Printing prime numbers from 1 through 100

Question

This c++ code prints out the following prime numbers: 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97.

But I don't think that's the way my book wants it to be written. It mentions something about square root of a number. So I did try changing my 2nd loop to for (int j=2; j<sqrt(i); j++) but it did not give me the result I needed.

How would I need to change this code to the way my book wants it to be?

int main () 
{
    for (int i=2; i<100; i++) 
        for (int j=2; j<i; j++)
        {
            if (i % j == 0) 
                break;
            else if (i == j+1)
                cout << i << " ";

        }   
    return 0;
}

A prime integer number is one that has exactly two different divisors, namely 1 and the number itself. Write, run, and test a C++ program that finds and prints all the prime numbers less than 100. (Hint: 1 is a prime number. For each number from 2 to 100, find Remainder = Number % n, where n ranges from 2 to sqrt(number). \ If n is greater than sqrt(number), the number is not equally divisible by n. Why? If any Remainder equals 0, the number is no a prime number.)

Answer 1

Three ways:

1.

int main () 
{
    for (int i=2; i<100; i++) 
        for (int j=2; j*j<=i; j++)
        {
            if (i % j == 0) 
                break;
            else if (j+1 > sqrt(i)) {
                cout << i << " ";

            }

        }   

    return 0;
}

2.

int main () 
{
    for (int i=2; i<100; i++) 
    {
        bool prime=true;
        for (int j=2; j*j<=i; j++)
        {
            if (i % j == 0) 
            {
                prime=false;
                break;    
            }
        }   
        if(prime) cout << i << " ";
    }
    return 0;
}

3.

#include <vector>
int main()
{
    std::vector<int> primes;
    primes.push_back(2);
    for(int i=3; i < 100; i++)
    {
        bool prime=true;
        for(int j=0;j<primes.size() && primes[j]*primes[j] <= i;j++)
        {
            if(i % primes[j] == 0)
            {
                prime=false;
                break;
            }
        }
        if(prime) 
        {
            primes.push_back(i);
            cout << i << " ";
        }
    }

    return 0;
}

Edit: In the third example, we keep track of all of our previously calculated primes. If a number is divisible by a non-prime number, there is also some prime <= that divisor which it is also divisble by. This reduces computation by a factor of primes_in_range/total_range.

Answer 2

If j is equal to sqrt(i) it might also be a valid factor, not only if it's smaller.

To iterate up to and including sqrt(i) in your inner loop, you could write:

for (int j=2; j*j<=i; j++)

(Compared to using sqrt(i) this has the advantage to not need conversion to floating point numbers.)

Answer 3

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.

Answer 4

This is my very simple c++ program to list down the prime numbers in between 2 and 100.

for(int j=2;j<=100;++j)
{
    int i=2;
    for(;i<=j-1;i++)
    {
        if(j%i == 0)
            break;
    }

    if(i==j && i != 2)
        cout<<j<<endl;
}

Answer 5

Using Sieve of Eratosthenes logic, I am able to achieve the same results with much faster speed.

My code demo VS accepted answer.

Comparing the count, my code takes significantly lesser iteration to finish the job. Checkout the results for different N values in the end.

Why this code performs better than already accepted ones:

- the even numbers are not checked even once throughout the process.

- both inner and outer loops are checking only within possible limits. No extraneous checks.

Code:

int N = 1000; //Print primes number from 1 to N
vector<bool> primes(N, true);
for(int i = 3; i*i < N; i += 2){    //Jump of 2
    for(int j = 3; j*i < N; j+=2){  //Again, jump of 2
        primes[j*i] = false;
    }
}
if(N >= 2) cout << "2 ";
for(int i = 3; i < N; i+=2){        //Again, jump of 2
    if(primes[i] == true) cout << i << " "; 
}

For N = 1000, my code takes 1166 iterations, accepted answer takes 5287 (4.5 times slower)

For N = 10000, my code takes 14637 iterations, accepted answer takes 117526 (8 times slower)

For N = 100000, my code takes 175491 iterations, accepted answer takes 2745693 (15.6 times slower)

Answer 6

actually the better solution is to use "A prime sieve or prime number sieve" which "is a fast type of algorithm for finding primes" .. wikipedia

The simple (but not faster) algorithm is called "sieve of eratosthenes" and can be done in the following steps(from wikipedia again):

1. Create a list of consecutive integers from 2 to n: (2, 3, 4, ..., n).
2. Initially, let p equal 2, the first prime number.
3. Starting from p, count up in increments of p and mark each of these numbers greater than p itself in the list. These numbers will be 2p, 3p, 4p, etc.; note that some of them may have already been marked.
4. Find the first number greater than p in the list that is not marked. If there was no such number, stop. Otherwise, let p now equal this number (which is the next prime), and repeat from step 3.

