You are given two numbers a and b (1 <= a,b <= 10^8 ) and n. The task is to find all numbers between a and b inclusively having exactly n distinct prime factors. The solution should be designed in a way that it efficiently handles multiple queries for different values of a and b like in Competitive Programming.
Examples:
Input : a = 1, b = 10, n = 2 Output : 2 // Only 6 = 2*3 and 10 = 2*5 have exactly two // distinct prime factors Input : a = 1, b = 100, n = 3 Output: 8 // only 30 = 2*3*5, 42 = 2*3*7, 60 = 2*2*3*5, 66 = 2*3*11, // 70 = 2*5*7, 78 = 2*3*13, 84 = 2*2*3*7 and 90 = 2*3*3*5 // have exactly three distinct prime factors
This problem is basically application of segmented sieve. As we know that all prime factors of a number are always less than or equal to square root of number i.e; sqrt(n). So we generate all prime numbers less than or equals to 10^8 and store them in an array. Now using this segmented sieve we check each number from a to b to have exactly n prime factors.
C++
// C++ program to find numbers with exactly n distinct// prime factor numbers from a to b#include<bits/stdc++.h>using namespace std;// Stores all primes less than and equals to sqrt(10^8) = 10000vector <int> primes;// Generate all prime numbers less than or equals to sqrt(10^8)// = 10000 using sieve of sundaramvoid segmentedSieve(){ int n = 10000; // Square root of 10^8 // In general Sieve of Sundaram, produces primes smaller // than (2*x + 2) for a number given number x. // Since we want primes smaller than n=10^4, we reduce // n to half int nNew = (n-2)/2; // This array is used to separate numbers of the form // i+j+2ij from others where 1 <= i <= j bool marked[nNew + 1]; // Initialize all elements as not marked memset(marked, false, sizeof(marked)); // Main logic of Sundaram. Mark all numbers of the // form i + j + 2ij as true where 1 <= i <= j for (int i=1; i<=nNew; i++) for (int j=i; (i + j + 2*i*j) <= nNew; j++) marked[i + j + 2*i*j] = true; // Since 2 is a prime number primes.push_back(2); // Remaining primes are of the form 2*i + 1 such that // marked[i] is false. for (int i=1; i<=nNew; i++) if (marked[i] == false) primes.push_back(2*i+1);}// Function to count all numbers from a to b having exactly// n prime factorsint Nfactors(int a, int b, int n){ segmentedSieve(); // result --> all numbers between a and b having // exactly n prime factors int result = 0; // check for each number for (int i=a; i<=b; i++) { // tmp --> stores square root of current number because // all prime factors are always less than and // equal to square root of given number // copy --> it holds the copy of current number int tmp = sqrt(i), copy = i; // count --> it counts the number of distinct prime // factors of number int count = 0; // check divisibility of 'copy' with each prime less // than 'tmp' and divide it until it is divisible by // current prime factor for (int j=0; primes[j]<=tmp; j++) { if (copy%primes[j]==0) { // increment count for distinct prime count++; while (copy%primes[j]==0) copy = copy/primes[j]; } } // if number is completely divisible then at last // 'copy' will be 1 else 'copy' will be prime, so // increment count by one if (copy != 1) count++; // if number has exactly n distinct primes then // increment result by one if (count==n) result++; } return result;}// Driver program to run the caseint main(){ int a = 1, b = 100, n = 3; cout << Nfactors(a, b, n); return 0;} |
Java
