[PAT]1015. Reversible Primes (20)@Java

1015. Reversible Primes (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 1023 223 10-2

Sample Output:

YesYesNo

分析：

判断是否是质数，以及加上数制转换。

这里的是指原数字与它在数制下的倒叙，如23 2，是10111（2）的倒叙11101的真实数字。判断原数与倒叙数字是否都是质数判别。

案例分析：

73 10，73是质数，10进制下的倒叙数是37，也是质数，Yes

23 2， 23是质数，2进制下表示为10111，倒叙数是11101，转换为10进制为29，也是质数，Yes

23 10，23是质数，10进制下的倒叙数是32，不是质数，No

package go.jacob.day919;import java.util.ArrayList;import java.util.Scanner;/** * [PAT]Reversible Primes (20) * @author Jacob * 两个知识点：1.将十进制转化为任意进制； * 2.判断是否为素数 */public class Demo1 {public static void main(String[] args) {Scanner sc=new Scanner(System.in);ArrayList<Integer[]> list=new ArrayList<Integer[]>();while(sc.hasNext()){int n=sc.nextInt();if(n<0)break;int d=sc.nextInt();list.add(toDecimal(n,d));}for(Integer[] nums:list){if(isPrime(nums[0])&&isPrime(nums[1]))System.out.println("Yes");elseSystem.out.println("No");}sc.close();}//转化为任意进制,返回原数和“逆”数//（逆数需要把原数转化成d进行，逆序，再转成十进制）private static Integer[] toDecimal(int n, int d) {Integer[] res=new Integer[2];res[0]=n;if(d==10){StringBuilder sb=new StringBuilder(n+"");res[1]=Integer.parseInt(sb.reverse().toString());return res;}String digits = "0123456789";  StringBuilder sb=new StringBuilder();while(n!=0){sb.append(digits.charAt(n%d));n=n/d;}int sum=0;for(int i=0;i<sb.length();i++){sum=sum*d+(sb.charAt(i)-'0');}res[1]=sum;return res;}//判断是否是质数private static boolean isPrime(int num) {if(num==1)return false;if(num==2)return true;int max=(int) Math.ceil(Math.sqrt(num));for(int i=2;i<=max;i++){if(num%i==0)return false;}return true;}}