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 ai as ak⋯a1a0 with 0 for all i and ak>0. Then N is palindromic if and only if ai=ak−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:

Sample Output 1:

97152 + 25179 = 122331

122331 + 133221 = 255552

255552 is a palindromic number.

Sample Input 2:

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 ;

 }

巴特西

PAT_A1136#A Delayed Palindrome

Source:

Description:

Input Specification:

Output Specification:

Sample Input 1:

Sample Output 1:

Sample Input 2:

Sample Output 2:

Keys:

Attention:

Code:

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