// Java program to find numbers with exactly n distinct// prime factor numbers from a to bimport java.util.*;class GFG{ // Stores all primes less than and // equals to sqrt(10^8) = 10000static ArrayList<Integer> primes = new ArrayList<Integer>();// Generate all prime numbers less // than or equals to sqrt(10^8)// = 10000 using sieve of sundaramstatic void segmentedSieve(){ int n = 10000; // Square root of 10^8 // In general Sieve of Sundaram, // produces primes smaller // than (2*x + 2) for a number // given number x. Since we want // primes smaller than n=10^4, // we reduce n to half int nNew = (n - 2)/2; // This array is used to separate // numbers of the form i+j+2ij // from others where 1 <= i <= j boolean[] marked=new boolean[nNew + 1]; // Main logic of Sundaram. Mark all // numbers of the form i + j + 2ij // as true where 1 <= i <= j for (int i = 1; i <= nNew; i++) for (int j = i; (i + j + 2 * i * j) <= nNew; j++) marked[i + j + 2 * i * j] = true; // Since 2 is a prime number primes.add(2); // Remaining primes are of the form 2*i + 1 such that // marked[i] is false. for (int i = 1; i <= nNew; i++) if (marked[i] == false) primes.add(2 * i + 1);}// Function to count all numbers from a to b having exactly// n prime factorsstatic int Nfactors(int a, int b, int n){ segmentedSieve(); // result --> all numbers between a and b having // exactly n prime factors int result = 0; // check for each number for (int i = a; i <= b; i++) { // tmp --> stores square root of current number because // all prime factors are always less than and // equal to square root of given number // copy --> it holds the copy of current number int tmp = (int)Math.sqrt(i), copy = i; // count --> it counts the number of distinct prime // factors of number int count = 0; // check divisibility of 'copy' with each prime less // than 'tmp' and divide it until it is divisible by // current prime factor for (int j = 0; primes.get(j) <= tmp; j++) { if (copy % primes.get(j) == 0) { // increment count for distinct prime count++; while (copy % primes.get(j) == 0) copy = copy/primes.get(j); } } // if number is completely divisible then at last // 'copy' will be 1 else 'copy' will be prime, so // increment count by one if (copy != 1) count++; // if number has exactly n distinct primes then // increment result by one if (count == n) result++; } return result;}// Driver codepublic static void main (String[] args) { int a = 1, b = 100, n = 3; System.out.println(Nfactors(a, b, n));}}// This code is contributed by chandan_jnu |
Python3
# Python3 program to find numbers with # exactly n distinct prime factor numbers# from a to bimport math# Stores all primes less than and # equals to sqrt(10^8) = 10000primes = [];# Generate all prime numbers less than # or equals to sqrt(10^8) = 10000# using sieve of sundaramdef segmentedSieve(): n = 10000; # Square root of 10^8 # In general Sieve of Sundaram, produces # primes smaller than (2*x + 2) for a # given number x. Since we want primes # smaller than n=10^4, we reduce n to half nNew = int((n - 2) / 2); # This array is used to separate # numbers of the form i+j+2ij # from others where 1 <= i <= j marked = [False] * (nNew + 1); # Main logic of Sundaram. Mark all # numbers of the form i + j + 2ij # as true where 1 <= i <= j for i in range(1, nNew + 1): j = i; while ((i + j + 2 * i * j) <= nNew): marked[i + j + 2 * i * j] = True; j += 1; # Since 2 is a prime number primes.append(2); # Remaining primes are of the # form 2*i + 1 such that # marked[i] is false. for i in range(1, nNew + 1): if (marked[i] == False): primes.append(2 * i + 1);# Function to count all numbers # from a to b having exactly n # prime factorsdef Nfactors(a, b, n): segmentedSieve(); # result --> all numbers between # a and b having exactly n prime # factors result = 0; # check for each number for i in range(a, b + 1): # tmp --> stores square root of # current number because all prime # factors are always less than and # equal to square root of given number # copy --> it holds the copy of # current number tmp = math.sqrt(i); copy = i; # count --> it counts the number of # distinct prime factors of number count = 0; # check divisibility of 'copy' with # each prime less than 'tmp' and # divide it until it is divisible # by current prime factor j = 0; while (primes[j] <= tmp): if (copy % primes[j] == 0): # increment count for # distinct prime count += 1; while (copy % primes[j] == 0): copy = (copy // primes[j]); j += 1; # if number is completely divisible # then at last 'copy' will be 1 else # 'copy' will be prime, so increment # count by one if (copy != 1): count += 1; # if number has exactly n distinct # primes then increment result by one if (count == n): result += 1; return result;# Driver Codea = 1;b = 100;n = 3;print(Nfactors(a, b, n));# This code is contributed# by chandan_jnu |
C#
