1015 Reversible Primes (20分)

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本文介绍了一种算法，用于判断一个数是否为特定进制下的可逆素数，即该数及其在指定进制下的反转数均为素数。通过将输入的整数转换为其在给定进制下的反转十进制数，再判断原数和反转数是否同时为素数，从而实现可逆素数的检测。

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:

Sample Output:

Yes
Yes
No

#include<iostream>
#include<vector>
using namespace std;
vector<int> v;

long long int to_radix(int n, int d){

    long long int rev_dec_num = 0;
    long long int exp = 1;
    while (n)
    {

        v.push_back(n % d);
        n = n / d;

    }
    for(int i = v.size() - 1 ; i >= 0; i--){


        rev_dec_num = rev_dec_num + v[i] * exp;
        exp *= d;


    }
    
    return rev_dec_num;

}
void judge_prime(long long int num1, long long int num2){

    bool flag = true;
    long long int max_num;
    long long int min_num;
    max_num = max(num1, num2);
    min_num = min(num1, num2);
    for(long long int i = 2; i < min_num; i ++)
        if(num1 % i == 0 || num2 % i == 0)

            flag = false;

    for(long long int i = min_num; i < max_num; i ++)
        if(max_num % i == 0)

            flag = false;
    
    if(flag)

        cout << "Yes" << endl;
    else
    {
        cout << "No" << endl;
    }
    
}

int main(){

    long long int n,d,r_n;

    while (true)
    {
        cin >> n;
        if(n < 0)
            break;
        cin >> d;

        r_n = to_radix(n, d);

        judge_prime(n, r_n);
        v.clear();


    }
    
    


    return 0;
}