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 }
原文地址：https://www.cnblogs.com/zllwxm123/p/11296164.html