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 = p1k1 * p2k2 * … * pnkn (where p1, p2, …, pn are prime numbers) has; the answer is (k1 + 1)*(k2 + 1)*…*(kn + 1). How? For each prime number in the prime factorization, we can have its ki + 1 possible powers in a divisor (0, 1, 2,…, ki).
- 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 ith 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 ki for the ith 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.
C++
// 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 99999999int 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 ivoid 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 hasvoid 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[] arrayvoid 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 codeint 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;} |
Java
// Java implementation of the above idea to process// queries of finding a number with maximum divisors.import java.util.*;class GFG{static int maxn = 10005;static int INF = 999999;static int []smallest_prime = new int[maxn];static int []divisors = new int[maxn];static int []segmentTree = new int[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 istatic 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 (int 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 (int 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 hasstatic 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[] arraystatic 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] = Math.max(segmentTree[2 * node], segmentTree[2 *node + 1]);}// returns the maximum number of divisors in [l, r]static 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 Math.max(query(2 * node, a, (a + b) / 2, l, r), query(2 * node + 1, ((a + b) / 2) + 1, b, l, r));}// Driver Codepublic static void main(String[] args) { // 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); System.out.println("Maximum divisors that a number " + "has in [1, 100] are " + query(1, 1, maxn - 1, 1, 100)); System.out.println("Maximum divisors that a number " + "has in [10, 48] are " + query(1, 1, maxn - 1, 10, 48)); System.out.println("Maximum divisors that a number " + "has in [1, 10] are " + query(1, 1, maxn - 1, 1, 10)); }}// This code is contributed by PrinciRaj1992 |
Python 3
# Python 3 implementation of the above # idea to process queries of finding a# number with maximum divisors.maxn = 1000005INF = 99999999smallest_prime = [0] * maxndivisors = [0] * maxnsegmentTree = [0] * (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 idef findSmallestPrimeFactors(): # Initialize the smallest_prime # factors of all to infinity for i in range(maxn ): smallest_prime[i] = INF # to be built like eratosthenes sieve for i in range(2, maxn): if (smallest_prime[i] == INF): # prime number will have its # smallest_prime equal to itself smallest_prime[i] = i for j in range(i * i, maxn , 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 hasdef buildDivisorsArray(): for i in range(1, maxn): divisors[i] = 1 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 += 1 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[] arraydef buildSegtmentTree( node, a, 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]def query(node, a, b, l, r): # If current node's range is disjoint # with query range if (l > b or a > r): return -1 # If the current node stores information # for the range that is completely inside # the query range if (a >= l and 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 codeif __name__ == "__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) print("Maximum divisors that a number has ", " in [1, 100] are ", query(1, 1, maxn - 1, 1, 100)) print("Maximum divisors that a number has", " in [10, 48] are ", query(1, 1, maxn - 1, 10, 48)) print( "Maximum divisors that a number has", " in [1, 10] are ", query(1, 1, maxn - 1, 1, 10))# This code is contributed by ita_c |
C#
// C# implementation of the above idea // to process queries of finding a number// with maximum divisors.using System; class GFG{static int maxn = 10005;static int INF = 999999;static int []smallest_prime = new int[maxn];static int []divisors = new int[maxn];static int []segmentTree = new int[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 istatic 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 (int 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 (int 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 hasstatic 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[] arraystatic 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] = Math.Max(segmentTree[2 * node], segmentTree[2 *node + 1]);}// returns the maximum number of divisors in [l, r]static 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 Math.Max(query(2 * node, a, (a + b) / 2, l, r), query(2 * node + 1, ((a + b) / 2) + 1, b, l, r));}// Driver Codepublic static void Main(String[] args) { // 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); Console.WriteLine("Maximum divisors that a number " + "has in [1, 100] are " + query(1, 1, maxn - 1, 1, 100)); Console.WriteLine("Maximum divisors that a number " + "has in [10, 48] are " + query(1, 1, maxn - 1, 10, 48)); Console.WriteLine("Maximum divisors that a number " + "has in [1, 10] are " + query(1, 1, maxn - 1, 1, 10)); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript implementation of the above idea to process// queries of finding a number with maximum divisors. let maxn = 10005; let INF = 999999; let smallest_prime = new Array(maxn); for(let i=0;i<maxn;i++) { smallest_prime[i]=0; } let divisors = new Array(maxn); for(let i=0;i<maxn;i++) { divisors[i]=0; } let segmentTree = new Array(4 * maxn); for(let i=0;i<4*maxn;i++) { segmentTree[i]=0; } // 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 function findSmallestPrimeFactors() { // Initialize the smallest_prime factors // of all to infinity for (let i = 0 ; i < maxn ; i ++ ) smallest_prime[i] = INF; // to be built like eratosthenes sieve for (let 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 (let 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 function buildDivisorsArray() { for (let i = 1; i < maxn; i++) { divisors[i] = 1; let n = i; let p = smallest_prime[i] let 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 = Math.floor(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 function buildSegtmentTree(node,a,b) { // leaf node if (a == b) { segmentTree[node] = divisors[a]; return ; } //build left and right subtree buildSegtmentTree(2 * node, a, Math.floor((a + b) / 2)); buildSegtmentTree((2 * node) + 1, Math.floor((a + b) / 2) + 1, b); //combine the information from left //and right subtree at current node segmentTree[node] = Math.max(segmentTree[2 * node], segmentTree[(2 *node) + 1]); } // returns the maximum number of divisors in [l, r] function query(node,a,b,l,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 Math.max(query(2 * node, a, Math.floor((a + b) / 2), l, r), query(2 * node + 1, Math.floor((a + b) / 2) + 1, b, l, r)); } // Driver Code // 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); document.write("Maximum divisors that a number " + "has in [1, 100] are " + query(1, 1, maxn - 1, 1, 100)+"<br>"); document.write("Maximum divisors that a number " + "has in [10, 48] are " + query(1, 1, maxn - 1, 10, 48)+"<br>"); document.write("Maximum divisors that a number " + "has in [1, 10] are " + query(1, 1, maxn - 1, 1, 10)+"<br>"); // This code is contributed by avanitrachhadiya2155 </script> |
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: O((maxn + Q) * log(maxn))
- For sieve: O(maxn * log(log(maxn)) )
- For calculating divisors of each number: O(k1 + k2 + … + kn) < O(log(maxn))
- For querying each range: O(log(maxn))
Auxiliary Space: O(n)
Related Topic: Segment Tree
This article is contributed by Saumye Malhotra. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
