1015 Reversible Primes （20 分)

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

 

•   对于每行输入的两个数据N和D 
•              将N转化为D进制的数（字符串），各位数字（字符）倒叙
•              倒叙后按进制D还原为10进制，得到数字rN
•              判断条件：如果rN和N都为素数，输出Yes，否则输出No
•              结束条件：N为负数

#include<iostream>
#include<string>
#include<stdio.h>
#include<math.h>
using namespace std;
bool judge(int x)//判断素数
{
	if (x == 1)
		return false;
	for (int i = 2; i <= sqrt(x); i++)
	{
		if (x%i == 0)
			return false;
	}
	return true;
}
int trans(int n, int d)//数字转化为D进制--逆转--还原10进制
{
	string s;
	while (n)
	{
		s += to_string(n%d);
		//printf("s=%d\n", n%d);
		n /= d;
	}
	//cout << s << endl;
	return stoi(s, 0, d);
}
int main()
{
	int n, d;
	while (scanf("%d%d", &n, &d) != EOF)
	{
		if (n < 0)
			break;
		if (judge(n) && judge(trans(n, d)))
			printf("Yes\n");
		else printf("No\n");

	}
}