How do you write a program in c to print the first five prime numbers?

Answer 1

In order to print the first 10 prime numbers you need a function to determine if a given value is prime or not. The following algorithm is the standard method of doing so:

1. Let n be the value.

2. If n < 2 then return false.

3. If n is even then return true if n is 2, otherwise return false.

4. Let divisor = 3.

5. If divisor is greater than the square root of n then return true.

6. If divisor is a factor of value then return false.

7. Let divisor = divisor + 2.

8. Go to step 5.

This algorithm can be efficiently implemented in C++ as follows:

bool is_prime (const unsigned n)

{

if (n<2) return false;

if (!(n&1)) return n==2;

const unsigned m = static_cast<unsigned>(std::sqrt (static_cast<double>(n)));

for (unsigned d=3; d<=m; d+=2)

if (!(n%d)) return false;

return true;

}

With this function defined, we can now go ahead and write the complete program:

#include<iostream>

bool is_prime (const unsigned n) {/*as above*/}

int main()

{

std::cout << "First 10 primes:\n";

unsigned primes = 0;

unsigned num = 0;

while (primes < 10)

{

if (is_prime (num))

{

std::cout << num << std::endl;

++primes;

}

++num;

}

}

Answer 2

#include<stdio.h>

#include<math.h>

bool is_prime (const unsigned);

unsigned next_prime (unsigned);

int main (void) {

unsigned count=0, num=0;

printf ("The first 100 prime numbers\n");

while (count<100) {

num = next_prime (num);

printf ("%d\t", num);

++count;

}

printf ("\n");

return 0;

}

unsigned next_prime (unsigned num) {

while (!is_prime (++num));

return num;

}

bool is_prime (const unsigned num) {

if (num<2) return false;

if (!(num%2)) return num==2;

const unsigned max = (unsigned) sqrt (num) + 1;

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

return true;

}

Answer 3

First, you need a function that can determine if a given number is prime or not:

#include<math.h>

bool is_prime (unsigned num) {

if (num < 2) return false;

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

unsigned max = (unsigned) sqrt ((double) num) + 1;

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

return true;

}

Next, a function that will return the next prime after a given number:

unsigned next_prime (unsigned num) {

while (!is_prime (++num));

return num;

}

Finally, a function that counts the number of primes in a given range:

unsigned count_primes (unsigned min, unsigned max) {

unsigned count = is_prime (num) ? 1 : 0;

while ((min = next_prime (min)) <= max) ++count;

return count;

}

To count the number of primes in the range 1 to 1,000,000:

#include<stdio.h>

int main (void) {

unsigned count = count_primes (1, 1000000);

printf ("There are %d primes in the range [1:1,000,000]\n", count);

return 0;

}

Answer 4

#include<iostream>

#include<cmath>

bool is_prime (unsigned n) {

if (n<2) return false;

if (!(n%2)) return n==2;

unsigned max {sqrt (n)};

for (unsigned factor {3}; factor<=max; factor+=2)

{

if (!(num%factor)) return false;

}

return true;

}

int main (void)

{

for (unsigned x=100; x<=200; ++x) if (is_prime (x)) std::cout<<x<<std::endl;

}

Answer 5

#include<stdio.h>

int main (void) {

printf ("The first 5 prime numbers are 2, 3, 5, 7 and 11.\n");

return 0;

}