1136 A Delayed Palindrome (20 分)

Posted 2022-06-08 zllwxm123

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 iand 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.

 

字符串相加，再来个迭代几次就成。

 

 1 #include <bits/stdc++.h>
 2 #define ll long long int
 3 using namespace std;
 4 string s,st;
 5 
 6 
 7 string add(string s, string ss)
 8     string st;
 9     int an = 0;
10     for(int i = s.length()-1 ; i >= 0; --i)
11         int x = s[i]-‘0‘;
12         int y = ss[i]-‘0‘;
13         int ans = x+y+an;
14         if(ans < 10)
15             st += ans+‘0‘;
16             an = 0;
17         else
18             st += (ans%10) + ‘0‘;
19             an = 1;
20         
21     
22     if(an) st += ‘1‘;
23     reverse(st.begin(), st.end());
24     return st;
25 
26 
27 
28 bool ishui(string s)
29     for(int i = 0; i < s.length(); i++)
30         if(s[i] != s[s.length() -i - 1])
31             return false;
32     
33     return true;
34 
35 
36 int main()
37     cin >> s;
38     st = s;
39     reverse(st.begin(), st.end());
40     int cnt = 10;
41     while(!ishui(s) && cnt--)
42         string ans = add(s,st);
43         cout <<s<<" + "<<st<<" = "<<ans<<endl;
44         s = ans;
45         st = s;
46         reverse(st.begin(), st.end());
47     
48     if(cnt == -1)
49         cout <<"Not found in 10 iterations."<<endl;
50     else
51         cout << s<<" is a palindromic number."<<endl;
52     
53     return 0;
54