1024. Palindromic Number (25)

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本文介绍了一种算法，用于找到一个正整数及其反转数之和的回文数，并记录达到回文数所需的步骤数。若在指定步数内未找到，则输出最后结果及步数。

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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：求一个数和自己反过来的数之和，求和之后如果是回文数，输出结果和循环次数。如果到k次都不是回文数，输出最后的结果和k

第一次用的long long，没过。10^10进行100次求和，超出long long 范围

还是大数

tip：有可能本身这个数就是回文数

#include<iostream>  
#include<cstring>  
#include<cstdio>  
#include<queue>  
#include<stack>  
#include<algorithm>  
#include<vector> 
#include<set>
using namespace std;
int add(int num[],int len){
	int num2[100]={0};
	for(int i=0;i<len;i++) num2[i]=num[len-i-1];
	/*for(int i=0;i<len;i++){
		cout<<num[i];
	}
	cout<<endl;
	for(int i=0;i<len;i++){
		cout<<num2[i];
	}
	cout<<endl;*/
	int r=0;
	for(int i=0;i<len;i++){
		int fz=num[i]+num2[i]+r;
		num[i]=fz%10;
		r=fz/10;
	}
	num[len]=r;
	if(r==0) return len;
	return len+1;
}
int pd(int num[],int len){
	for(int i=0;i<len;i++){
		if(num[i]!=num[len-i-1]) return 0;
	}
	return 1;
}
int main(){
	char n[100];
	int k;
	cin>>n>>k;
	int num[100]={0};
	for(int i=0;i<strlen(n);i++) num[i]=n[strlen(n)-i-1]-'0';
	if(pd(num,strlen(n))){
		cout<<n<<endl<<0<<endl;
		return 0;
	}
	int len=strlen(n);
	for(int i=1;i<=k;i++){
		len=add(num,len);
		if(pd(num,len)){
			for(int i=0;i<len;i++) cout<<num[i];
			cout<<endl<<i<<endl;
			return 0;
		}
	}
	for(int i=len-1;i>=0;i--) cout<<num[i];
		cout<<endl<<k<<endl;
	return 0;
}