Print all prime factors and their powers - GeeksforGeeks

04 Mar, 2022
Given a number N, print all its unique prime factors and their powers in N. Examples:  
Input: N = 100
Output: Factor Power
          2      2
          5      2

Input: N = 35
Output: Factor  Power
          5      1
          7      1
A Simple Solution is to first find prime factors of N. Then for every prime factor, find the highest power of it that divides N and print it.An Efficient Solution is to use Sieve of Eratosthenes.  
1) First compute an array s[N+1] using Sieve of Eratosthenes.

s[i] = Smallest prime factor of "i" that
       divides "i".

For example let N  = 10
  s[2] = s[4] = s[6] = s[8] = s[10] = 2;
  s[3] = s[9] = 3;
  s[5] = 5;
  s[7] = 7;


2) Using the above computed array s[],
   we can find all powers in O(Log N) time.

    curr = s[N];  // Current prime factor of N
    cnt = 1;   // Power of current prime factor

    // Printing prime factors and their powers
    while (N > 1)
    {
        N /= s[N];

        // N is now N/s[N].  If new N also has its 
        // smallest prime factor as curr, increment 
        // power and continue
        if (curr == s[N])
        {
            cnt++;
            continue;
        }

        // Print prime factor and its power
        print(curr, cnt);

        // Update current prime factor as s[N] and
        // initializing count as 1.
        curr = s[N];
        cnt = 1;
    }
Below is the implementation of above steps. 
C++
Java
Python3
C#
PHP
Javascript
// C++ Program to print prime factors and their// powers using Sieve Of Eratosthenes#include<bits/stdc++.h>using namespace std; // Using SieveOfEratosthenes to find smallest prime// factor of all the numbers.// For example, if N is 10,// s[2] = s[4] = s[6] = s[10] = 2// s[3] = s[9] = 3// s[5] = 5// s[7] = 7void sieveOfEratosthenes(int N, int s[]){    // Create a boolean array "prime[0..n]" and    // initialize all entries in it as false.    vector <bool> prime(N+1, false);     // Initializing smallest factor equal to 2    // for all the even numbers    for (int i=2; i<=N; i+=2)        s[i] = 2;     // For odd numbers less then equal to n    for (int i=3; i<=N; i+=2)    {        if (prime[i] == false)        {            // s(i) for a prime is the number itself            s[i] = i;             // For all multiples of current prime number            for (int j=i; j*i<=N; j+=2)            {                if (prime[i*j] == false)                {                    prime[i*j] = true;                     // i is the smallest prime factor for                    // number "i*j".                    s[i*j] = i;                }            }        }    }} // Function to generate prime factors and its powervoid generatePrimeFactors(int N){    // s[i] is going to store smallest prime factor    // of i.    int s[N+1];     // Filling values in s[] using sieve    sieveOfEratosthenes(N, s);     printf("Factor Power\n");     int curr = s[N];  // Current prime factor of N    int cnt = 1;   // Power of current prime factor     // Printing prime factors and their powers    while (N > 1)    {        N /= s[N];         // N is now N/s[N].  If new N als has smallest        // prime factor as curr, increment power        if (curr == s[N])        {            cnt++;            continue;        }         printf("%d\t%d\n", curr, cnt);         // Update current prime factor as s[N] and        // initializing count as 1.        curr = s[N];        cnt = 1;    }} //Driver Programint main(){    int N = 360;    generatePrimeFactors(N);    return 0;}
// Java Program to print prime// factors and their powers using// Sieve Of Eratosthenesclass GFG{// Using SieveOfEratosthenes// to find smallest prime// factor of all the numbers.// For example, if N is 10,// s[2] = s[4] = s[6] = s[10] = 2// s[3] = s[9] = 3// s[5] = 5// s[7] = 7static void sieveOfEratosthenes(int N,                                int s[]){    // Create a boolean array    // "prime[0..n]"  and initialize    // all entries in it as false.    boolean[] prime = new boolean[N + 1];     // Initializing smallest    // factor equal to 2    // for all the even numbers    for (int i = 2; i <= N; i += 2)        s[i] = 2;     // For odd numbers less    // then equal to n    for (int i = 3; i <= N; i += 2)    {        if (prime[i] == false)        {            // s(i) for a prime is            // the number itself            s[i] = i;             // For all multiples of            // current prime number            for (int j = i; j * i <= N; j += 2)            {                if (prime[i * j] == false)                {                    prime[i * j] = true;                     // i is the smallest prime                    // factor for number "i*j".                    