PAT甲级 1015 Reversible Primes

A reversible prime in any number system is a prime whose "reverse" in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime.

Now given any two positive integers N (< 10^5^) and D (1 < D <= 10), you are supposed to tell if N is a reversible prime with radix D.

Input Specification:

The input file consists of several test cases. Each case occupies a line which contains two integers N and D. The input is finished by a negative N.

Output Specification:

For each test case, print in one line "Yes" if N is a reversible prime with radix D, or "No" if not.

Sample Input:

73 10
23 2
23 10
-2

Sample Output:

Yes
Yes
No

题目大意：如果一个数本身是素数，而且在d进制下反转后的数在十进制下也是素数，就输出Yes，否则就输出No
解法一:

#include<stdio.h>
using namespace std;

bool isprimer(int n){
	if(n <= 1)	return false;
	for(int i = 2;i < n/2;i++)
		if(n%i == 0){
			return false;
		}	
	return true;
}

int main(){
	int n,d;
	
	while(1){
		scanf("%d",&n);
		if(n < 0)
			return 0;
		scanf("%d",&d);
		
		if( isprimer(n) == false){
			printf("No\n");
			continue;
		}
			
		int len = 0,arr[100] = {0};
		do{
			arr[len++] = n % d;
			n /= d;
		}while(n>0);
		
		for(int i=0;i<len;i++)
			n = n * d + arr[i];//计算反转的 
		if( isprimer(n)){
			printf("Yes\n");
		}else
			printf("No\n");
			
	}
	
	return 0;
} 

解法二:

#include<stdio.h>
using namespace std;

bool isprimer(int n){
	if(n <= 1)	return false;
	for(int i = 2;i < n/2;i++)
		if(n%i == 0){
			return false;
		}	
	return true;
}

int rever(int n,int d){//将指定进制的数进行反转 
	int sum = 0;
	while(n > 0){
		sum *= d;
		sum += (n % d);
		n /= d;
	}
	return sum;
}

int main(){
	int n,d;
	
	while(1){
		scanf("%d",&n);
		if(n < 0)
			return 0;
		scanf("%d",&d);
		
		if( isprimer(n) && isprimer(rever(n,d)) )
			printf("Yes\n");
		else
			printf("No\n");
			
	}
	
	return 0;
}