在本文中，我们研究了一种使用O O(n * log n)时间复杂度并允许预先计算的方法来计算高达n个自然数的独特素数分解的一种优化方法。
先决条件： Eratosthenes筛，直到n的数字的最小素因数。
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.
为了计算每个数字的最小素因数，我们将使用eratosthenes的筛子。在原始的Sieve中，每当我们将数字标记为非质数时，我们都会为该数字存储相应的最小质数(请参阅本文以获得更好的理解)。
上述方法的实现如下：
C++
// C++ program to find prime factorization upto n natural number
// O(n*Log n) time with precomputation
#include
using namespace std;
#define MAXN 100001
// Stores smallest prime factor for every number
int spf[MAXN];
// Adjacency vector to store distinct prime factors
vectoradj[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 Java
// 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> adj =
new ArrayList>();
// 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());
}
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
# 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)
# Prthe 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 29C#
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> adj = new List>();
// 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());
}
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 输出
Distinct prime factor for first 10 natural number : 0 1 1 1 1 2 1 1 1 2