//I have a problem, i trying to get the total prime numbers for each row on the output.
//I jus getting the total prime numbers from the firs line 1 to 100
// my task is to get the total prime number from 1 to 100, 1001 to 2000.....49001 to 5000
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;

bool find_prime(long);
//long primeCount (long x, long y);

int main ()
{
	int i, counter =0;
	bool test;


	for (i =1; i < 1000; i++ )
	{
        
		test = find_prime(i);
		if ( test == true )
		{
		
			counter++;
		
		}
		
}	
	cout << "Start" <<setw(6) << "End" << setw(24) << "Number of Primes" << endl;
  
  
  for (i; i <= 50000; i = i + 1000)
 
{  
      	
    cout << i-999 <<setw(10) << i << setw(11) << counter << endl;
 
  	
}
   cout << "Total primes in the first 50 chiliads:" << counter << endl;
    
    cout << "Average number per chiliad:" << counter/1 << endl;
    system("pause");
}

bool find_prime (long num)
{
 int j;
 
 if(num <= 1)
 return false;
 
 for ( j = 2; j*j <= num; j++ )
 
	
	if (num%j == 0)
	
	return false; 
 return true;
 

 }

Here is u r answer..

#include<stdio.h>
#include<conio.h>
#include<iostream>
using namespace std;
void TestPrime(int );
void TestPrime(int n)
{
     int no=n;
     int i,j,p,flag=0,k;
     if(no==1)
     {
        printf("1 is prime");      
     }
     for ( i = 2; i<no && !flag; i++)
     {
          if(no%i==0)
          {
             flag=1;
          }
     }
        if(flag==0)
       {       
         cout<<"\nno is"<<i<<"is Prime No";
          
         }
        else
        {
            //cout<<"\nno is"<<i;
              // cout<<"\nNot Prime";    
         }
}
int main()
{
    cout<<"Hello";
    int i,j,n;
    int a[10];
    for (i=1; i<100; i++)
    {
        TestPrime(i);
      
    }
    getch();	
  
}
#include<stdio.h>
#include<conio.h>
#include<iostream>
using namespace std;
void TestPrime(int );
int total=0;
void TestPrime(int n)
{
     int no=n;
     int i,j,p,flag=0,k;
    if(no==1)
     {
         
     }
     for ( i = 2; i<no && !flag; i++)
     {
          if(no%i==0)
          {
             flag=1;
          }
     }
        if(flag==0)
       {       
        // cout<<"\nno is"<<i<<"is Prime No";
         total++;
          
         }
        else
        {
            //cout<<"\nno is"<<i;
              // cout<<"\nNot Prime";    
         }
}
int main()
{
    
    int i,j,n;
    int a[10];
    for (i=1; i<100; i++)
    {
        TestPrime(i);
      
    }
    printf("Total Prime no upto 100 is:-> %d",total-1);
    getch();	
  
}

Here is u r answer..

L33t speak is against the rules, and so is giving away free code. Read this.

Now for your code...

void TestPrime(int );
void TestPrime(int n)
{

Why is the first line there? You don't have to declare the function if you're going to define it right away. Lose the first line.

if(no==1)
     {
        printf("1 is prime");      
     }

I wouldn't call 1 a prime, but there's still somewhat discussion on the subject. Anyway: why didn't you just include the number 1 is this loop:

for ( i = 2; i<no && !flag; i++)
     {

By changing the '2' to '1', the first if statement is obsolete.

if(no%i==0)
          {
             flag=1;
          }

If this function isn't going to return anything, but only print if it's a prime or not; why not print it here right away? Saves you an extra if statement.

[edit] And don't get me started on your second code

So at the OP: You might want to have a look at the logic from the post above, but ctrl-c ctrl-v is not a good idea

at OP: You need nested loops. One that counts from 0-1000 and one that run the first loop 50 times (50 * 1000 = 50000)

Using your own code, it would be something like:

cout << "Start" <<setw(6) << "End" << setw(24) << "Number of Primes\n";
for ( int k = 0; k < 50000; k+=1000)
{
   for (i =1; i < 1000; i++ )
   {
	test = find_prime(i+k);
   	if ( test == true )
	{
	    counter++;
	}
   }
   cout << i+k-999 <<setw(10) << i+k << setw(11) << counter << endl;
   counter = 0;
}

You'll have to declare one more var, to hold the total number of primes

To check Wheather a no 'n' is prime or not why not check if 'n' is divided by the prime numbers from 2 to sqrt of the number 'n'?
It would prove better

This article has been dead for over six months. Start a new discussion instead.