s[i * j] = i;                }            }        }    }} // Function to generate prime// factors and its powerstatic void generatePrimeFactors(int N){    // s[i] is going to store    // smallest prime factor of i.    int[] s = new int[N + 1];     // Filling values in s[] using sieve    sieveOfEratosthenes(N, s);     System.out.println("Factor Power");     int curr = s[N]; // Current prime factor of N    int cnt = 1; // Power of current prime factor     // Printing prime factors    // and their powers    while (N > 1)    {        N /= s[N];         // N is now N/s[N]. If new N        // also has smallest prime        // factor as curr, increment power        if (curr == s[N])        {            cnt++;            continue;        }         System.out.println(curr + "\t" + cnt);         // Update current prime factor        // as s[N] and initializing        // count as 1.        curr = s[N];        cnt = 1;    }} // Driver Codepublic static void main(String[] args){    int N = 360;    generatePrimeFactors(N);}} // This code is contributed by mits
# Python3 program to print prime# factors and their powers# using Sieve Of Eratosthenes # Using SieveOfEratosthenes to# find smallest prime factor# of all the numbers. # For example, if N is 10,# s[2] = s[4] = s[6] = s[10] = 2# s[3] = s[9] = 3# s[5] = 5# s[7] = 7def sieveOfEratosthenes(N, s):         # Create a boolean array    # "prime[0..n]" and initialize    # all entries in it as false.    prime = [False] * (N+1)     # Initializing smallest factor    # equal to 2 for all the even    # numbers    for i in range(2, N+1, 2):        s[i] = 2     # For odd numbers less then    # equal to n    for i in range(3, N+1, 2):        if (prime[i] == False):                         # s(i) for a prime is            # the number itself            s[i] = i             # For all multiples of            # current prime number            for j in range(i, int(N / i) + 1, 2):                if (prime[i*j] == False):                    prime[i*j] = True                     # i is the smallest                    # prime factor for                    # number "i*j".                    s[i * j] = i # Function to generate prime# factors and its powerdef generatePrimeFactors(N):     # s[i] is going to store    # smallest prime factor    # of i.    s = [0] * (N+1)     # Filling values in s[]    # using sieve    sieveOfEratosthenes(N, s)     print("Factor Power")     # Current prime factor of N    curr = s[N]         # Power of current prime factor    cnt = 1     # Printing prime factors and    #their powers    while (N > 1):        N //= s[N]         # N is now N/s[N]. If new N        # also has smallest prime        # factor as curr, increment        # power        if (curr == s[N]):            cnt += 1            continue         print(str(curr) + "\t" + str(cnt))         # Update current prime factor        # as s[N] and initializing        # count as 1.        curr = s[N]        cnt = 1 #Driver ProgramN = 360generatePrimeFactors(N) # This code is contributed by Ansu Kumari
// C# Program to print prime// factors and their powers using// Sieve Of Eratosthenesclass GFG{// Using SieveOfEratosthenes// to find smallest prime// factor of all the numbers.// For example, if N is 10,// s[2] = s[4] = s[6] = s[10] = 2// s[3] = s[9] = 3// s[5] = 5// s[7] = 7static void sieveOfEratosthenes(int N, int[] s){    // Create a boolean array    // "prime[0..n]" and initialize    // all entries in it as false.    bool[] prime = new bool[N + 1];     // Initializing smallest    // factor equal to 2    // for all the even numbers    for (int i = 2; i <= N; i += 2)        s[i] = 2;     // For odd numbers less    // then equal to n    for (int i = 3; i <= N; i += 2)    {        if (prime[i] == false)        {            // s(i) for a prime is            // the number itself            s[i] = i;             // For all multiples of            // current prime number            for (int j = i; j * i <= N; j += 2)            {                if (prime[i * j] == false)                {                    prime[i * j] = true;                     // i is the smallest prime                    // factor for number "i*j".                    s[i * j] = i;                }            }        }    }} // Function to generate prime// factors and its powerstatic void generatePrimeFactors(int N){    // s[i] is going to store    // smallest prime factor of i.    int[] s = new int[N + 1];     // Filling values in s[] using sieve    sieveOfEratosthenes(N, s);     System.Console.WriteLine("Factor Power");     int curr = s[N]; // Current prime factor of N    int cnt = 1; // Power of current prime factor     // Printing prime factors    // and their powers    while (N > 1)    {        N /= s[N];         // N is now N/s[N]. If new N        // also has smallest prime        // factor as curr, increment power        if (curr == s[N])        {            cnt++;            continue;        }         System.Console.WriteLine(curr + "\t" + cnt);         // Update current prime factor        // as s[N] and initializing        // count as 1.        curr = s[N];        cnt = 1;    }} // Driver Codestatic void Main(){    int N = 360;    generatePrimeFactors(N);}} // This code is contributed by mits
