pat 1015 Reversible Primes（20 分）

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

 

 #include <map>
 #include <set>
 #include <queue>
 #include <cmath>
 #include <stack>
 #include <vector>
 #include <string>
 #include <cstdio>
 #include <cstring>
 #include <climits>
 #include <iostream>
 #include <algorithm>
 #define wzf ((1 + sqrt(5.0)) / 2.0)
 #define INF 0x3f3f3f3f
 #define LL long long
 using namespace std;

 const int MAXN = 1e4 + ;

 int n, r;

 bool is_prime(int n)
 {
     if (n ==  || n == ) return false;
     for (int i = ; i * i <= n; ++ i)
         if (n % i == ) return false;
     return true;
 }

 int main()
 {
     while (scanf("%d", &n), n >= )
     {
         scanf("%d", &r);
         if (!is_prime(n))
         {
             printf("No\n");
             continue;
         }
         int cnt = , ans = ;
         stack <int> my_stack;
         while (n)
         {
             my_stack.push(n % r);
             n /= r;
         }
         while (my_stack.size())
         {
             ans += my_stack.top() * (pow(r, cnt));
             ++ cnt;
             my_stack.pop();
         }
         if (!is_prime(ans))
             printf("No\n");
         else
             printf("Yes\n");
     }
     return ;
 }