PAT Grade --A1104 Sum of Number Segments [20]

Consider a positive integer N written in standard notation with k+1 digits ai​​ as ak​​a1​​a0​​ with 0 for all i and ak​​>0. Then N is palindromic if and only if ai​​=aki​​ 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 Cbecomes 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.
 1 #include <iostream>
 2 #include <string>
 3 #include <algorithm>
 4 using namespace std;
 5 int main()
 6 {
 7     int k = 0;
 8     string str1, str2, str;
 9     cin >> str;
10     while (k < 10)
11     {
12         int c = 0;
13         str1 = str2 = str;
14         reverse(str2.begin(), str2.end());
15         if (str1 == str2)
16             break;
17         cout << str1 << " + " << str2 << " = ";
18         for (int i = str1.length()-1; i>=0; --i)
19         {
20             str[i] = (str1[i] - '0' + str2[i] - '0' + c) % 10 + '0';
21             c = (str1[i] - '0' + str2[i] - '0' + c) / 10;
22         }
23         if (c > 0)
24             str.insert(str.begin(), 1, c + '0');
25         cout << str << endl;
26         k++;
27     }
28     if (k == 10)
29         cout << "Not found in 10 iterations." << endl;
30     else
31         cout << str1 << " is a palindromic number." << endl;
32     return 0;
33 }

 

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Origin www.cnblogs.com/zzw1024/p/11493903.html