1136 A Delayed Palindrome

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.

题意:

  寻找回文数字(关于中心对称)

思路:

  大数加法模拟。

Code:

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 string addFunction(string num1, string num2) {
 6     int carry = 0, n1, n2, temp;
 7     int len = num1.length();
 8     reverse(num2.begin(), num2.end());
 9     string ans = "";
10     while (len >= 1) {
11         len--;
12         n1 = num1[len] - '0';
13         n2 = num2[len] - '0';
14         temp = n1 + n2 + carry;
15         carry = temp / 10;
16         ans = to_string(temp % 10) + ans;
17     }
18     if (carry == 1) ans = to_string(carry) + ans;
19     return ans;
20 }
21 
22 bool isPalindrome(string str) {
23     int i = 0, j = str.length() - 1;
24     while (i < j) {
25         if (str[i++] != str[j--]) return false;
26     }
27     return true;
28 }
29 
30 int main() {
31     string str, sum, temp;
32     cin >> str;
33     for (int i = 0; i < 10; ++i) {
34         if (isPalindrome(str)) {
35             cout << str << " is a palindromic number." << endl;
36             break;
37         }
38         temp = str;
39         sum = addFunction(str, str);
40         reverse(temp.begin(), temp.end());
41         cout << str << " + " << temp << " = " << sum << endl;
42         if (i == 9) cout << "Not found in 10 iterations." << endl;
43         str = sum;
44     }
45 
46     return 0;
47 }

  刚开始因为求进位的时候不小心算错了,结果最后一组数据没有通过。

永远渴望,大智若愚(stay hungry, stay foolish)
原文地址:https://www.cnblogs.com/h-hkai/p/12728768.html