1136 A Delayed Palindrome（20 分）

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

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.

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 <map>
 2 #include <set>
 3 #include <queue>
 4 #include <cmath>
 5 #include <stack>
 6 #include <vector>
 7 #include <string>
 8 #include <cstdio>
 9 #include <cstring>
10 #include <climits>
11 #include <iostream>
12 #include <algorithm>
13 #define wzf ((1 + sqrt(5.0)) / 2.0)
14 #define INF 0x3f3f3f3f
15 #define LL long long
16 using namespace std;
17 
18 const int MAXN = 2e3 + 10;
19 
20 char A[MAXN], B[MAXN], C[MAXN];
21 
22 void calcB()
23 {
24     int len = strlen(A), a = len - 1, b = 0;
25         for (a ,b ; b < len; ++ b, -- a)
26             B[b] = A[a];
27 }
28 
29 void calcC()
30 {
31     int len1 = strlen(A), len = len1, b = 0;
32     int temp[MAXN];
33     for (int i = 0, j = len1 - 1; i < len; ++ i, -- j)
34     {
35         if (j != -1) b += int(A[j] - ‘0‘) + int(B[j] - ‘0‘);
36         temp[i] = b % 10;
37         b /= 10;
38         if (i == len - 1 && b > 0) ++ len;
39     }
40 
41     for (int i = 0, j = len - 1; i < len; ++ i, -- j)
42         C[i] = char (‘0‘ + temp[j]);
43 }
44 
45 int main()
46 {
47     scanf("%s", &A);
48     int len = strlen(A), a = len - 1, b = 0;
49         for (a ,b ; b < len; ++ b, -- a)
50             B[b] = A[a];
51     for (int i = 0; ; ++ i)
52     {
53         if (i == 10)
54         {
55             printf("Not found in 10 iterations.\n");
56             break;
57         }
58         calcB();
59         if (strcmp(A, B) == 0)
60         {
61             printf("%s is a palindromic number.\n", A);
62             break;
63         }
64         calcC();
65         printf("%s + %s = %s\n", A, B, C);
66         strcpy(A, C);
67     }
68     return 0;
69 }

pat 1136 A Delayed Palindrome（20 分）