Source:

PAT_A1136 A Delayed Palindrome (20 分)

Description:

Consider a positive integer N written in standard notation with k+1 digits a​i​​ as a​k​​⋯a​1​​a​0​​ with 0 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.

Keys:

• 快乐模拟

Attention:

• under algorithm, reverse(s.begin(),s.end());

Code:

 1 /*
 2 Data: 2019-05-24 20:58:33
 3 Problem: PAT_A1136#A Delayed Palindrome
 4 AC: 25:40
 5 
 6 题目大意：
 7 非回文数转化为回文数；
 8 while(! palindrome){
 9 1.Reverse
10 2.add
11 输入：
12 给一个不超过1000位的正整数
13 输出：
14 给出每次循环的加法操作，最多10次循环
15 */
16 
17 #include<cstdio>
18 #include<string>
19 #include<iostream>
20 #include<algorithm>
21 using namespace std;
22 
23 int IsPal(string s)
24 {
25     for(int i=0; i<s.size()/2; i++)
26         if(s[i]!=s[s.size()-1-i])
27             return 0;
28     return 1;
29 }
30 
31 string Add(string s1, string s2)
32 {
33     string s="";
34     int num=0;
35     for(int i=s1.size()-1; i>=0; i--)
36     {
37         num += (s1[i]-'0')+(s2[i]-'0');
38         s += (num%10+'0');
39         num /= 10;
40     }
41     while(num != 0)
42     {
43         s += (num%10+'0');
44         num /= 10;
45     }
46     reverse(s.begin(),s.end());
47     return s;
48 }
49 
50 int main()
51 {
52 #ifdef    ONLINE_JUDGE
53 #else
54     freopen("Test.txt", "r", stdin);
55 #endif
56 
57     string s,t;
58     cin >> s;
59     int pt=0;
60     while(pt<10)
61     {
62         if(IsPal(s))
63             break;
64         t=s;
65         reverse(t.begin(),t.end());
66         cout << s << " + "<< t << " = ";
67         s=Add(s,t);
68         cout << s << endl;
69         pt++;
70     }
71     if(pt<10)
72         cout << s << " is a palindromic number.";
73     else
74         printf("Not found in 10 iterations.");
75 
76     return 0;
77 }