PAT (Advanced Level) Practice 1136 A Delayed Palindrome (20 分) 凌宸1642

PAT (Advanced Level) Practice 1136 A Delayed Palindrome (20 分) 凌宸1642

题目描述：

Consider a positive integer N written in standard notation with k+1 digits a_ias a_k ⋯a₁ a₀ 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.

译：考虑一个用标准符号写成的正整数 N，其中 k+1 个数字 a_i 为 a_k ⋯a₁ a₀ 0 ≤ a_i < 10 对于所有 i 和 a_k >0。 那么 N 是回文的当且仅当 a_i =a_k−i 对于所有 i。 零写为 0，根据定义也是回文。

非回文数可以通过一系列操作与回文数配对。 首先将非回文数取反，将结果加到原数上。 如果结果不是回文数，则重复此操作直到给出回文数。 这样的数字称为延迟回文数。 （引自 https://en.wikipedia.org/wiki/Palindromic_number ）

给定任何正整数，你应该找到它的成对回文数。

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Input Specification (输入说明):

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

译：每个输入文件包含一个测试用例，它给出一个不超过 1000 位的正整数。

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output Specification (输出说明):

对于每个测试用例，逐行打印查找回文数的过程。 每行的格式如下：

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.

译：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

其中 A 是原始数，B 是反转的 A，C 是它们的总和。 A 开始是输入数字，这个过程结束直到 C 变成一个回文数——在这种情况下，我们在最后一行打印C is a palindromic number. 。 或者如果在 10 次迭代中找不到回文数，则打印 Not found in 10 iterations.。

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Sample Input1 (样例输入1):

97152

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Sample Output1 (样例输出1):

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

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Sample Input2 (样例输入2):

196

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Sample Output2 (样例输出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.

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The Idea：

• 回文数的判断。
• 大整数的加法。

The Codes:

#include<bits/stdc++.h>
using namespace std ;
int n = 10 ;
string a , b , c ;
string add(string a , string b){ // 得到的是 a + b 的结果的字符串的逆序
	string res = "" ;
	int temp = 0  ; // 存储对应位置上的相加的结果以及进位
	for(int i = a.size() - 1 ; i >= 0 ; i --){
		temp += (a[i] - '0') + (b[i] - '0') ;
		res += (temp % 10) + '0' ;
		temp /= 10 ;
	}
	if(temp) res += temp + '0' ;
	return res ;
}
int main(){
	cin >> a ;
	b = a ;
	reverse(b.begin() , b.end()) ;
	while(n -- > 0){ // 如果 10 次操作之后，还没找到的话，则 n 会变为 -1
		if(a == b) { 
			cout << a << " is a palindromic number."<<endl ;
			break ;
		}
		cout << a << " + " << b << " = " ;
		c = add(a , b) ;
		b = c ;
		reverse(c.begin() , c.end()) ; // 逆序之后得到正确的 a + b 的结果
		cout << c << endl ;
		a = c ; 
	}
	if(n == -1) cout << "Not found in 10 iterations." << endl ;
	return 0 ;
}