// C# program to find numbers with exactly n// distinct prime factor numbers from a to busing System;using System.Collections;class GFG{ // Stores all primes less than and // equals to sqrt(10^8) = 10000static ArrayList primes = new ArrayList();// Generate all prime numbers less // than or equals to sqrt(10^8)// = 10000 using sieve of sundaramstatic void segmentedSieve(){ int n = 10000; // Square root of 10^8 // In general Sieve of Sundaram, produces // primes smaller than (2*x + 2) for a number // given number x. Since we want primes // smaller than n=10^4, we reduce n to half int nNew = (n - 2) / 2; // This array is used to separate // numbers of the form i+j+2ij // from others where 1 <= i <= j bool[] marked = new bool[nNew + 1]; // Main logic of Sundaram. Mark all // numbers of the form i + j + 2ij // as true where 1 <= i <= j for (int i = 1; i <= nNew; i++) for (int j = i; (i + j + 2 * i * j) <= nNew; j++) marked[i + j + 2 * i * j] = true; // Since 2 is a prime number primes.Add(2); // Remaining primes are of the form // 2*i + 1 such that marked[i] is false. for (int i = 1; i <= nNew; i++) if (marked[i] == false) primes.Add(2 * i + 1);}// Function to count all numbers from // a to b having exactly n prime factorsstatic int Nfactors(int a, int b, int n){ segmentedSieve(); // result --> all numbers between a and b // having exactly n prime factors int result = 0; // check for each number for (int i = a; i <= b; i++) { // tmp --> stores square root of current // number because all prime factors are // always less than and equal to square // root of given number // copy --> it holds the copy of current number int tmp = (int)Math.Sqrt(i), copy = i; // count --> it counts the number of // distinct prime factors of number int count = 0; // check divisibility of 'copy' with each // prime less than 'tmp' and divide it until // it is divisible by current prime factor for (int j = 0; (int)primes[j] <= tmp; j++) { if (copy % (int)primes[j] == 0) { // increment count for distinct prime count++; while (copy % (int)primes[j] == 0) copy = copy / (int)primes[j]; } } // if number is completely divisible then // at last 'copy' will be 1 else 'copy' // will be prime, so increment count by one if (copy != 1) count++; // if number has exactly n distinct // primes then increment result by one if (count == n) result++; } return result;}// Driver codepublic static void Main() { int a = 1, b = 100, n = 3; Console.WriteLine(Nfactors(a, b, n));}}// This code is contributed by mits |
PHP
<?php// PHP program to find numbers with exactly n // distinct prime factor numbers from a to b// Stores all primes less than and equals // to sqrt(10^8) = 10000$primes = array();// Generate all prime numbers less than or // equals to sqrt(10^8) = 10000 using// sieve of sundaramfunction segmentedSieve(){ global $primes; $n = 10000; // Square root of 10^8 // In general Sieve of Sundaram, produces // primes smaller than (2*x + 2) for a // given number x. Since we want primes // smaller than n=10^4, we reduce n to half $nNew = (int)(($n-2)/2); // This array is used to separate numbers of // the form i+j+2ij from others where 1 <= i <= j $marked = array_fill(0, $nNew + 1, false); // Main logic of Sundaram. Mark all numbers of the // form i + j + 2ij as true where 1 <= i <= j for ($i = 1; $i <= $nNew; $i++) for ($j = $i; ($i + $j + 2 * $i * $j) <= $nNew; $j++) $marked[$i + $j + 2 * $i * $j] = true; // Since 2 is a prime number array_push($primes, 2); // Remaining primes are of the form 2*i + 1 // such that marked[i] is false. for ($i = 1; $i <= $nNew; $i++) if ($marked[$i] == false) array_push($primes, 2 * $i + 1);}// Function to count all numbers from a to b // having exactly n prime factorsfunction Nfactors($a, $b, $n){ global $primes; segmentedSieve(); // result --> all numbers between a and b // having exactly n prime factors $result = 0; // check for each number for ($i = $a; $i <= $b; $i++) { // tmp --> stores square root of current // number because all prime factors are // always less than and equal to square // root of given number // copy --> it holds the copy of current number $tmp = sqrt($i); $copy = $i; // count --> it counts the number of // distinct prime factors of number $count = 0; // check divisibility of 'copy' with each // prime less than 'tmp' and divide it until // it is divisible by current prime factor for ($j = 0; $primes[$j] <= $tmp; $j++) { if ($copy % $primes[$j] == 0) { // increment count for distinct prime $count++; while ($copy % $primes[$j] == 0) $copy = (int)($copy / $primes[$j]); } } // if number is completely divisible then // at last 'copy' will be 1 else 'copy' // will be prime, so increment count by one if ($copy != 1) $count++; // if number has exactly n distinct primes // then increment result by one if ($count == $n) $result++; } return $result;}// Driver Code$a = 1;$b = 100;$n = 3;print(Nfactors($a, $b, $n));// This code is contributed by chandan_jnu?> |
Javascript
<script> // JavaScript program to find numbers with exactly n // distinct prime factor numbers from a to b // Stores all primes less than and // equals to sqrt(10^8) = 10000 let primes = []; // Generate all prime numbers less // than or equals to sqrt(10^8) // = 10000 using sieve of sundaram function segmentedSieve() { let n = 10000; // Square root of 10^8 // In general Sieve of Sundaram, produces // primes smaller than (2*x + 2) for a number // given number x. Since we want primes // smaller than n=10^4, we reduce n to half let nNew = parseInt((n - 2) / 2, 10); // This array is used to separate // numbers of the form i+j+2ij // from others where 1 <= i <= j let marked = new Array(nNew + 1); marked.fill(false); // Main logic of Sundaram. Mark all // numbers of the form i + j + 2ij // as true where 1 <= i <= j for (let i = 1; i <= nNew; i++) for (let j = i; (i + j + 2 * i * j) <= nNew; j++) marked[i + j + 2 * i * j] = true; // Since 2 is a prime number primes.push(2); // Remaining primes are of the form // 2*i + 1 such that marked[i] is false. for (let i = 1; i <= nNew; i++) if (marked[i] == false) primes.push(2 * i + 1); } // Function to count all numbers from // a to b having exactly n prime factors function Nfactors(a, b, n) { segmentedSieve(); // result --> all numbers between a and b // having exactly n prime factors let result = 0; // check for each number for (let i = a; i <= b; i++) { // tmp --> stores square root of current // number because all prime factors are // always less than and equal to square // root of given number // copy --> it holds the copy of current number let tmp = parseInt(Math.sqrt(i), 10), copy = i; // count --> it counts the number of // distinct prime factors of number let count = 0; // check divisibility of 'copy' with each // prime less than 'tmp' and divide it until // it is divisible by current prime factor for (let j = 0; primes[j] <= tmp; j++) { if (copy % primes[j] == 0) { // increment count for distinct prime count++; while (copy % primes[j] == 0) copy = parseInt(copy / primes[j], 10); } } // if number is completely divisible then // at last 'copy' will be 1 else 'copy' // will be prime, so increment count by one if (copy != 1) count++; // if number has exactly n distinct // primes then increment result by one if (count == n) result++; } return result; } let a = 1, b = 100, n = 3; document.write(Nfactors(a, b, n)); </script> |
Output:
8
If you have another approach to solve this problem then please share in comments.
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