Palindromic Number（注意数字过大的情况）

最新推荐文章于 2022-03-18 12:34:16 发布

原创最新推荐文章于 2022-03-18 12:34:16 发布 · 657 阅读

2 ·

CC 4.0 BY-SA版权

文章标签：

#PAT

PAT题目练习专栏收录该内容

154 篇文章

订阅专栏

该博客探讨了一个编程问题，涉及找到一个非回文数的配对回文数。通过一系列操作，如反转数字并相加，直到得到回文数。程序处理了数字位数过大可能导致的溢出问题，使用字符串存储并进行逐位操作。文章还提醒注意数字位数过大的情况，防止在计算过程中出现错误。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目

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

答案

#include<iostream>
using namespace std;
int main()
{
	string num;
	int k,cnt=0;
	cin>>num>>k;
	while(k--)
	{
		int len=num.length(),flag=0;
		string tmp1="",tmp2="";
		for(int i=0;i<len/2;i++)//判断是否为回文数 
		{
			if(num[i]!=num[len-1-i])
			{
				flag=1;break;
			}
		}
		if(!flag) break;
		cnt++;
		for(int i=0;num[i];i++)//正序 
		{
			tmp1+=num[i];
		}
		for(int i=len-1;i>=0;i--)//逆序 
		{
			tmp2+=num[i];
		}
		int t1=0,t2=0;
		for(int i=len-1;i>=0;i--)//诸位求和，保留进位 
		{
			t1=tmp1[i]-'0'+tmp2[i]-'0'+t2;
			num[i]=t1%10+'0';
			t2=t1/10;
		}
		if(t2!=0) num=to_string(t2)+num;
	}
	cout<<num<<endl;
	cout<<cnt;
}