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 + 10;

int n, r;

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

int main()
{
    while (scanf("%d", &n), n >= 0)
    {
        scanf("%d", &r);
        if (!is_prime(n))
        {
            printf("No
");
            continue;
        }
        int cnt = 0, ans = 0;
        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
");
        else
            printf("Yes
");
    }
    return 0;
}