+ 1

# Prime numbers

How do I find prime numbers less than a known number?

5 Answers

+ 4

Roman J I haven't considered time complexity. If that's so, then for loop to sqrt(n) will be even more efficient
This will give time complexity of O(sqrt(n))

+ 3

Look here, this should work perfectly:
https://www.sololearn.com/learn/969/?ref=app

+ 2

#include <iostream>
using namespace std;
int main()
{
int n, i;
bool isPrime = true;
cout << "Enter a positive integer: ";
cin >> n;
for(i = 2; i <= n / 2; ++i)
{
if(n % i == 0)
{
isPrime = false;
break;
}
}
if (isPrime)
cout << "This is a prime number";
else
cout << "This is not a prime number";
return 0;
}

+ 1

Make a Prime() function which contains a for loop i in the range between 2 and a num (say n). It will return False if n%i = = 0.
and then
return True. // will execute only if the above for loop doesn't return anything //
In the main function use another for loop in the range 2 to the input num (say p) . so primes are present between 2 and p.
int main()
{
for (i= 2 ; i<p;i++)
{
if( Prime(i) ==True)
cout<

0

1 is noy a prime number , but in ur code the opposit

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