In this article, we study an optimized way to calculate the distinct prime factorization up to n natural number using O O(n*log n) time complexity with pre-computation allowed.
Prerequisites: Sieve of Eratosthenes, Least prime factor of numbers till n.
Key Concept: Our idea is to store the Smallest Prime Factor(SPF) for every number. Then to calculate the distinct prime factorization of the given number by dividing the given number recursively with its smallest prime factor till it becomes 1.
To calculate to smallest prime factor for every number we will use the sieve of eratosthenes. In original Sieve, every time we mark a number as not-prime, we store the corresponding smallest prime factor for that number (Refer this article for better understanding).
The implementation for the above method is given below :
// C++ program to find prime factorization upto n natural number // O(n*Log n) time with precomputation #include <bits/stdc++.h> using namespace std;
#define MAXN 100001 // Stores smallest prime factor for every number int spf[MAXN];
// Adjacency vector to store distinct prime factors vector< int >adj[MAXN];
// Calculating SPF (Smallest Prime Factor) for every // number till MAXN. // Time Complexity : O(nloglogn) void sieve()
{ spf[1] = 1;
// marking smallest prime factor for every
// number to be itself.
for ( int i=2; i<MAXN; i++)
spf[i] = i;
for ( int i=2; i*i<MAXN; i++)
{
// checking if i is prime
if (spf[i] == i)
{
// marking SPF for all numbers divisible by i
for ( int j=i*i; j<MAXN; j+=i)
// marking spf[j] if it is not
// previously marked
if (spf[j]==j)
spf[j] = i;
}
}
} // A O(nlog n) function returning distinct primefactorization // upto n natural number by dividing by smallest prime factor // at every step void getdistinctFactorization( int n)
{ int index,x,i;
for ( int i=1;i<=n;i++)
{
index=1;
x=i;
if (x!=1)
adj[i].push_back(spf[x]);
x=x/spf[x];
// Push all distinct prime factor in adj
while (x != 1)
{
if (adj[i][index-1]!=spf[x])
{
adj[i].push_back(spf[x]);
index+=1;
}
x = x / spf[x];
}
}
} // Driver code int main()
{ // Precalculating smallest prime factor
sieve();
int n = 10;
getdistinctFactorization(n);
// Print the prime count
cout << "Distinct prime factor for first " << n
<< " natural number" << " : " ;
for ( int i=1; i<=n; i++)
cout << adj[i].size() << " " ;
return 0;
} |
// Java program to find prime factorization upto n natural number // O(n*Log n) time with precomputation import java.io.*;
import java.util.*;
class GFG
{ static int MAXN = 100001 ;
// Stores smallest prime factor for every number
static int [] spf = new int [MAXN];
// Adjacency vector to store distinct prime factors
static ArrayList<ArrayList<Integer>> adj =
new ArrayList<ArrayList<Integer>>();
// Calculating SPF (Smallest Prime Factor) for every
// number till MAXN.
// Time Complexity : O(nloglogn)
static void sieve()
{
for ( int i = 0 ; i < MAXN; i++)
{
adj.add( new ArrayList<Integer>());
}
spf[ 1 ] = 1 ;
// marking smallest prime factor for every
// number to be itself.
for ( int i = 2 ; i < MAXN; i++)
{
spf[i] = i;
}
for ( int i = 2 ; i * i < MAXN; i++)
{
// checking if i is prime
if (spf[i] == i)
{
// marking SPF for all numbers divisible by i
for ( int j = i * i; j < MAXN; j += i)
{
// marking spf[j] if it is not
// previously marked
if (spf[j] == j)
spf[j] = i;
}
}
}
}
// A O(nlog n) function returning distinct primefactorization
// upto n natural number by dividing by smallest prime factor
// at every step
static void getdistinctFactorization( int n)
{
int index, x, i;
for (i = 1 ; i <= n; i++)
{
index = 1 ;
x = i;
if (x != 1 )
adj.get(i).add(spf[x]);
x = x / spf[x];
// Push all distinct prime factor in adj
while (x != 1 )
{
if (adj.get(i).get(index - 1 ) != spf[x])
{
adj.get(i).add(spf[x]);
index += 1 ;
}
x = x / spf[x];
}
}
}
// Driver code
public static void main (String[] args)
{
// Precalculating smallest prime factor
sieve();
int n = 10 ;
getdistinctFactorization(n);
// Print the prime count
System.out.print( "Distinct prime factor for first " +
n + " natural number" + " : " );
for ( int i = 1 ; i <= n; i++)
{
System.out.print(adj.get(i).size()+ " " );
}
}
} // This code is contributed by avanitrachhadiya2155 |
# Python3 program to find prime factorization upto n natural number # O(n*Log n) time with precomputation # Calculating SPF (Smallest Prime Factor) for every # number till MAXN. # Time Complexity : O(nloglogn) def sieve():