<?php// PHP Program to print prime factors and// their powers using Sieve Of Eratosthenes // Using SieveOfEratosthenes to find smallest// prime factor of all the numbers.// For example, if N is 10,// s[2] = s[4] = s[6] = s[10] = 2// s[3] = s[9] = 3// s[5] = 5// s[7] = 7function sieveOfEratosthenes($N, &$s){    // Create a boolean array "prime[0..n]" and    // initialize all entries in it as false.    $prime = array_fill(0, $N + 1, false);     // Initializing smallest factor equal    // to 2 for all the even numbers    for ($i = 2; $i <= $N; $i += 2)        $s[$i] = 2;     // For odd numbers less then equal to n    for ($i = 3; $i <= $N; $i += 2)    {        if ($prime[$i] == false)        {            // s(i) for a prime is the            // number itself            $s[$i] = $i;             // For all multiples of current            // prime number            for ($j = $i; $j * $i <= $N; $j += 2)            {                if ($prime[$i * $j] == false)                {                    $prime[$i * $j] = true;                     // i is the smallest prime factor                    // for number "i*j".                    $s[$i * $j] = $i;                }            }        }    }} // Function to generate prime factors// and its powerfunction generatePrimeFactors($N){    // s[i] is going to store smallest    // prime factor of i.    $s = array_fill(0, $N + 1, 0);     // Filling values in s[] using sieve    sieveOfEratosthenes($N, $s);     print("Factor Power\n");     $curr = $s[$N]; // Current prime factor of N    $cnt = 1; // Power of current prime factor     // Printing prime factors and their powers    while ($N > 1)    {        if($s[$N])        $N = (int)($N / $s[$N]);         // N is now N/s[N]. If new N als has smallest        // prime factor as curr, increment power        if ($curr == $s[$N])        {            $cnt++;            continue;        }         print($curr . "\t" . $cnt . "\n");         // Update current prime factor as s[N]        // and initializing count as 1.        $curr = $s[$N];        $cnt = 1;    }} // Driver Code$N = 360;generatePrimeFactors($N); // This code is contributed by mits?>
<script> // javascript Program to print prime// factors and their powers using// Sieve Of Eratosthenes // Using SieveOfEratosthenes// to find smallest prime// factor of all the numbers.// For example, if N is 10,// s[2] = s[4] = s[6] = s[10] = 2// s[3] = s[9] = 3// s[5] = 5// s[7] = 7function sieveOfEratosthenes(N,  s){    // Create a boolean array    // "prime[0..n]"  and initialize    // all entries in it as false.    prime = Array.from({length: N+1}, (_, i) => false);     // Initializing smallest    // factor equal to 2    // for all the even numbers    for (i = 2; i <= N; i += 2)        s[i] = 2;     // For odd numbers less    // then equal to n    for (i = 3; i <= N; i += 2)    {        if (prime[i] == false)        {            // s(i) for a prime is            // the number itself            s[i] = i;             // For all multiples of            // current prime number            for (j = i; j * i <= N; j += 2)            {                if (prime[i * j] == false)                {                    prime[i * j] = true;                     // i is the smallest prime                    // factor for number "i*j".                    s[i * j] = i;                }            }        }    }} // Function to generate prime// factors and its powerfunction generatePrimeFactors(N){    // s[i] is going to store    // smallest prime factor of i.    var s = Array.from({length: N+1}, (_, i) => 0);     // Filling values in s using sieve    sieveOfEratosthenes(N, s);     document.write("Factor Power");     var curr = s[N]; // Current prime factor of N    var cnt = 1; // Power of current prime factor     // Printing prime factors    // and their powers    while (N > 1)    {        N /= s[N];         // N is now N/s[N]. If new N        // also has smallest prime        // factor as curr, increment power        if (curr == s[N])        {            cnt++;            continue;        }         document.write("<br>"+curr + "\t" + cnt);         // Update current prime factor        // as s[N] and initializing        // count as 1.        curr = s[N];        cnt = 1;    }} // Driver Codevar N = 360;generatePrimeFactors(N);  // This code contributed by Princi Singh</script>
Output:  
Factor  Power
  2      3
  3      2
  5      1
The above algorithm finds all powers in O(Log N) time after we have filled s[]. This can be very useful in competitive environment where we have an upper limit and we need to compute prime factors and their powers for many test cases. In this scenario, the array needs to be s[] filled only once.This article is contributed by Rahul Agrawal. If you like GeeksforGeeks 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 GeeksforGeeks 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. 
Mithun Kumar
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