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第六章第二十七题(反素数)(Emirp) - 编程练习题答案

**6.27(反素数)反素数(反转拼写的素数)是指一个非回文素数,将其反转之后也是一个素数。例如:17是一个素数,而31也是一个素数,所以17和71是反素数。编写程序,显示前100个反素数。每行显示10个,并且数字间用空格隔开,如下所示:

13 17 31 37 71 73 79 97 107 113

149 157 167 179 199 311 337 347 359 389

**6.27(Emirp)An emirp (prime spelled backward) is a nonpalindromic prime number whose reversal is also a prime. For example, 17 is a prime and 71 is a prime, so 17 and 71 are emirps. Write a program that displays the first 120 emirps. Display 10 numbers per line, separated by exactly one space, as follows:

13 17 31 37 71 73 79 97 107 113

149 157 167 179 199 311 337 347 359 389

下面是参考答案代码:

public class Ans6_27_page202 {
    public static void main(String[] args) {
        ReverseePrime(100,10);
    }
    public static void ReverseePrime(int n, int line) {
        int count = 0; // Count the number of prime numbers
        int number = 2; // A number to be tested for primeness

        System.out.println("The first "+n+" numbers are \n");

        // Repeatedly find prime numbers
        while (count < n) {
            String strNumber = number+"";
            // Assume the number is prime
            boolean isPrime = true; // Is the current number prime?

            // Test if number is prime
            for (int divisor = 2; divisor <= number / 2; divisor++) {
                if (number % divisor == 0) { // If true, number is not prime
                    isPrime = false; // Set isPrime to false
                    break; // Exit the for loop
                }
            }

            // Test if number is palindrome
            boolean isPalindrome = true;
            //String strNumber = number+"";
            int low = 0;
            int high = strNumber.length() - 1;
            while (low < high) {
                if (strNumber.charAt(low) != strNumber.charAt(high)) {
                    isPalindrome = false;
                    break;
                }
                low++;
                high--;
            }

            // reverseNumber
            int tempNumber = number;
            String strReverseNumber = "";
            while (tempNumber != 0) {
                strReverseNumber += tempNumber % 10;
                tempNumber /=10;
            }
            int reverseNumber = Integer.parseInt(strReverseNumber);
            // ReverseePrime
            boolean isReverseePrime = true;
            for (int divisor = 2; divisor <= reverseNumber / 2; divisor++) {
                if (reverseNumber % divisor == 0) {
                    isReverseePrime = false;
                    break;
                }
            }

            // Print the prime number and increase the count
            if (isPrime && !isPalindrome && isReverseePrime) {
                count++; // Increase the count

                if (count % line == 0) {
                    // Print the number and advance to the new line
                    System.out.println(number);
                }
                else
                    System.out.print(number + " ");
            }

            // Check if the next number is prime
            number++;
        }
    }
}

适用Java语言程序设计与数据结构(基础篇)(原书第11版)Java语言程序设计(基础篇)(原书第10/11版)

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