pat 1136 A Delayed Palindrome（20 分）

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.

 

#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <climits>
#include <iostream>
#include <algorithm>
#define wzf ((1 + sqrt(5.0)) / 2.0)
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;

const int MAXN = 2e3 + 10;

char A[MAXN], B[MAXN], C[MAXN];

void calcB()
{
    int len = strlen(A), a = len - 1, b = 0;
        for (a ,b ; b < len; ++ b, -- a)
            B[b] = A[a];
}

void calcC()
{
    int len1 = strlen(A), len = len1, b = 0;
    int temp[MAXN];
    for (int i = 0, j = len1 - 1; i < len; ++ i, -- j)
    {
        if (j != -1) b += int(A[j] - '0') + int(B[j] - '0');
        temp[i] = b % 10;
        b /= 10;
        if (i == len - 1 && b > 0) ++ len;
    }

    for (int i = 0, j = len - 1; i < len; ++ i, -- j)
        C[i] = char ('0' + temp[j]);
}

int main()
{
    scanf("%s", &A);
    int len = strlen(A), a = len - 1, b = 0;
        for (a ,b ; b < len; ++ b, -- a)
            B[b] = A[a];
    for (int i = 0; ; ++ i)
    {
        if (i == 10)
        {
            printf("Not found in 10 iterations.
");
            break;
        }
        calcB();
        if (strcmp(A, B) == 0)
        {
            printf("%s is a palindromic number.
", A);
            break;
        }
        calcC();
        printf("%s + %s = %s
", A, B, C);
        strcpy(A, C);
    }
    return 0;
}