Exactly n distinct prime factor numbers from a to b

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: 2
// 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

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

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++ program to find numbers with exactly n distinct // prime factor numbers from a to b #include using namespace std;    // Stores all primes less than and equals to sqrt(10^8) = 10000 vector primes;    // Generate all prime numbers less than or equals to sqrt(10^8) // = 10000 using sieve of sundaram 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[nNew + 1];        // Initalize 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 factors 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 = 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 case int main() {     int a = 1, b = 100, n = 3;     cout << Nfactors(a, b, n);     return 0; }

 // Java program to find numbers with exactly n distinct // prime factor numbers from a to b import java.util.*;    class GFG {        // Stores all primes less than and  // equals to sqrt(10^8) = 10000 static ArrayList primes = new ArrayList();    // Generate all prime numbers less  // than or equals to sqrt(10^8) // = 10000 using sieve of sundaram static 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 factors static 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 code public 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 program to find numbers with  # exactly n distinct prime factor numbers # from a to b import math    # Stores all primes less than and  # equals to sqrt(10^8) = 10000 primes = [];    # Generate all prime numbers less than  # or equals to sqrt(10^8) = 10000 # using sieve of sundaram def 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 factors def 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 Code a = 1; b = 100; n = 3; print(Nfactors(a, b, n));    # This code is contributed # by chandan_jnu

 // C# program to find numbers with exactly n // distinct prime factor numbers from a to b using System; using System.Collections;    class GFG {        // Stores all primes less than and  // equals to sqrt(10^8) = 10000 static ArrayList primes = new ArrayList();    // Generate all prime numbers less  // than or equals to sqrt(10^8) // = 10000 using sieve of sundaram static 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 factors static 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 code public static void Main()  {     int a = 1, b = 100, n = 3;     Console.WriteLine(Nfactors(a, b, n)); } }    // This code is contributed by mits

 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 ?>

Output:
8

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