1024 Palindromic Number

1024 Palindromic Number (25 point(s))

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 (≤10​10​​) 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

 知识点：

1.大数加法（这里采用string完成，而不再转换到数组，更加减少了代码量）；

需要再注意的是，读取到的数值高位反而在string的低位，因此需要先分别翻转两个加数（但是这里不需要）。同理，最后的结果和也需要翻转。

2.回文数判断（同样不用数组解决）。

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
bool isPal(string s){
	for(int i=0;i<s.length()/2;i++){
		if(s[i]!=s[s.length()-1-i]) return false;
	}
	return true;
}
string add(string s){
	string res = ""; 
	int carry = 0;
	int len = s.length();
	for(int i=0;i<len;i++){
		int temp = (s[i]-'0')+(s[len-1-i]-'0')+carry;
		res += temp%10+'0';
		carry = temp/10; 
	}
	if(carry!=0) res+= carry+'0';
	reverse(res.begin(),res.end());//重要 
	return res;
}
int main(void){
	string N;
	int K;
	cin>>N>>K;
	bool flag = false;
	int i=0;
	while(i<K){
		if(isPal(N)){
			cout<<N<<endl<<i<<endl;
			flag=true;
			break;
		}
		N = add(N);
		i++;
	}
	if(!flag) cout<<N<<endl<<K<<endl;
	return 0;
}