Answer 7

Finding primes up to a 100 is especially nice and easy:

    printf("2 3 ");                        // first two primes are 2 and 3
    int m5 = 25, m7 = 49, i = 5, d = 4;
    for( ; i < 25; i += (d=6-d) )
    {
        printf("%d ", i);                  // all 6-coprimes below 5*5 are prime
    }
    for( ; i < 49; i += (d=6-d) )
    {
        if( i != m5) printf("%d ", i);
        if( m5 <= i ) m5 += 10;            // no multiples of 5 below 7*7 allowed!
    }
    for( ; i < 100; i += (d=6-d) )         // from 49 to 100,
    {
        if( i != m5 && i != m7) printf("%d ", i);
        if( m5 <= i ) m5 += 10;            //   sieve by multiples of 5,
        if( m7 <= i ) m7 += 14;            //                       and 7, too
    }

The square root of 100 is 10, and so this rendition of the sieve of Eratosthenes with the 2-3 wheel uses the multiples of just the primes above 3 that are not greater than 10 -- viz. 5 and 7 alone! -- to sieve the 6-coprimes below 100 in an incremental fashion.

Answer 8

It's fine to change your for loop to for (int j=2; j<=sqrt(i); j++) but then you also need to change something else. Looking specifically at your print condition,

else if (i == j+1) {
      cout << i << " ";
}

why will that never be triggered if you only iterate up to sqrt(i)? Where can you move the cout to to change this? (Hint: you may want to move the print out of the loop and then make use of some type of flag variable)

Answer 9

I check if a number is prime or not with the following code( of course using sqrt ):

bool IsPrime(const unsigned int x)
{
  const unsigned int TOP
  = static_cast<int>(
      std::sqrt( static_cast<double>( x ) )
    ) + 1;

  for ( int i=2; i != TOP; ++i )
  {
    if (x % i == 0) return false;
  }
  return true;
}

I use this method to determine the primes:

#include <iostream>

using std::cout;
using std::cin;
using std::endl;

#include <cmath>

void initialize( unsigned int *, const unsigned int );
void show_list( const unsigned int *, const unsigned int );
void criba( unsigned int *, const unsigned int );
void setItem ( unsigned int *, const unsigned int, const unsigned int );

bool IsPrime(const unsigned int x)
{
  const unsigned int TOP
  = static_cast<int>(
      std::sqrt( static_cast<double>( x ) )
    ) + 1;

  for ( int i=2; i != TOP; ++i )
  {
    if (x % i == 0) return false;
  }
  return true;
}

int main()
{

    unsigned int *l;
    unsigned int n;

    cout << "Ingrese tope de criba" << endl;
    cin >> n;

    l = new unsigned int[n];

    initialize( l, n );

    cout << "Esta es la lista" << endl;
    show_list( l, n );

    criba( l, n );  

    cout << "Estos son los primos" << endl;
    show_list( l, n );
}

void initialize( unsigned int *l, const unsigned int n)
{
    for( int i = 0; i < n - 1; i++ )
        *( l + i ) = i + 2;
}

void show_list( const unsigned int *l, const unsigned int n)
{
    for( int i = 0; i < n - 1; i++ )
    {
        if( *( l + i ) != 0)
            cout << l[i] << " - ";
    }
    cout << endl;
}

void setItem( unsigned int *l, const unsigned int n, const unsigned int p)
{
    unsigned int i = 2;
    while( p * i <= n)
    {
        *( l + (i * p - 2) ) = 0;
        i++;
    }
}

void criba( unsigned int *l, const unsigned int n)
{
    for( int i = 0;  i * i <= n ; i++ )
     if( IsPrime ( *( l + i) ) )
        setItem( l, n, *(l + i) );      
}

Answer 10

The book seems to be "C++ for Engineers and Scientists" written by Gary Bronson (googled it).
Is this a possible answer? IMHO it's surprising.

I had to read the question (from the book) a few times. My interpretation:
For each number N: 2 <= N < 100 check whether it's prime.
How? For each divisor D: 2 <= D < sqrt(N) ,
if D divides N, N is not prime, if D > sqrt(N), N is prime.

Give it a try:

N = 2, sqrt(2) ≈ 1.41, D = 2, 2 < 1.41 ?  no 2 > 1.41 ? yes 2 is prime.  
N = 3, sqrt(3) ≈ 1.73, D = 2, 2 < 1.73 ?  no 2 > 1.73 ? yes 3 is prime.  
N = 4, sqrt(4) = 2.00, D = 2, 2 < 2.00 ?  no 2 > 2.00 ?  no 4 is not prime.  
N = 5, sqrt(5) ≈ 2.24, D = 2, 2 < 2.24 ? yes 5 % 2 > 0? yes  
                       D = 3, 3 < 2.24 ?  no 3 > 2.24 ? yes 5 is prime.    
N = 6, sqrt(6) ≈ 2.45, D = 2, 2 < 2.45 ? yes 6 % 2 = 0  2 > 2.45 ? no 6 is not prime.

