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Find count of Almost Prime numbers from 1 to N in C++
Suppose we have a number N. We have to find almost prime numbers in range 1 to N. A number is called almost prime when it has exactly two distinct factors. The numbers can have any number of non-prime factors, but should be two prime factors. So if N is 2, then output will be 2. There are two numbers 6 and 10.
Here we will use the Sieve of Eratosthenes approach. Please check the following implementation to get better idea.
Example
#include<iostream>
#define N 100005
using namespace std;
bool prime[N];
void SieveOfEratosthenes() {
for(int i = 0; i<N; i++)
prime[i] = true;
prime[1] = false;
for (int i = 2; i * i < N; i++) {
if (prime[i] == true) {
for (int j = i * 2; j < N; j += i)
prime[j] = false;
}
}
}
int countAlmostPrime(int n) {
int result = 0;
for (int i = 6; i <= n; i++) {
int div_count = 0;
for (int j = 2; j * j <= i; j++) {
if (i % j == 0) {
if (j * j == i) {
if (prime[j])
div_count++;
}else {
if (prime[j])
div_count++;
if (prime[i / j])
div_count++;
}
}
}
if (div_count == 2)
result++;
}
return result;
}
int main() {
SieveOfEratosthenes();
int n = 21;
cout << "Number of almost primes in range 1 to "<<n << " is: " << countAlmostPrime(n);
}Output
Number of almost primes in range 1 to 21 is: 8
Updated on: 18-Dec-2019
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