# Print all prime factors and their powers

Given a number N, print all its unique prime factors and their powers in N.

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 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 powerswhile(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 continueif(curr == s[N]) { cnt++;continue;} // Print prime factor and its power

Below is C++ implementation of above steps.

// 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] = 7 void 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 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; }

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.

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below

source and credits.

source and credits:http://www.geeksforgeeks.org/print-all-prime-factors-and-their-powers/

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