[C][瘋狂程設][CPE考題20141223] 14G2.UVA10235：Simply Emirp

Problem G: Simply Emirp

An integer greater than 1 is called a prime number if its only positive divisors (factors) are 1 and itself. Prime numbers have been studied over the years by a lot of mathematicians. Applications of prime numbers arise in Cryptography and Coding Theory among others.

Have you tried reversing a prime ? For most primes, you get a composite (43 becomes 34). An Emirp (Prime spelt backwards) is a Prime that gives you a different Prime when its digits are reversed. For example, 17 is Emirp because 17 as well as 71 are Prime. In this problem, you have to decide whether a number N is Non-prime or Prime or Emirp. Assume that 1< N< 1000000.

Interestingly, Emirps are not new to NTU students. We have been boarding 199 and 179 buses for quite a long time!

Input

Input consists of several lines specifying values for N.

Output

For each N given in the input, output should contain one of the following:
    1. "N is not prime.", if N is not a Prime number.
    2. "N is prime.", if N is Prime and N is not Emirp.
    3. "N is emirp.", if N is Emirp.

Sample Input

17
18
19
179
199

Sample Output

17 is emirp.
18 is not prime.
19 is prime.
179 is emirp.
199 is emirp.

Notes:

1.判斷是否為質數，若為質數則回傳1，不是質數則回傳0

/*return value: 1=prime, 0=not prime*/
int is_prime(int *n){

 int num,i,result=0;
 num=*n; 
  
 // num == 2: 2 is prime.
 // num >  2: num is even, num is not prime.
 if(num%2==0){
  //printf("**[%d]\n",__LINE__);
  (num==2)?(result=1):(result=0);
  return result;
 }
 
 for(i=3;i*i<=num;i+=2){
  //printf("**[%d]\n",__LINE__);
  if(num%i==0){
   //printf("**[%d]\n",__LINE__);
   result=0;
   return result;
  }
  //printf("**[%d]\n",__LINE__);
 }
 result=1;
 return result;
}

2.整數轉字串、字串反轉、字串轉整數

#include <string.h>
int input,rev_input;
int value;
char temp[7];
char *rev_temp;

value=is_prime(&input);
sprintf(temp,"%d",input);
rev_temp=strrev(temp);
rev_input=atoi(rev_temp);

3.debug專用，顯示行號的標籤"__LINE__"

printf("**[%d]\n",__LINE__);

4.利用兩個三元運算子(條件運算子 conditional operator)，化簡三選一的if else

三選一:

  if(value==1)
   printf("%d %s\n",input,PRIME);
  else if(value==2)
   printf("%d %s\n",input,EMIRP);
  else
   printf("%d %s\n",input,NOT_PRIME);

利用兩個條件運算子化簡:

  printf("%d %s\n",input,(value>1)?(EMIRP):((value<1)?NOT_PRIME:PRIME));

Code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define PRIME "is prime."
#define NOT_PRIME "is not prime."
#define EMIRP "is emirp."

/*return value: 1=prime, 0=not prime*/
int is_prime(int *n){

 int num,i,result=0;
 num=*n; 
  
 // num == 2: 2 is prime.
 // num >  2: num is even, num is not prime.
 if(num%2==0){
  (num==2)?(result=1):(result=0);
  return result;
 }
 
 for(i=3;i*i<=num;i+=2){
  if(num%i==0){
   result=0;
   return result;
  }
 }
 result=1;
 return result;
}

int main(void){

 int input,rev_input;
 int value;
 char temp[7];
 char *rev_temp;
 
 while(scanf("%d",&input)!=EOF){

  value=0;  
  value=is_prime(&input);
  sprintf(temp,"%d",input);
 
  if(value>0){
   rev_temp=strrev(temp);
   rev_input=atoi(rev_temp);
   value+=is_prime(&rev_input);
  }
 
  if(value!=0 && input==rev_input)
   value--;
  
  printf("%d %s\n",input,(value>1)?(EMIRP):((value<1)?NOT_PRIME:PRIME));