ` `" in [10, 48] are "` ` ` `<< query(1, 1, maxn - 1, 10, 48) << endl; ` ` ` ` ` ` ` `cout << ` `"Maximum divisors that a number has"` ` ` `<< ` `" in [1, 10] are "` ` ` `<< query(1, 1, maxn - 1, 1, 10) << endl; ` ` ` ` ` `return` `0; ` `} ` |

**Output:**

Maximum divisors that a number has in [1, 100] are 12 Maximum divisors that a number has in [10, 48] are 10 Maximum divisors that a number has in [1, 10] are 4

**Time Complexity:**

For sieve:O(maxn * log(log(maxn)) )For calculating divisors of each number:O(k<_{1}+ k_{2}+ ... + k_{n})O(log(maxn))For querying each range:O(log(maxn))Total:O((maxn + Q) * log(maxn))

This article is contributed by **Saumye Malhotra**. 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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