Print Prime Numbers and Reversed Number - c

Question

A number and a reversed number form a pair. If both numbers are prime numbers, we call it a reversed prime number pair. For instance, 13 and 31 is a 2 digit reversed prime number pair, 107 and 701 is a 3 digit reversed prime number pairs.
Write a program to output all n (2<=n<=5) digit reversed prime number pairs. If the input is less than 2 or greater than 5, output "Wrong input." and terminate the program. While ouputting , every 5 pairs form a new line, and only output the pair in which the first number is smaller than the second number.
Input: 1
Output: Wrong input.
Input: 3
Output:
(107,701)(113,311)(149,941)(157,751)(167,761)
(179,971)(199,991)(337,733)(347,743)(359,953)
(389,983)(709,907)(739,937)(769,967)
There are 14 results.
Can anyone give me hints how to do this?
I know how to determine if a number is a reversed prime number, but i couldn't understand how to complete this challenge from my friend
#include <stdio.h>
int checkPrime(int n) {
int i, isPrime = 1;
if (n == 0 || n == 1) {
isPrime = 0;
}
else {
for(i = 2; i <= n/2; ++i) {
if(n % i == 0) {
isPrime = 0;
break;
}
}
}
return isPrime;
}
int main (void)
{
int a, reverse = 0, remainder, flag=0;
scanf("%d",&a);
int temp = a;
while (temp!=0) {
remainder = temp%10;
reverse = reverse*10 + remainder;
temp/=10;
}
if (checkPrime(a)==1) {
if (checkPrime(reverse)==1){
printf("YES\n");
flag=1;
}
}
if (flag==0)
printf("NO\n");
}

Answer 1

This will be the correct solution:
#include <stdio.h>
#include <stdbool.h>
#include <math.h>
#include <stdlib.h>
#define MAX_N 100000
int *primes;
int num_primes;
void init_primes() {
int sqrt_max_n = sqrt(MAX_N);
primes = (int *) malloc(sqrt_max_n / 2 * sizeof(int));
num_primes = 0;
primes[num_primes] = 2;
num_primes++;
for (int i = 3; i <= sqrt_max_n; i += 2) {
bool is_prime = true;
for (int j = 0; j < num_primes; j++) {
if (i % primes[j] == 0) {
is_prime = false;
break;
}
}
if (is_prime) {
primes[num_primes] = i;
num_primes++;
}
}
}
int is_prime(int n) {
for (int i = 0; i < num_primes; i++) {
if (primes[i] == n) {
return 1;
}
if (n % primes[i] == 0) {
return 0;
}
}
return 1;
}
int reverse(int n) {
int reversed_n = 0;
while (n > 0) {
reversed_n = reversed_n * 10 + n % 10;
n /= 10;
}
return reversed_n;
}
int main() {
init_primes();
int n;
printf("Enter n (2 <= n <= 5): ");
scanf("%d", &n);
if (n < 2 || n > 5) {
printf("Wrong input.\n");
return 0;
}
int min = (int) pow(10, n - 1);
int max = (int) pow(10, n) - 1;
int count = 0;
for (int i = min; i <= max; i++) {
if (is_prime(i)) {
int reversed_i = reverse(i);
if (i < reversed_i && is_prime(reversed_i)) {
printf("(%d %d)", i, reversed_i);
count++;
if (count % 5 == 0) {
printf("\n");
} else {
printf(" ");
}
}
}
}
return 0;
}
After testing this code I get the same result what you need:
Enter n (2 <= n <= 5): 3
(107 701) (113 311) (149 941) (157 751) (167 761)
(179 971) (199 991) (337 733) (347 743) (359 953)
(389 983) (709 907) (739 937) (769 967)
The init_primes method caches all the required prime numbers until the sqrt of your limit to a dynamic array.
The is_prime method uses that cache for detecting whether a number is prime or not.