1136 A Delayed Palindrome (20 分)
Consider a positive integer N written in standard notation with k+1 digits ai as ak⋯a1a0 with 0≤ai<10 for all i and ak>0. Then N is palindromic if and only if ai=ak−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.
#include<bits/stdc++.h>
using namespace std;
string a,b,c;
void calc()
{
int cat=0;
int ba;
c.clear();
for(int i=0;i<a.size();i++)
{
ba=(a[i]-'0'+b[i]-'0'+cat)%10;
cat=(a[i]-'0'+b[i]-'0'+cat)/10;
c.push_back(ba+'0');
}
if(cat>0)
c.push_back('1');
reverse(c.begin(),c.end());
}
int main()
{
cin>>a;
b=a;
reverse(b.begin(),b.end());
int cnt=0;
while(cnt<=10)
{
if(a==b)
{
cout<<a<<" is a palindromic number.";
return 0;
}
else if(cnt==10)
{
cout<< "Not found in 10 iterations." ;
return 0;
}
calc();
cout<<a<<" + "<<b<<" = "<<c<<endl;
a=b=c;
reverse(b.begin(),b.end());
cnt++;
}
}