number of prime numbers till n java code example
Example 1: java prime numbers from 1 to 100
public class Main {
public static void main(String[] args) {
for (int i = 2; i < 100; ++i) {
boolean isPrime = true;
int sqrt = (int)Math.ceil(Math.sqrt(i));
for (int divisor = 2; divisor <= sqrt; ++divisor) {
if (i % divisor == 0) {
isPrime = false;
break;
}
}
if (isPrime) {
System.out.println(i);
}
}
}
}
Example 2: generate all prime number less than n java
/**
Author: Jeffrey Huang
As far as I know this is almost the fastest method in java
for generating prime numbers less than n.
A way to make it faster would be to implement Math.sqrt(i)
instead of i/2.
I don't know if you could implement sieve of eratosthenes in
this, but if you could, then it would be even faster.
If you have any improvements please email me at
[email protected].
*/
import java.util.*;
public class Primecounter {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
//int i =0;
int num =0;
//Empty String
String primeNumbers = "";
boolean isPrime = true;
System.out.print("Enter the value of n: ");
System.out.println();
int n = scanner.nextInt();
for (int i = 2; i < n; i++) {
isPrime = true;
for (int j = 2; j <= i/2; j++) {
if (i%j == 0) {
isPrime = false;
}
}
if (isPrime)
System.out.print(" " + i);
}
}
}
Example 3: Java program to display prime numbers from 1 to 100
public class PrimeNumbersFrom1To100
{
public static void main(String[] args)
{
int maxNumber = 100;
boolean prime = true;
String primeNumbers = "";
for(int a = 1; a <= maxNumber; a++)
{
prime = checkPrime(a);
if(prime)
{
primeNumbers = primeNumbers + a + " ";
}
}
System.out.println("Display prime numbers from 1 to " + maxNumber + ": ");
System.out.println(primeNumbers);
}
public static boolean checkPrime(int number)
{
int remainder;
for(int a = 2; a <= number / 2; a++)
{
remainder = number % a;
// if remainder is equal to 0 then number is not prime
if(remainder == 0)
{
return false;
}
}
return true;
}
}