## UVA Problem 10235 ( Simply Emirp ) Solution

UVA Problem 10235 ( Simply Emirp ) Solution:

Solving Technique:

Rank 115, Run time 0.012 s ( 12 ms ).

I have solved this problem in two ways first one is optimized primality test and second one is Sieve of Eratosthenes.

Problem is simple with a is simple we need to check if it’s prime or not. If it is then is it emrip or just prime.

There is one critical example that wasn’t given, its given below. If a prime is reversed and the reversed number is also prime then it is emrip. Ex, 17 is prime and reverse of 17 is 71 is also prime, then 17 is emrip.

If the number is emrip we should print emrip instead prime.

Critical example: 383 is prime but not emrip. If the number is prime and reversed number is equal to original number then it is not emrip.

Important:  Be sure to add or print a new line after each output. The outputs should match exactly because sometimes even a space character causes the answer to be marked as wrong answer.

Code Explanation:

I’ve written a custom check prime function to check for primes since it’ll be called twice. If the number is prime then I reversed the number and checked if that number is also prime. Lastly I checked if the reversed number is prime and also reversed number not equal to original.

Critical Input:

```999
481184
373
998001
998857
753257
823455
999999
850058
78887
999983```

Output:

```999 is not prime.
481184 is not prime.
373 is prime.
998001 is not prime.
998857 is emirp.
753257 is prime.
823455 is not prime.
999999 is not prime.
850058 is not prime.
78887 is prime.
999983 is emirp.```

### Code:

```/*
* @author  Quickgrid ( Asif Ahmed )
* Problem: UVA 10235 ( Simply Emrip optimized primality )
*/

#include<stdio.h>
#include<math.h>

unsigned int checkPrime(unsigned int n){
if (n == 2)
return 1;

if (n % 2 == 0 || n < 2)
return 0;

register unsigned int i;
unsigned int len = sqrt(n);

for (i = 3; i <= len; i += 2){
if (n % i == 0)
return 0;
}

return 1;
}

int main(){
register unsigned int n;
unsigned int p, prime, mod;

while (scanf("%u", &n) == 1){

/**
* check if number is prime
*/
p = checkPrime(n);

if (p){
/**
* Reverse prime
*/
prime = n;
mod = 1;

while (prime / mod){
mod *= 10;
}
mod /= 10;

unsigned int reversePrime = 0;
while (prime){
/**
* Here reversePrime is the reversed prime
*/
reversePrime += mod * (prime % 10);
prime /= 10;
mod /= 10;
}
/**
* Reverse prime end
*/

if (checkPrime(reversePrime) && reversePrime != n){  /* check if reverse is prime but also check the reversed number does not match the original */
printf("%u is emirp.\n", n);
}else{
printf("%u is prime.\n", n);
}
}else{
printf("%u is not prime.\n", n);
}
}
return 0;
}
```

### Simply Emrip with Sieve of Eratosthenes:

```/*
* @author  Quickgrid ( Asif Ahmed )
* Problem: UVA 10235 ( Simply Emrip Sieve of Eratosthenes)
*/

#include<stdio.h>
#include<math.h>

#define N 1000000

/**
* Since value can be up to 1000000
*/
unsigned int *primes = new unsigned int[N];

unsigned int *SieveOfEratosthenes(unsigned int n){
register unsigned int i = 2;

for (; i <= n; ++i)
primes[i] = 1;

primes[0] = primes[1] = 0;
unsigned int len = sqrt(n);

for (i = 2; i <= len; ++i){
if (primes[i]){
for (unsigned int k = i * i; k <= n; k += i)
primes[k] = 0;
}
}
return primes;
}

int main(){
register unsigned int n;
unsigned int prime, mod;

/**
* Pre calculate sieve for every value
*/
primes = SieveOfEratosthenes(N);

while (scanf("%u", &n) == 1){
if (primes[n]){                  /* Since sieve is calculated and stored on table just do a look up */

prime = n;
mod = 1;

while (prime / mod){
mod *= 10;
}
mod /= 10;

unsigned int reversePrime = 0;
while (prime){
reversePrime += mod * (prime % 10);
prime /= 10;
mod /= 10;
}

/**
* Again access table to see if it is a prime
*/
if (primes[reversePrime] && reversePrime != n)
printf("%u is emirp.\n", n);
else
printf("%u is prime.\n", n);
}else
printf("%u is not prime.\n", n);
}
return 0;
}
```