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 (< 10⁵) 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

先判断该十进制数N是否为素数，再将其转为r进制，逆序之后化为十进制，在判断这个数是否为素数

源码：

#include<bits/stdc++.h>

using namespace std;
stack<int> s;

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

int rev(int n, int r) {
    while (n != 0) {
        s.push(n % r);
        n /= r;
    }
    int sum = 0, cnt = 0;
    while (!s.empty()) {
        sum += pow(r, cnt++) * s.top();
        s.pop();
    }
    return sum;
}

int main() {
    string t;
    int n, m, r;
    while (1) {
        cin >> t;
        n = atoi(t.c_str());
        if (n < 0)break;
        cin >> r;
        //将十进制n转化为r进制，颠倒，再转为十进制 
        m = rev(n, r);
        if (is_prime(n) && is_prime(m))
            cout << "Yes" << endl;
        else
            cout << "No" << endl;
    }
}