Reputation: 23
Im currently trying to create a program that prints the prime numbers from 0 to 10,000 using only for,do while and ifs. I created this program but it doesn't runs
static void Main(string[] args)
{
for (int x = 2; x < 10000; x++)
{
for (int y = 1; y < x; y++)
{
if (x % y != 0)
{
Console.WriteLine(x);
}
}
Console.ReadKey();
}
I don't know where the problem is and also if the for inside resets.
Upvotes: 2
Views: 31304
Reputation: 17
Here is my code where you can generate and print the prime numbers between two numbers (in between string_starting_number and string_last_number). The lowest possible value for string_starting_number is 0 and the highest possible value for string_last_number is decimal.MaxValue-1=79228162514264337593543950334 and not 79228162514264337593543950335 because of the decimal_a++ command inside a for loop which will result to an overflow error.
Take note that you should input the values in string type in string_starting_number and in string_last_number.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace GeneratingPrimeNumbers
{
class Program
{
static void Main(string[] args)
{
string string_starting_number = "1"; //input here your choice of starting number
string string_last_number = "10"; //input here your choice of last number
decimal decimal_starting_number = Convert.ToDecimal(string_starting_number);
decimal decimal_last_number = Convert.ToDecimal(string_last_number);
string primenumbers = "";
ulong ulong_b;
ulong ulong_c;
if (decimal_starting_number <= ulong.MaxValue)
{
ulong ulong_starting_number = Convert.ToUInt64(decimal_starting_number);
ulong ulong_last_number;
if (decimal_last_number > ulong.MaxValue)
{
ulong_last_number = ulong.MaxValue;
}
else
{
ulong_last_number = Convert.ToUInt64(decimal_last_number);
}
if (ulong_starting_number == 0 || ulong_starting_number == 1 || ulong_starting_number == 2 || ulong_starting_number == 3)
{
primenumbers = 2 + " " + 3;
ulong_starting_number = 5;
}
if (ulong_starting_number % 2 == 0)
{
ulong_starting_number++;
}
ulong ulong_a;
for (ulong_a = ulong_starting_number; ulong_a <= ulong_last_number; ulong_a += 2)
{
ulong_b = Convert.ToUInt64(Math.Ceiling(Math.Sqrt(ulong_a)));
for (ulong_c = 3; ulong_c <= ulong_b; ulong_c += 2)
{
if (ulong_a % ulong_c == 0)
{
goto next_value_of_ulong_a;
}
}
primenumbers = primenumbers + " " + ulong_a;
next_value_of_ulong_a:
{
}
}
}
if (decimal_last_number > ulong.MaxValue)
{
string ulong_maximum_value_plus_two = "18446744073709551617";
if (decimal_starting_number <= ulong.MaxValue)
{
decimal_starting_number = Convert.ToDecimal(ulong_maximum_value_plus_two);
}
if (decimal_starting_number % 2 == 0)
{
decimal_starting_number++;
}
decimal decimal_a;
for (decimal_a = decimal_starting_number; decimal_a <= decimal_last_number; decimal_a += 2)
{
ulong_b = Convert.ToUInt64(Math.Ceiling(Math.Sqrt(ulong.MaxValue) * Math.Sqrt(Convert.ToDouble(decimal_a / ulong.MaxValue))));
for (ulong_c = 3; ulong_c <= ulong_b; ulong_c += 2)
{
if (decimal_a % ulong_c == 0)
{
goto next_value_of_decimal_a;
}
}
primenumbers = primenumbers + " " + decimal_a;
next_value_of_decimal_a:
{
}
}
}
Console.WriteLine(primenumbers);
Console.ReadKey();
}
}
}
Upvotes: 0
Reputation: 69
Here is simple logic to Print Prime No for any upper limit.
Input : 10 Output : 2 , 3 , 5 ,7
namespace PurushLogics
{
class Purush_PrimeNos
{
static void Main()
{
//Prime No Program
bool isPrime = true;
Console.WriteLine("Enter till which number you would like print Prime Nos\n");
int n = int.Parse(Console.ReadLine());
Console.WriteLine("Prime Numbers : ");
for (int i = 2; i <= n; i++)
{
for (int j = 2; j <= n; j++)
{
if (i != j && i % j == 0)
{
isPrime = false;
break;
}
}
if (isPrime)
{
Console.Write("\t" + i);
}
isPrime = true;
}
Console.ReadKey();
}
}
}
Upvotes: 0
Reputation: 3834
Try this with no bool
variable!!!:
static void Main(string[] args)
{
for (int x = 2; x < 10000; x++)
{
int isPrime = 0;
for (int y = 1; y < x; y++)
{
if (x % y == 0)
isPrime++;
if(isPrime == 2) break;
}
if(isPrime != 2)
Console.WriteLine(x);
isPrime = 0;
}
Console.ReadKey();
}
Check Console.ReadKey();
it should be after upper for
loop, you can even change condition for upper for
loot with <=
since 10000
also need to check for prime
condition.
Upvotes: 5
Reputation: 2355
Below is the efficient way to print prime numbers between 0 and 10000
using System.IO;
using System;
class Program
{
static void Main()
{
Console.WriteLine("Below are prime numbers between 0 and 10000!");
Console.WriteLine(2);
for(int i=3;i<=10000;i++)
{
bool isPrime=true;
for(int j=2;j<=Math.Sqrt(i);j++)
{
if(i%j==0)
{
isPrime=false;
break;
}
}
if(isPrime)
{
Console.WriteLine(i);
}
}
}
}
Upvotes: 3
Reputation: 882626
The first problem is that x % 1
will always be zero, at least for non-zero x
. You need to start the test (inner) loop at one and, for efficiency, stop when you've exceeded the square root of the number itself - if n
has a factor f
where f > sqrt(n)
, you would have already found the factor n / f
.
The second problem is that you will write out a candidate number every time the remainder is non-zero. So, because 15 % 4
is three, it will be output despite the fact that fifteen is very much a non-prime. It will also be output at 15 % 2
, 15 % 4
, 15 % 6
, 15 % 7
, and so on.
The normal (naive) algorithm for prime testing is:
# All numbers to test.
foreach number 2..whatever:
# Assume prime, check all numbers up to squareroot(number).
isPrime = true
foreach test 2..infinity until test * test > number:
# If a multiple, flag as composite and stop inner loop.
if number % test == 0:
isPrime = false
exit foreach
end
end
# If never flagged as composite, output as prime.
if isPrime:
output number
end
Upvotes: 2
Reputation: 21
Is there any reason that you put Console.ReadKey(); inside of loop?
You should put that out of the loop unless press key during loop.
static void Main(string[] args)
{
for (int x = 2; x < 10000; x++)
{
for (int y = 1; y < x; y++)
{
if (x % y != 0)
{
Console.WriteLine(x);
}
}
}
Console.ReadKey();
}
And probably that code is just print lots of x. You should to fix it.
Upvotes: 2