As far as I can see, that's how the primes should be found,
not with a sieve (much, much faster),
but with: the answer is in the question! Surprising?

Speed? primes < 400,000 : less than 10 seconds (on my watch, a rolex, I bought it on the market, the seller said it was a real one, a real one for the price of two baguettes, with 12 real diamonds as well).
Let's count the primes (I'm not going to show code ;) : 664579 primes < 10,000,000 : 5 seconds.

#include "stdafx.h"
#include <math.h>
#include <iostream>
using namespace std;
int main()
{
    double rt;
    for (int d = 2, n = 2; n < 100; d = 2, n++)
    {
        for (rt = sqrt(n); d < rt; d++)
            if (n % d == 0) break;
        if (d > rt) cout << n << " ";
    }
    getchar();  // 25 primes :-)
}

Deleted an earlier answer with (like other answers) a prime-sieve.
Hopefully I get my next "Necromancer" badge soon.

I asked the author: In your book: "C++ for E&S" is an exercise about prime numbers,[xrcs]...[/xrcs]. Seven years ago it was asked at: SO/q/5200879
A few days ago I gave an answer: SO/a/49199435
Do you think it is a reasonable solution, or perhaps the solution.

He replied: Peter, I never really have a specific solution in mind when I am making up the exercises,
so I can’t say I had your exact solution in mind. The joy of C++ is that one can come up with really creative solutions and great code, as, on first glance, it looks like you have done.
Thanks for sending it!
Dr. Bronson

I went to https://youtu.be/1175axY2Vvw

PS. A sieve: https://pastebin.com/JMdTxbeJ

Answer 11

Here is my implementation of Sieve of Eratosthenes (for primes between 2 & n)

#include <iostream>

int main (){
int n=0;
std::cout << "n = ";
std::cin >> n;
std::cout << std::endl;

if (n==0 || n==1){
    std::cout << "No primes in this range" << std::endl;
    return 0;
}


const int array_len = n-2+1;

int the_int_array[array_len];
for (int counter=2; counter <=n; counter++)
    the_int_array[counter-2]=counter;


int runner = 0;
int new_runner = 0;

while (runner < array_len ){
    if (the_int_array[runner]!=0){
        new_runner = runner;
        new_runner = new_runner + the_int_array[runner];

        while (new_runner < array_len){
           the_int_array[new_runner] = 0;
           new_runner = (new_runner + the_int_array[runner]);
        }
    }
runner++;
}

runner = 0;

while (runner < array_len ){
    if (the_int_array[runner]!=0)
        std::cout << the_int_array[runner] << " ";
    runner++;
}

std::cout << std::endl;
return 0;

}

Answer 12

#include "stdafx.h"
#include<iostream>
using namespace std;
void main()
{ int f =0;
 for(int i=2;i<=100;i++)
  {
   f=0;
   for(int j=2;j<=i/2;j++)
   { 
     if(i%j==0)
     { f=1;
       break;
     }
   }
 if (f==0)
  cout<<i<<" ";
}
 system("pause");
}

Answer 13

this is my approach from a simple blog:

//Prime Numbers generation in C++
//Using for loops and conditional structures
#include <iostream>
using namespace std;

int main()
{
int a = 2;       //start from 2
long long int b = 1000;     //ends at 1000

for (int i = a; i <= b; i++)
{

 for (int j = 2; j <= i; j++)
 {
    if (!(i%j)&&(i!=j))    //Condition for not prime
        {
            break;
        }

    if (j==i)             //condition for Prime Numbers
        {
              cout << i << endl;

        }
 }
}
}

- See more at: http://www.programmingtunes.com/generation-of-prime-numbers-c/#sthash.YoWHqYcm.dpuf

Answer 14

I always use this one (it's easy and fast) :

#include <iostream>
using namespace std;

int i,j;
bool b[101];

int main( )
{
    for(i=2;i<101;i++){
        b[i]=true;
    }
    for(i=1;i<101;i++){
        if(b[i]){
            cout<<i<<" ";
            for(j=i*2;j<101;j+=i) b[j]=false;
        }
    }
}

Here is output of this code: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

Answer 15

To find whether no. is prime or not C++:

#include<iostream>
#include<cmath>

using namespace std;
int main(){

int n, counter=0;

cout <<"Enter a number to check whether it is prime or not \n";
cin>>n;

  for(int i=2; i<=n-1;i++) {
    if (n%i==0) {
      cout<<n<<" is NOT a prime number \n";
      break;
    }
    counter++;
                }
    //cout<<"Value n is "<<n<<endl;
    //cout << "number of times counter run is "<<counter << endl;
    if (counter == n-2)
      cout << n<< " is prime \n";
   return 0;
}

Answer 16

I did it in perl based on the most popular answer by ProdigySim's 2nd method. I had to add a break equivalent in perl, last, right after the print $i . " \n"; to avoid outputting the primes twice.

