PAT_A1136#A Delayed Palindrome

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:

 /*
 Data: 2019-08-07 19:32:34
 Problem: PAT_A1136#A Delayed Palindrome
 AC: 17:12

 题目大意：
 非回文数转化为回文数；
 while(! palindrome){
 1.Reverse
 2.add
 输入：
 给一个不超过1000位的正整数
 输出：
 给出每次循环的加法操作，最多10次循环
 */
 #include<cstdio>
 #include<string>
 #include<iostream>
 #include<algorithm>
 using namespace std;

 bool IsPali(string s)
 {
     int len=s.size();
     for(int i=; i<len/; i++)
         if(s[i] != s[len--i])
             return false;
     return true;
 }

 string Func(string s1)
 {
     string s,s2=s1;
     reverse(s2.begin(),s2.end());
     int carry=;
     for(int i=; i<s1.size(); i++)
     {
         carry += (s1[i]-''+s2[i]-'');
         s.insert(s.end(),''+carry%);
         carry /= ;
     }
     while(carry!=)
     {
         s.insert(s.end(),''+carry%);
         carry /= ;
     }
     reverse(s.begin(),s.end());
     printf("%s + %s = %s\n", s1.c_str(),s2.c_str(),s.c_str());
     return s;
 }

 int main()
 {
 #ifdef ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDGE

     string s;
     cin >> s;
     for(int i=; i<; i++)
     {
         if(IsPali(s))
         {
             printf("%s is a palindromic number.\n", s.c_str());
             s.clear();break;
         }
         s = Func(s);
     }
     if(s.size())
         printf("Not found in 10 iterations.");

     return ;
 }