PAT (Advanced Level) Practice 1136 A Delayed Palindrome (20分)

1.题目

Consider a positive integer N written in standard notation with k+1 digits a​i​​ as a​k​​⋯a​1​​a​0​​ with 0≤a​i​​<10 for all i and a​k​​>0. Then N is palindromic if and only if a​i​​=a​k−i​​ for all i. Zero is written 0 and is also palindromic by definition.

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. Such number is called a delayed palindrome. (Quoted from https://en.wikipedia.org/wiki/Palindromic_number )

Given any positive integer, you are supposed to find its paired palindromic number.

Input Specification:

Each input file contains one test case which gives a positive integer no more than 1000 digits.

Output Specification:

For each test case, print line by line the process of finding the palindromic number. The format of each line is the following:

A + B = C

where A is the original number, B is the reversed A, and C is their sum. A starts being the input number, and this process ends until C becomes a palindromic number -- in this case we print in the last line C is a palindromic number.; or if a palindromic number cannot be found in 10 iterations, print Not found in 10 iterations. instead.

Sample Input 1:

97152

Sample Output 1:

97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.

Sample Input 2:

196

Sample Output 2:

196 + 691 = 887
887 + 788 = 1675
1675 + 5761 = 7436
7436 + 6347 = 13783
13783 + 38731 = 52514
52514 + 41525 = 94039
94039 + 93049 = 187088
187088 + 880781 = 1067869
1067869 + 9687601 = 10755470
10755470 + 07455701 = 18211171
Not found in 10 iterations.

2.代码

#include<iostream>
#include<string>
#include<algorithm>
#include<cstring>
using namespace std;
int main()
{
	string num;
	cin >> num;
	string answer,reanswer;
	int count = 10;
	reanswer.assign(num.rbegin(), num.rend());
	if (num == reanswer)
	{
		printf("%s is a palindromic number.
", num.c_str());
		return 0;
	}

	while (count--)
	{
		int add = 0; char c;
		for (int i = 0,j=num.length()-1; i <num.length(); i++,j--)
		{
			c = (((num[i] - 48) + (num[j] - 48) + add) % 10) + '0';
			add = ((num[i] - 48) + (num[j] - 48)+add) / 10;
			answer.push_back(c);
		}
		if (add > 0)answer.push_back(add + '0');
		reverse(answer.begin(), answer.end());
		printf("%s", num.c_str());
		reverse(num.begin(), num.end());
		printf(" + %s = %s
", num.c_str(), answer.c_str());
		reanswer.assign(answer.rbegin(), answer.rend());
		if (answer == reanswer)
		{
			printf("%s is a palindromic number.
", answer.c_str()); break;
		}
		num = answer; answer = "";
	}
	if (count == -1)printf("Not found in 10 iterations.
");

}