Print all prime factors and their powers

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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 = s = s = s = s = 2;
s = s = 3;
s = 5;
s = 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++ Program to print prime factors and their // powers using Sieve Of Eratosthenes #include using namespace std;    // Using SieveOfEratosthenes to find smallest prime // factor of all the numbers. // For example, if N is 10, // s = s = s = s = 2 // s = s = 3 // s = 5 // s = 7 void sieveOfEratosthenes(int N, int s[]) {     // Create a boolean array "prime[0..n]" and     // initialize all entries in it as false.     vector 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 power void 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 Program int main() {     int N = 360;     generatePrimeFactors(N);     return 0; }

 // Java Program to print prime  // factors and their powers using // Sieve Of Eratosthenes class GFG { // Using SieveOfEratosthenes  // to find smallest prime // factor of all the numbers. // For example, if N is 10, // s = s = s = s = 2 // s = s = 3 // s = 5 // s = 7 static 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 power static 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 Code public 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 = s = s = s = 2 # s = s = 3 # s = 5 # s = 7 def 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 power def generatePrimeFactors(N):        # s[i] is going to store     # smallest prime factor      # of i.     s =  * (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          # als 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 Program N = 360 generatePrimeFactors(N)    # This code is contributed by Ansu Kumari

 // C# Program to print prime  // factors and their powers using // Sieve Of Eratosthenes class GFG { // Using SieveOfEratosthenes  // to find smallest prime // factor of all the numbers. // For example, if N is 10, // s = s = s = s = 2 // s = s = 3 // s = 5 // s = 7 static 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 power static 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 Code static void Main() {     int N = 360;     generatePrimeFactors(N); } }    // This code is contributed by mits

 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 ?>

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.

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Improved By : Mithun Kumar, nidhi_biet