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 fromhttps://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.
输入数的长度最长有1000位，应该用string来保存输入数据
1 #include<iostream>
2 #include<algorithm>
3 #include<string>
4 using namespace std;
5 string add(string a, string b){
6 int l=a.size(), carry=0;
7 for(int i=l-1; i>=0; i--){
8 a[i]=b[i]+a[i]+carry-‘0‘;
9 if(a[i]>‘9‘){
10 carry=1;
11 a[i] -= 10;
12 }else carry=0;
13 }
14 if(carry==1) a = ‘1‘+a;
15 return a;
16 }
17 int main(){
18 int i;
19 bool flag=true;
20 string n, num, copy;
21 cin>>n;
22 for(i=0; i<10; i++){
23 copy=num=n;
24 reverse(num.begin(), num.end());
25 if(num==n){ cout<<num<<" is a palindromic number."<<endl; flag=false; break;}
26 n=add(n, num);
27 cout<<copy<<" + "<<num<<" = "<<add(copy, num)<<endl;
28 }
29 if(flag) cout<<"Not found in 10 iterations."<<endl;
30 return 0;
31 }
原文地址：https://www.cnblogs.com/mr-stn/p/9161612.html