PAT-1015 Reversible Primes

part 3, 3.1

1不是质数

#include 
using namespace std;
#include 
#include 
#include 

string transfrom(int N, int D)
{
    string m = "";
    while (N >= D)
    {
        m.append(to_string(N % D));
        N = N / D;
    }
    m.append(to_string(N));
    return m;
}

bool isPrime(string n, int D)
{
    int num = 0;
    for (int i = 0; i < n.size(); i++)
        num += (n[i] - '0') * pow(D, n.size() - 1 - i);
    if (num == 1)
        return false;
    for (int i = 2; i < num; i++)
        if (num % i == 0)
            return false;
    return true;
}

int main()
{
    int N, D;
    while (1)
    {
        cin >> N;
        if (N < 0)
            break;
        cin >> D;
        int flag = 1;
        string resN_D = transfrom(N, D);
        string N_D = resN_D;
        reverse(N_D.begin(), N_D.end());
        if (isPrime(N_D, D) && isPrime(resN_D, D))
            cout << "Yes" << endl;
        else
            cout << "No" << endl;
    }
    system("pause");
    return 0;
}

柳神

#include 
#include 
using namespace std;
bool isprime(int n) {
    if(n <= 1) return false;
    int sqr = int(sqrt(n * 1.0));
    for(int i = 2; i <= sqr; i++) {
        if(n % i == 0)
            return false;
    }
    return true;
}
int main() {
    int n, d;
    while(scanf("%d", &n) != EOF) {
        if(n < 0) break;
        scanf("%d", &d);
        if(isprime(n) == false) {
            printf("Non");
            continue;
        }
        int len = 0, arr[100];
        do{
            arr[len++] = n % d;
            n = n / d;
        }while(n != 0);
        for(int i = 0; i < len; i++)
            n = n * d + arr[i];
        printf("%s", isprime(n) ? "Yesn" : "Non");
    }
    return 0;
}