给定一个整数N ，任务是找到数字1到N的素因子的指数和。
例子：
Input: N = 4
Output: 4
Explanation:
Numbers up to 4 are 1, 2, 3, 4 where
The exponent of 1 in the prime factorization of 1 is 0 (20),
For 2 it is 1 (21),
For 3 it is 1 (31), and
For 4 it is 2 (22).
The sum of the exponent of prime factors of each number up to 4 is 0 + 1 + 1 + 2 = 4.
Input: N = 10
Output: 15
Explanation:
sum of the exponent of prime factors of each number up to 10 is 15.
方法：想法是使用素数因子及其幂的概念。步骤如下：
- 为每个数字从2到N进行迭代，并对每个数字执行以下操作：
- 求出每个数N的素因子的幂。
- 在上述步骤中找到每个功率的总和
- 打印N的素数的所有幂的和并打印和。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to implement sieve of
// erastosthenes
void sieveOfEratosthenes(int N, int s[])
{
// Create a boolean array and
// initialize all entries 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;
// Iterate 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
int generatePrimeFactors(int N)
{
// s[i] is going to store
// smallest prime factor of i
int s[N + 1];
int sum = 0;
sieveOfEratosthenes(N, s);
// Current prime factor of N
int curr = s[N];
// Power of current prime factor
int cnt = 1;
// Calculating prime factors
// and their powers sum
while (N > 1) {
N /= s[N];
if (curr == s[N]) {
// Increment the count and
// continue the process
cnt++;
continue;
}
// Add count to the sum
sum = sum + cnt;
curr = s[N];
// Reinitialize count
cnt = 1;
}
// Return the result
return sum;
}
// Function to find the sum of all the
// power of prime factors of N
void findSum(int N)
{
int sum = 0;
// Iterate for in [2, N]
for (int i = 2; i <= N; i++) {
sum += generatePrimeFactors(i);
}
cout << sum << endl;
}
// Driver Code
int main()
{
// Given Number N
int N = 4;
// Function Call
findSum(N);
return 0;
} Java
// Java program for the above approach
class GFG{
// Function to implement sieve of
// erastosthenes
static void sieveOfEratosthenes(int N, int s[])
{
// Create a boolean array and
// initialize all entries 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;
// Iterate 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 int generatePrimeFactors(int N)
{
// s[i] is going to store
// smallest prime factor of i
int []s = new int[N + 1];
int sum = 0;
sieveOfEratosthenes(N, s);
// Current prime factor of N
int curr = s[N];
// Power of current prime factor
int cnt = 1;
// Calculating prime factors
// and their powers sum
while (N > 1)
{
N /= s[N];
if (curr == s[N])
{
// Increment the count and
// continue the process
cnt++;
continue;
}
// Add count to the sum
sum = sum + cnt;
curr = s[N];
// Reinitialize count
cnt = 1;
}
// Return the result
return sum;
}
// Function to find the sum of all the
// power of prime factors of N
static void findSum(int N)
{
int sum = 0;
// Iterate for in [2, N]
for(int i = 2; i <= N; i++)
{
sum += generatePrimeFactors(i);
}
System.out.print(sum + "\n");
}
// Driver Code
public static void main(String[] args)
{
// Given Number N
int N = 4;
// Function Call
findSum(N);
}
}
// This code is contributed by Ajay KumarPython3
# Python3 program for the above approach
# Function to implement sieve of
# erastosthenes
def sieveOfEratosthenes(N, s):
# Create a boolean array and
# initialize all entries 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
# Iterate 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
j = i
while (j * i <= N):
if (prime[i * j] == False):
prime[i * j] = True
# i is the smallest
# prime factor for
# number "i*j"
s[i * j] = i
j += 2
# Function to generate prime
# factors and its power
def generatePrimeFactors(N):
# s[i] is going to store
# smallest prime factor of i
s = [0] * (N + 1)
sum = 0
sieveOfEratosthenes(N, s)
# Current prime factor of N
curr = s[N]
# Power of current prime factor
cnt = 1
# Calculating prime factors
# and their powers sum
while (N > 1):
N //= s[N]
if (curr == s[N]):
# Increment the count and
# continue the process
cnt += 1
continue
# Add count to the sum
sum = sum + cnt
curr = s[N]
# Reinitialize count
cnt = 1
# Return the result
return sum
# Function to find the sum of all the
# power of prime factors of N
def findSum (N):
sum = 0
# Iterate for in [2, N]
for i in range(2, N + 1):
sum += generatePrimeFactors(i)
print(sum)
# Driver Code
if __name__ == '__main__':
# Given number N
N = 4
# Function call
findSum(N)
# This code is contributed by himanshu77C#
// C# program for the above approach
using System;
class GFG{
// Function to implement sieve of
// erastosthenes
static void sieveOfEratosthenes(int N, int []s)
{
// Create a bool array and
// initialize all entries 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;
// Iterate 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 int generatePrimeFactors(int N)
{
// s[i] is going to store
// smallest prime factor of i
int []s = new int[N + 1];
int sum = 0;
sieveOfEratosthenes(N, s);
// Current prime factor of N
int curr = s[N];
// Power of current prime factor
int cnt = 1;
// Calculating prime factors
// and their powers sum
while (N > 1)
{
N /= s[N];
if (curr == s[N])
{
// Increment the count and
// continue the process
cnt++;
continue;
}
// Add count to the sum
sum = sum + cnt;
curr = s[N];
// Reinitialize count
cnt = 1;
}
// Return the result
return sum;
}
// Function to find the sum of all the
// power of prime factors of N
static void findSum(int N)
{
int sum = 0;
// Iterate for in [2, N]
for(int i = 2; i <= N; i++)
{
sum += generatePrimeFactors(i);
}
Console.Write(sum + "\n");
}
// Driver Code
public static void Main(String[] args)
{
// Given number N
int N = 4;
// Function call
findSum(N);
}
}
// This code is contributed by 29AjayKumar输出：
4
时间复杂度： O(N * log 2 N)
辅助空间： O(N)