How do you write a c program to print all prime numbers from 1 to n by using while loop with output?

Answer 1

100% working

void main()

{

int n,i=1,j,c;

clrscr();

printf("Enter Range To Print Prime Numbers")

scanf("%d",&n);

printf("Prime Numbers Are Following;

while(i<=n)

{

c=0;

for(j=1;j<=i;j++)

{

if(i%j==0)

c++;

}

if(c==2)

printf("%d ",i)

i++;

}

getch();

}

Answer 2

#include<stdio.h>

void main()

{

int i, prime, lim_up, lim_low, n;

clrscr();

printf("\n\n\t ENTER THE LOWER LIMIT…: ");

scanf("%d", &lim_low);

printf("\n\n\t ENTER THE UPPER LIMIT…: ");

scanf("%d", &lim_up);

printf("\n\n\t PRIME NUMBERS ARE…: ");

for(n=lim_low+1; n<lim_up; 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);

}

getch();

}

Answer 3

#include<iostream>

bool is_prime (const size_t num)

{

// The first two primes are 2 and 3

switch (num)

{

case (0): case (1): return false;

case (2): case (3): return true;

}

// All even numbers > 3 are non-prime.

if (num%2==0) return false;

// Initialise a maximum divisor (the square root of num).

size_t max = (size_t) std::sqrt ((double) num);

// Add 1 if even, 2 if odd (so that max is odd).

max += (max%2) ? 2 : 1;

// Test every divisor in the half-closed range [3:max)

for (size_t div=3; div!=max; div+=2)

// Is the number evenly divisible?

if (num%div==0)

// Yes, it is non-prime.

return false;

// The number is definitely prime.

return true;

}

int main()

{

using std::cout;

cout << "Prime numbers in closed range [0:100]:\n";

for (size_t num=0; num<=100; ++num)

if (is_prime (num))

cout << num << '\n';

cout << std::endl;

}

Answer 4

#include<stdio.h>

#include<conio.h>

void main ()

{

int i,j,k;

clrscr();

for(i=50;i<100;i++)

{

k=1;

for(j=1;j<i;j++)

{

if(i%j==0)

{

k++;

}

}

if(k==2)

{

printf("\n %d is prime",i);

}

}

getch ();

}

Answer 5

#include

#include

#include

void main()

{

int i,j;

clrscr();

for(i=3;i<=1000;i++)

{

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

{

if(i%j==0)

break;

}

if(i==j)

cout<

}

getch(); // this is the easiest method to print prime nos made by Taabi

}

Answer 6

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

int main(int argc, char **argv)

{

int iNum,iRem,iCount=2;

printf("enter the no");

scanf("%d",&iNum);

if(iNum==1)

{

printf("The no. is not prime");

exit(0);

}

if(iNum==2)

{

printf("The no. is prime");

exit(0);

}

do

{

iRem=iNum%iCount;

if(iRem==0)

{

printf("the given no is not prime");

exit(0);

}

iCount++;

}

while(iCount<=(iNum/2));

printf("the given no is prime");

}

Answer 7

bool is_prime (unsigned);

for (unsigned n=1; n<=50; ++n) if (is_prime (n)) std::cout << n << " is prime" << std::endl;

You will have to implement the is_prime() function yourself. The following is a reasonably efficient but simple implementation given the narrow range of values to be considered. A larger range will require a more efficient implementation.

bool is_prime (unsigned n) {

if (n<2) return false; // 0 and 1 are not prime

if !(n%2)) return n==2; // 2 is the only even prime

// test all odd divisors from 3 up to the square root of n

unsigned max {(unsigned) std::sqrt ((double) n))};

for (unsigned div=3; div<=max; div+=2) if (!(n%div)) return false;

// if we get this far, n is prime

return true;

}

Answer 8

// Returns true if n is prime, false if n is composite (non-prime). bool is_prime (unsigned n) {

if (n<2) return false; // 2 is the first prime

if (!(n%2)) return n==2; // 2 is the only even prime

unsigned max = (unsigned) sqrt ((double) n); // calculate max factor

for (unsigned f=3; f<=max; f+=2) if (!(n%f)) return false; // n is composite (it has a prime factor)

return true; // n is prime

}

// Returns the next prime greater than n.

unsigned next_prime (unsigned n) {

if (n<2) return 2; // 2 is the first (and only) even prime

if (n==2) return 3; // 3 is the first odd prime

if (!(n%2))++n; else n+=2; // increment n, ensuring n is odd

while (!is_prime(n)) n+=2; // check every odd value until n is prime

return n;

}

void print_primes (unsigned max) {

unsigned n=0;

printf ("List of primes up to %u\n", max);

while ((n=next_prime (n))<=max) printf ("%u\n", n);

}