# Count all the numbers less than 10^6 whose minimum prime factor is N

Given a number N which is prime. The task is to find all the numbers less than or equal to 10^6 whose minimum prime factor is N.

Examples:

```Input: N = 2
Output: 500000

Input: N = 3
Output: 166667
```

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

Approach: Use sieve of Eratosthenes to find the solution of the problem. Store all the prime numbers less than 10^6 . Form another sieve which will store the count of all the numbers whose minimum prime factor is the index of the sieve. Then display the count of the prime number N (i.e. sieve_count[n]+1), where n is the prime number .

Below is the implementation of above approach:

## C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` `#define MAX 1000000 ` ` `  `// the sieve of prime number and ` `// count of minimum prime factor ` `int` `sieve_Prime[MAX + 4] = { 0 }, ` `                      ``sieve_count[MAX + 4] = { 0 }; ` ` `  `// form the prime sieve ` `void` `form_sieve() ` `{ ` `    ``// 1 is not a prime number ` `    ``sieve_Prime = 1; ` ` `  `    ``// form the sieve ` `    ``for` `(``int` `i = 2; i <= MAX; i++) { ` ` `  `        ``// if i is prime ` `        ``if` `(sieve_Prime[i] == 0) { ` `            ``for` `(``int` `j = i * 2; j <= MAX; j += i) { ` ` `  `                ``// if i is the least prime factor ` `                ``if` `(sieve_Prime[j] == 0) { ` ` `  `                    ``// mark the number j as non prime ` `                    ``sieve_Prime[j] = 1; ` ` `  `                    ``// count the numbers whose least prime factor is i ` `                    ``sieve_count[i]++; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// form the sieve ` `    ``form_sieve(); ` ` `  `    ``int` `n = 2; ` ` `  `    ``// display ` `    ``cout << ``"Count = "` `<< (sieve_count[n] + 1) << endl; ` ` `  `    ``n = 3; ` ` `  `    ``// display ` `    ``cout << ``"Count = "` `<< (sieve_count[n] + 1) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of above approach ` `import` `java.io.*; ` ` `  `class` `GFG { ` `     `  `static` `int` `MAX = ``1000000``; ` ` `  `// the sieve of prime number and ` `// count of minimum prime factor ` `static` `int` `sieve_Prime[] = ``new` `int``[MAX + ``4``]; ` `static` `int` `sieve_count[] =  ``new` `int``[MAX + ``4``]; ` ` `  `// form the prime sieve ` `static` `void` `form_sieve() ` `{ ` `    ``// 1 is not a prime number ` `    ``sieve_Prime[``1``] = ``1``; ` ` `  `    ``// form the sieve ` `    ``for` `(``int` `i = ``2``; i <= MAX; i++) { ` ` `  `        ``// if i is prime ` `        ``if` `(sieve_Prime[i] == ``0``) { ` `            ``for` `(``int` `j = i * ``2``; j <= MAX; j += i) { ` ` `  `                ``// if i is the least prime factor ` `                ``if` `(sieve_Prime[j] == ``0``) { ` ` `  `                    ``// mark the number j as non prime ` `                    ``sieve_Prime[j] = ``1``; ` ` `  `                    ``// count the numbers whose least prime factor is i ` `                    ``sieve_count[i]++; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` ` `  `    ``public` `static` `void` `main (String[] args) { ` `        ``// form the sieve ` `    ``form_sieve(); ` ` `  `    ``int` `n = ``2``; ` ` `  `    ``// display ` `    ``System.out.println( ``"Count = "` `+ (sieve_count[n] + ``1``)); ` ` `  `    ``n = ``3``; ` ` `  `    ``// display ` `    ``System.out.println (``"Count = "`  `+(sieve_count[n] + ``1``)); ` `    ``} ` `} ` `// This code was contributed ` `// by inder_mca `

## Python3

 `# Python3 implementation of ` `# above approach ` ` `  `MAX` `=` `1000000` ` `  `# the sieve of prime number and ` `# count of minimum prime factor ` `sieve_Prime ``=` `[``0` `for` `i ``in` `range``(``MAX` `+` `4``)] ` `sieve_count ``=` `[``0` `for` `i ``in` `range``(``MAX` `+` `4``)] ` ` `  `# form the prime sieve ` `def` `form_sieve(): ` `     `  `    ``# 1 is not a prime number ` `    ``sieve_Prime[``1``] ``=` `1` ` `  `    ``# form the sieve ` `    ``for` `i ``in` `range``(``2``, ``MAX` `+` `1``): ` ` `  `        ``# if i is prime ` `        ``if` `sieve_Prime[i] ``=``=` `0``: ` `            ``for` `j ``in` `range``(i ``*` `2``, ``MAX` `+` `1``, i): ` ` `  `                ``# if i is the least prime factor ` `                ``if` `sieve_Prime[j] ``=``=` `0``: ` ` `  `                    ``# mark the number j  ` `                    ``# as non prime ` `                    ``sieve_Prime[j] ``=` `1` ` `  `                    ``# count the numbers whose  ` `                    ``# least prime factor is i ` `                    ``sieve_count[i] ``+``=` `1` ` `  `# Driver code ` ` `  `# form the sieve ` `form_sieve() ` ` `  `n ``=` `2` ` `  `# display ` `print``(``"Count ="``, sieve_count[n] ``+` `1``) ` ` `  `n ``=` `3` ` `  `# display ` `print``(``"Count ="``, sieve_count[n] ``+` `1``) ` ` `  `# This code was contributed ` `# by VishalBachchas `

## C#

 `// C# implementation of above approach ` `using` `System; ` ` `  `class` `GFG { ` `     `  `static` `int` `MAX = 1000000; ` ` `  `// the sieve of prime number and ` `// count of minimum prime factor ` `static` `int` `[]sieve_Prime = ``new` `int``[MAX + 4]; ` `static` `int` `[]sieve_count = ``new` `int``[MAX + 4]; ` ` `  `// form the prime sieve ` `static` `void` `form_sieve() ` `{ ` `    ``// 1 is not a prime number ` `    ``sieve_Prime = 1; ` ` `  `    ``// form the sieve ` `    ``for` `(``int` `i = 2; i <= MAX; i++) { ` ` `  `        ``// if i is prime ` `        ``if` `(sieve_Prime[i] == 0) { ` `            ``for` `(``int` `j = i * 2; j <= MAX; j += i) { ` ` `  `                ``// if i is the least prime factor ` `                ``if` `(sieve_Prime[j] == 0) { ` ` `  `                    ``// mark the number j as non prime ` `                    ``sieve_Prime[j] = 1; ` ` `  `                    ``// count the numbers whose least prime factor is i ` `                    ``sieve_count[i]++; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` ` `  `    ``public` `static` `void` `Main () { ` `        ``// form the sieve ` `    ``form_sieve(); ` ` `  `    ``int` `n = 2; ` ` `  `    ``// display ` `    ``Console.WriteLine( ``"Count = "` `+ (sieve_count[n] + 1)); ` ` `  `    ``n = 3; ` ` `  `    ``// display ` `    ``Console.WriteLine (``"Count = "` `+(sieve_count[n] + 1)); ` `    ``} ` `} ` `// This code was contributed ` `// by shs `

## PHP

 ` `

Output:

```Count = 500000
Count = 166667
```

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