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Lychrel numbers 题号:55 难度: 5 中英对照

If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.

Not all numbers produce palindromes so quickly. For example,

349 + 943 = 1292,
1292 + 2921 = 4213
4213 + 3124 = 7337

That is, 349 took three iterations to arrive at a palindrome.

Although no one has proved it yet, it is thought that some numbers, like 196, never produce a palindrome. A number that never forms a palindrome through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem, we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either (i) become a palindrome in less than fifty iterations, or, (ii) no one, with all the computing power that exists, has managed so far to map it to a palindrome. In fact, 10677 is the first number to be shown to require over fifty iterations before producing a palindrome: 4668731596684224866951378664 (53 iterations, 28-digits).

Surprisingly, there are palindromic numbers that are themselves Lychrel numbers; the first example is 4994.

How many Lychrel numbers are there below ten-thousand?

NOTE: Wording was modified slightly on 24 April 2007 to emphasise the theoretical nature of Lychrel numbers.

Solution

要求10000以内有多少个Lychrel数。题目已告知10000以内50次迭代即可判断是否为Lychrel数,因此可以采用暴力法。 > 判断一个数是否回文数,只要把数转换成字符串,比较其正反即可。

Code

import java.math.BigInteger;

public final class p055 {

    public static void main(String[] args) {
        long start = System.nanoTime();
        long result = run();
        long end = System.nanoTime();
        System.out.println(result);
        System.out.println((end - start) / 1000000 + "ms");
    }

    public static long run() {
        int count = 0;
        for (int i = 0; i < 10000; i++) {
            if (isLychrel(i)) {
                count++;
            }
        }
        return count;
    }

    private static boolean isLychrel(int n) {
        BigInteger temp = BigInteger.valueOf(n);
        for (int i = 0; i < 49; i++) {
            temp = temp.add(new BigInteger(reverse(temp.toString())));
            if (isPalindrome(temp.toString())) {
                return false;
            }
        }
        return true;
    }

    private static boolean isPalindrome(String s) {
        return s.equals(reverse(s));
    }

    public static String reverse(String s) {
        return new StringBuilder(s).reverse().toString();
    }
}
249
173ms