Pat class A - 1015 reversible primes

Link

The question ： Judge a number N Whether it is D Inverted prime number in base , That is to say, it is in D It is still a prime number after the decimal inversion .

Ideas ： The idea is simple , But the meaning of the title is a little easy to be misunderstood , At present N It's a decimal number , First N Must be prime ; Furthermore, it is converted into D Flip after hexadecimal number , When it becomes a decimal number, it still has to be a prime number to be satisfied .

Code implementation ：

#include<bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
const int N = 1e5+10;

int x, d;

string radix_to_ten(int radix, string s){  // radix->10
    int res = s[0]-'0';
    for(int i = 1; i < s.size(); i ++){
        res *= radix;
        res += s[i]-'0';
    }
    return to_string(res);
}

string ten_to_radix(string s, int radix){  // 10->radix
    int x = stoi(s);
    string res;
    while(x){
        res += (x%radix)+'0';
        x /= radix;
    }
    reverse(res.begin(), res.end());
    return res;
}
;
bool prime[N];
void is_prime() {
    memset(prime, true, sizeof(prime));
    prime[0] = prime[1] = false;
    for(int i = 2; i < N; i++) {
        if(prime[i]) {
            for(int j = i + i; j < N; j += i) {
                prime[j] = false;
            }
        }
    }
}

signed main()
{
    is_prime();
    while(cin >> x){
        if(x<0) break;
        cin >> d;
        bool a = prime[x];
        string x_tmp = ten_to_radix(to_string(x), d); //  Again into d Base number 
        reverse(x_tmp.begin(), x_tmp.end()); //  Flip again 
        int xx = stoi(radix_to_ten(d, x_tmp));  //  Again into 10 Base number 
        bool b = prime[xx];
        cout << (a&&b?"Yes":"No") << endl;
    }

    return 0;
}