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 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.

这题考察大数加法，我们已知一个数字，反转求和，判断是否回文数，10次循环退出。

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
string add(string a, string b) {
    int c = 0;
    int i = 0;
    string ans;
    while(i < a.length() && i < b.length()) {
        c += (a[i] - '0' + b[i] - '0');
        ans += (c % 10 + '0');
        c /= 10;
        i++;
    }
    while(i < a.length()) {
        c += (a[i] - '0');
        ans += (c % 10 + '0');
        c /= 10;
        i++;
    }
    while(i < b.length()) {
        c += (b[i] - '0');
        ans += (c % 10 + '0');
        c /= 10;
        i++;
    }
    if(c != 0) ans += (c + '0');
    reverse(ans.begin(), ans.end());
    return ans;
}
bool isPalindrome(string s) {
    for(int i = 0; i < s.length() >> 1; i++) 
        if(s[i] != s[s.length() - i - 1]) return false;
    return true;
}
int main() {
    char asc[10000], desc[10000];
    string tmp;
    cin >> tmp;
    if(isPalindrome(tmp)) {
        printf("%s is a palindromic number.\n", tmp.c_str());
        exit(0);
    }
    for(int i = 0; i < 10; i++) {
        strcpy(asc, tmp.c_str());
        strcpy(desc, asc);
        reverse(desc, desc + strlen(desc));
        tmp = add(asc, desc);
        printf("%s + %s = %s\n", asc, desc, tmp.c_str());
        if(isPalindrome(tmp)) {
            printf("%s is a palindromic number.\n", tmp.c_str());
            exit(0);
        }
    }
    printf("Not found in 10 iterations.\n");
    return 0;
}