#!/bin/perl
use strict;

for(my $i=2; $i < 100; $i++){

    my $prime = 1;

    for (my $j=2; $j*$j<=$i; $j++){
        if ($i % $j == 0){
            $prime = 0;
            last;
        }
        if($prime) {
            print $i . " \n";
            last;
        }
    }

}

Answer 17

A simple program to print "N" prime numbers. You can use N value as 100.

    #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 18

here is a simple code for printing all the prime numbers until given number n,

#include<iostream.h>
#include<conio.h>

void main()
{
clrscr();
int n,i,j,k;
cout<<"Enter n\n";
cin>>n;

for(i=1;i<=n;i++)
{   k=0;
  for(j=1;j<=i;j++)
  {
    if((i%j)==0)
    k++;
   }
  if(k==2)
  cout<<i<<endl;
}
getch();
}

Answer 19

Using the rules of divisibility prime numbers can be found out in O(n) and it`s really effecient Rules of Divisibility

The solution would be based on the individual digits of the number...

Answer 20

While this is relatively more production grade prime number generator, it is still valid to use for finding prime numbers from 1 through 100. The code uses Miller-Rabin Primality Test to achieve calculate prime numbers. Since it is probabilistic method, the accuracy increases with value of k. While the focus is on readability of code rather than speed, on AWS r5.2xlarge instance it took 3.791s for prime number until 1,000,000.

// C++ program to print all primes smaller than or equal to 
// n using Miller-Rabin Primality Test
// Reference: https://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test
// It is not particularly to optimized 
// since focus is readability
// Compile: g++  -std=c++17 -o prime prime.c++ -ltbb 

#include <execution>
#include <iostream>
#include <math.h>
using namespace std; 

int power(unsigned long int x, unsigned long y, unsigned long p){
    int res = 1;
    x = x % p;
    while (y > 0) {
        if (y & 1)
            res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}

bool millerTest(unsigned long d, unsigned long n) {
    unsigned long a = 2 + rand () % (n - 4);

    unsigned long x = power(a, d, n);

    if (x == 1  || x == n - 1)
        return true;

    while (d != n - 1){
        x = (x * x) % n;
        d *= 2;
        if (x == 1) return false;
        if (x == n - 1) return true;

    }
    return false;
}

bool isPrime(unsigned long n, int k) {
    if (n <= 1 || n == 4) return false;
    if (n <= 3) return true;

    unsigned long int d = n - 1;
    while (d % 2 == 0)
        d /= 2;
    for(int i = 0; i < k; i++){
        if (!millerTest(d, n))
            return false;
    }
    return true;
}




int main() 
{ 
    int n = 1000000;
    int k = 200; 
    vector<unsigned long> primeN(n);
    iota(primeN.begin(), primeN.end(), 1);

    vector<bool> isPrimeV(n);



    transform(execution::par,
        primeN.begin(), primeN.end(), 
        isPrimeV.begin(), 
        [k](unsigned long x) -> bool {
            return isPrime(x, k);
        });

    int count = accumulate(isPrimeV.begin(), isPrimeV.end(), 0, [](int d, bool v){
        if (v == true) return d += 1; else return d;
    });
    cout << count << endl;

    return 0; 
} 

Answer 21

#include<iostream>
using namespace std;
void main()
{
        int num,i,j,prime;
    cout<<"Enter the upper limit :";
    cin>>num;
    cout<<"Prime numbers till "<<num<<" are :2, ";

    for(i=3;i<=num;i++)
    {
        prime=1;
        for(j=2;j<i;j++)
        {
            if(i%j==0)
            {
                prime=0;
                break;
            }
        }
        if(prime==1)
            cout<<i<<", ";
    }
}

Answer 22

Just try this. It's easy without any extra builtin functions.

#include <iostream>

int prime(int n,int r){

  for(int i=2;n<=r;i++){
    if(i==2 || i==3 || i==5 || i==7){
      std::cout<<i<<" ";
      n++;
    } else if(i%2==0 || i%3==0 || i%5==0 || i%7==0)
      continue;
    else {
      std::cout<<i<<" ";
      n++;
    }
  }

}

main(){

  prime(1,25);
}

Testing by 2,3,5,7 is good enough for up to 120, so 100 is OK.

There are 25 primes below 100, an 30 below 121 = 11*11.