global spf, adj, MAXN
spf[ 1 ] = 1
# marking smallest prime factor for every
# number to be itself.
for i in range ( 2 , MAXN):
spf[i] = i
for i in range ( 2 , MAXN):
if i * i > MAXN:
break
# checking if i is prime
if (spf[i] = = i):
# marking SPF for all numbers divisible by i
for j in range (i * i, MAXN, i):
# marking spf[j] if it is not
# previously marked
if (spf[j] = = j):
spf[j] = i
# A O(nlog n) function returning distinct primefactorization # upto n natural number by dividing by smallest prime factor # at every step def getdistinctFactorization(n):
global adj, spf, MAXN
index = 0
for i in range ( 1 , n + 1 ):
index = 1
x = i
if (x ! = 1 ):
adj[i].append(spf[x])
x = x / / spf[x]
# Push all distinct prime factor in adj
while (x ! = 1 ):
if (adj[i][index - 1 ] ! = spf[x]):
adj[i].append(spf[x])
index + = 1
x = x / / spf[x]
# Driver code if __name__ = = '__main__' :
MAXN = 100001
spf = [ 0 for i in range (MAXN)]
adj = [[] for i in range (MAXN)]
# Precalculating smallest prime factor
sieve()
n = 10
getdistinctFactorization(n)
# Print prime count
print ( "Distinct prime factor for first " , n, " natural number : " , end = "")
for i in range ( 1 , n + 1 ):
print ( len (adj[i]), end = " " )
# This code is contributed by mohit kumar 29 |
using System;
using System.Collections.Generic;
public class GFG
{ static int MAXN = 100001;
// Stores smallest prime factor for every number
static int [] spf = new int [MAXN];
// Adjacency vector to store distinct prime factors
static List<List< int >> adj = new List<List< int >>();
// Calculating SPF (Smallest Prime Factor) for every
// number till MAXN.
// Time Complexity : O(nloglogn)
static void sieve()
{
for ( int i = 0; i < MAXN; i++)
{
adj.Add( new List< int >());
}
spf[1] = 1;
// marking smallest prime factor for every
// number to be itself.
for ( int i = 2; i < MAXN; i++)
{
spf[i] = i;
}
for ( int i = 2; i * i < MAXN; i++)
{
// checking if i is prime
if (spf[i] == i)
{
// marking SPF for all numbers divisible by i
for ( int j = i * i; j < MAXN; j += i)
{
// marking spf[j] if it is not
// previously marked
if (spf[j] == j)
spf[j] = i;
}
}
}
}
// A O(nlog n) function returning distinct primefactorization
// upto n natural number by dividing by smallest prime factor
// at every step
static void getdistinctFactorization( int n)
{
int index, x, i;
for (i = 1; i <= n; i++)
{
index = 1;
x = i;
if (x != 1)
{
adj[i].Add(spf[x]);
}
x = x / spf[x];
// Push all distinct prime factor in adj
while (x != 1)
{
if (adj[i][index-1] != spf[x])
{
adj[i].Add(spf[x]);
index += 1;
}
x = x / spf[x];
}
}
}
// Driver code
static public void Main ()
{
// Precalculating smallest prime factor
sieve();
int n = 10;
getdistinctFactorization(n);
// Print the prime count
Console.Write( "Distinct prime factor for first " +
n + " natural number" + " : " );
for ( int i = 1; i <= n; i++)
{
Console.Write(adj[i].Count + " " );
}
}
} // This code is contributed by rag2127 |
<script> // Javascript program to find prime // factorization upto n natural number // O(n*Log n) time with precomputation let MAXN = 100001;
// Stores smallest prime factor
// for every number
let spf = new Array(MAXN);
// Adjacency vector to store distinct prime factors
let adj=[];
// Calculating SPF (Smallest Prime Factor) for every
// number till MAXN.
// Time Complexity : O(nloglogn)
function sieve()
{
for (let i = 0; i < MAXN; i++)
{
adj.push([]);
}
spf[1] = 1;
// marking smallest prime factor for every
// number to be itself.
for (let i = 2; i < MAXN; i++)
{
spf[i] = i;
}
for (let i = 2; i * i < MAXN; i++)
{
// checking if i is prime
if (spf[i] == i)
{
// marking SPF for all numbers divisible by i
for (let j = i * i; j < MAXN; j += i)
{
// marking spf[j] if it is not
// previously marked
if (spf[j] == j)
spf[j] = i;
}
}
}
}
// A O(nlog n) function returning
// distinct primefactorization
// upto n natural number by dividing by
// smallest prime factor
// at every step
function getdistinctFactorization(n)
{
let index, x, i;
for (i = 1; i <= n; i++)
{
index = 1;
x = i;
if (x != 1)
adj[i].push(spf[x]);
x = Math.floor(x / spf[x]);
// Push all distinct prime factor in adj
while (x != 1)
{
if (adj[i][index - 1] != spf[x])
{
adj[i].push(spf[x]);
index += 1;
}
x = Math.floor(x / spf[x]);
}
}
}
// Driver code
// Precalculating smallest prime factor
sieve();
let n = 10;
getdistinctFactorization(n);
// Print the prime count
document.write( "Distinct prime factor for first " +
n + " natural number" + " : " );
for (let i = 1; i <= n; i++)
{
document.write(adj[i].length+ " " );
}
// This code is contributed by unknown2108 </script> |
Distinct prime factor for first 10 natural number : 0 1 1 1 1 2 1 1 1 2