pat（甲）1024--（大数相加+string的应用（reverse函数））

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本文介绍了一种算法，用于找到一个给定正整数的配对回文数及其所需的步数。通过反转并累加非回文数来生成回文数，详细展示了实现过程中的关键步骤和技术细节。

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1024 Palindromic Number（25 分）

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

题意：判断回文数，回文数：所有个位数和例如1234321的数。

输入一个数n（表示要操作的数），和k（表示操作的次数）。将n先翻转变成t，再将n与t相加得到n，如果n是回文数，输出n和操作的次数。加入操作的次数超过k就输出k次操作后的数n，和次数k。

思路：每次判断是否是回文数如果是回文，输出结果，如果不是，进行操作。可以将每次操作和判断设为一个函数。

注意：1、不能用long long，因为会超限。

2、操作函数相加后要进行翻转，因为是反向相加。

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
using namespace std;
string n;

void add()
{
	string res=n;
	reverse(res.begin(),res.end());
	int l=n.length(),i,j=0,tp=0;
	for(i=0;i<l;i++)
	{
		tp=tp+(res[i]-'0')+(n[i]-'0');
		if(tp>9)
		{
			n[i]=tp%10+'0';
			tp/=10;
		}
		else
		{
			n[i]=tp+'0';
			tp=0;
		}
	}
	if(tp) n+='1';
	reverse(n.begin(),n.end());
}

int pd()
{
	string tp=n;
	reverse(tp.begin(),tp.end());
	if(tp==n) return 1;
	return 0;
}

int main(void)
{
	int k,i,j;
	while(cin>>n>>k)
	{
		int fg=0;
		for(i=0;i<k;i++)
		{
			if(fg==0&&pd()==1)
			{
				cout<<n<<endl<<i<<endl;
				fg=1;break;
			}
			add();
		}
		if(fg==0)
		{
			cout<<n<<endl<<i<<endl;
		}
	}
	return 0;
}

参考文章：https://www.liuchuo.net/archives/2329