C++ finding all prime numbers from 1 to a number entered - c++

Question

So the point is to have the program find and list all prime numbers between 1 and the number you enter. I'm using number_test as the number tested for prime, and divisor and the number to divide by.
I'm not sure what's wrong, as to me it looks functionally the same as the program posted here: Printing prime numbers from 1 through 100
with some minor changes (inputting a number, changing "i" to less than the number entered).
I've been looking for the past three or four days, and I haven't found anything that really answers this question fully, to the degree I need for class. Any help is much appreciated.
#include iostream
#include conio.h
using namespace std;
void main(void){
//Declare variables
int number_entered;
//Get inputs
cout << "This program lists all prime numbers from 1 through a positive number entered."
<< endl;
cout << "Please enter a positive integer."
<< endl;
cin >> number_entered;
cout << "Displaying all numbers from 1 to " << number_entered
<< endl
<< "Press any key to continue..."
<< endl;
getch();
for(int number_test = 2; number_test < number_entered; number_test++){
for(int divisor = 2; divisor < number_test; divisor++){
if(number_test % divisor == 0){
break;
}
else if(number_test % divisor != 0){
cout << number_test << " ";
break;
}
}
}
getch();
}

Answer 1

You should use the Sieve of Eratosthenes to compute the primes less than n. Begin by making a list of all numbers from 2 to the maximum desired prime n. Then, at each iterative step, the smallest remaining number that hasn't yet been considered is output and all of its multiples are crossed off the list.
function primes(n)
sieve := makeArray(2..n, True)
for p from 2 to n step 1
if sieve(p)
output p
for i from p*p to n step p
sieve[i] := False
This O(n log log n) algorithm is very fast; you should be able to compute the 78498 primes less than a million in less than a second.

Answer 2

A simple C++ Program to find the "N" prime numbers.
#include <iostream >
using namespace std;
int main()
{
int N;
cin >> N;
for (int i = 2; N > 0; ++i)
{
bool isPrime = true ;
for (int j = 2; j < i; ++j)
{
if (i % j == 0)
{
isPrime = false ;
break ;
}
}
if (isPrime)
{
--N;
cout << i << "\n";
}
}
return 0;
}

Answer 3

Just a small suggestion. Since prime numbers are odd, even numbers can be left out.
For example, in below loops, i and j increase by 2 (i +=2) instead of by 1 (i ++).
for (int i=3;i<=numberByUser; i+=2){
for (j=3;j<=i;j +=2){
if (i%j==0){
break;
}
}

Answer 4

i think in your answer any way one time the loop will terminated(i am talking about the loop checking the whether it is prime or not)once it comes out you don't know whether it made the break or not.So try to make a flag variable and check outside.I ope that will work
for(n=lower+1; n<upper; n++)
{
prime = 1;
for(i=2; i<n; i++)
if(n%i == 0)
{
prime = 0;
break;
}
if(prime)
printf("\n\n\t\t\t%d", n);
}

Answer 5

for(int number_test = 2; number_test < number_entered; number_test++){
for(int divisor = 2; divisor < number_test; divisor++){
if(number_test % divisor == 0){
break;
}
else if(number_test % divisor != 0){
cout << number_test << " ";
break;
}
}
}
The above code will not show you the prime numbers, it will just show you the number you entered if/when you run into a divisor that is not a factor of the number. For example, if you enter "9", you will start at 2, which is not a factor of 9, so you will show "9" (incorrectly) as a "prime", when it is not.
The easiest method for testing if a number is a prime is by checking all prime numbers below it's square root to see if they are factors of the given number. If none of them are (then none of the non-prime numbers below the given number will be either), the number is a prime number. If it has at least one prime factor less than or equal to it's square root, it is not prime.
Since you are looking to show all primes in a range of [0, X], you can simply check your list of factors as you go along (or do it in reverse, which is effectively what the Sieve of Eratosthenes does).

Answer 6

When my point was like your one, I wrote this code, it worked. Hope it will help you.
#include <cstdio>
#include <vector>
using namespace std;
vector <int> sn;
bool isPrime(int n) {
if (n <= 1) {
return 0;
}
if (n == 2) {
return true;
}
if (!(n % 2)) {
return false;
}
for (int i = 2; i*i <= n; i++) {
if (!(n % i)) {
return 0;
}
}
return 1;
}
void primeNumbers(int k) {
sn.push_back (2);
int i = 3, j = 1;
for ( ; j < k + 1; i += 2 && j++) {
if (isPrime(i)) {
sn.push_back(i);
}
}
}
int main() {
int i, k;
scanf("%d", &k);
primeNumbers(k);
for (i = 0; i < sn.size(); i++) {
printf("%d ", sn[i]);
}
return 0;
}

Answer 7

int getNumberOfPrimes(int N) {
bool *numbers = new bool[N-1]();
for (int i = 2; i <= N/2; ++i) {
if (numbers[i-2] == true) continue;
for (int j = i+i; j <= N; j = j+i) {
numbers[j-2] = true;
}
}
int count = 0;
for (int i = 0; i < (N-1); ++i) {
if (numbers[i] == false) ++count;
}
delete []numbers;
return(count);
}

Answer 8

Man I guess I have the simplest methode of this all. Hope it works for you!
#include < iostream >
using namespace std;
int main()
{
int n, i, j
cin>>n; //The max limith
for(i=2; i<=2; i++)
{
for(j=1; j<=i/2; j++)
if(i%j!=o)
cout<<i;
}
return 0;
}

Answer 9

If a number has divisors, at least one of them must be less than or equal to the square root of the number. When you check divisors, you only need to check up to the square root, not all the way up to the number being tested.