Palindromic Number（注意数字过大的情况）

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 (≤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

答案

#include
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;
}

注意

和这篇文章类似——Have Fun with Numbers（注意数字位数过大的问题），都是需要注意位数过大的情况

在计算机中，如果数字的位数超限，就会将其转为负数。为了避免这种情况，我们采用字符串存贮，相加时诸位相加、保留进位