1015. Reversible Primes (20)

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

思路：先说一下题目的意思，判断该十进制数N在D进制下的反转再转为十进制是否是质数，比如sample中的23在二进制的反转就是11101(本来是10111)，再转为10进制就是29，是质数。

所以我们要做的就是先把N转为D进制表示，反转，再转回十进制进行判断。

import java.math.BigInteger;
import java.util.*;

public class Main {

    private static int revInRadixD(int num, int D) {
        StringBuilder rev = new StringBuilder();
        int shang = num;
        while (shang != 0) {
            rev.append(shang % D);
            shang /= D;

        }
        int result = 0;
        int len = rev.length();
        for (int i = 0; i < len; i++) {
            result += (rev.charAt(i) - '0') * Math.pow(D, len - 1 - i);
        }
        return result;
    }

    private static boolean isPrime(int num) {
        if (num < 2) return false;
        for (int i = 2; i <= Math.sqrt(num); i++) {
            if (num % i == 0) {
                return false;
            }
        }
        return true;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while (in.hasNext()) {
            int N = in.nextInt();
            if (N < 0) return;
            int D = in.nextInt();
            boolean isPrime = isPrime(N) && isPrime(revInRadixD(N, D));
            System.out.println(isPrime ? "Yes" : "No");
        }

    }
}