1024 Palindromic Number (25 分) java 题解

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

Non-palindromic numbers can be paired with palindromic ones via a series of operations. First, the non-palindromic number is reversed and the result is added to the original number. If the result is not a palindromic number, this is repeated until it gives a palindromic number. For example, if we start from 67, we can obtain a palindromic number in 2 steps: 67 + 76 = 143, and 143 + 341 = 484.

Given any positive integer N, you are supposed to find its paired palindromic number and the number of steps taken to find it.

Input Specification:

Each input file contains one test case. Each case consists of two positive numbers N and K, where N (≤1010) is the initial numer and K (≤100) is the maximum number of steps. The numbers are separated by a space.

Output Specification:

For each test case, output two numbers, one in each line. The first number is the paired palindromic number of N, and the second number is the number of steps taken to find the palindromic number. If the palindromic number is not found after K steps, just output the number obtained at the Kth step and K instead.

Sample Input 1:

67 3

Sample Output 1:

484
2

Sample Input 2:

69 3

Sample Output 2:

1353
3

题目大意：

在给定的步骤内对给定的数字进行回文处理：处理成为回文数字时输出该数字和处理步骤，如果在有限步骤内不能成为回文数字，则输出最后一步的数字。

解题思路：

题目给出的数据范围在10的10次方之内，是一个很大的数字，可以直接用大整型类BigInteger搭配反转方法进行处理。

要注意for循环的连贯以及给出的数字本身就是回文数的特殊处理。

java代码:

import java.io.*;
import java.math.*;

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] split = br.readLine().split(" ");
		String num = split[0];
		int n = Integer.parseInt(split[1]);
		StringBuilder builderNum = new StringBuilder(num).reverse();
		if(builderNum.toString().equals(num)) {
			System.out.println(num);
			System.out.print("0");
			System.exit(0);
		}
		
		int i = 1;
		BigInteger temp = null;
		for(i = 1;i <= n;i++) {
			BigInteger bigNum = new BigInteger(num);
			builderNum = new StringBuilder(num).reverse();
			BigInteger bigNum1 = new BigInteger(builderNum.toString());
			bigNum = bigNum.add(bigNum1);
			temp = bigNum;
			num = new StringBuilder(bigNum.toString()).toString();
			
			if(new StringBuilder(temp.toString()).reverse().toString().equals(bigNum.toString())) {
				break;
			}
		}
		System.out.println(temp.toString());
		if(i > n) {
			System.out.print(n);
		}else
			System.out.print(i);
		
	}
}

PTA提交截图：