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; `
`} ` |

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