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. (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.
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 }
原文地址：https://www.cnblogs.com/GetcharZp/p/9580080.html