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 <stdio.h>
#include <string>
#include <iostream>
#include <algorithm>
#include <vector>
#include <string.h>
using namespace std;
string add(string s1,string s2){
string res="";
int carry=;
for(int i=s1.length()-;i>=;i--){
int tmp=s2[i]-''+s1[i]-''+carry;
res+=(tmp%+'');
carry=tmp/;
}
if(carry!=)res += (carry+'');
reverse(res.begin(),res.end());
return res;
}
bool ispali(string s){
string s2;
s2=s;
reverse(s2.begin(),s2.end());
if(s==s2) return true;
else return false;
}
int main(){
string s1,s2,s3;
cin>>s1;
int i;
if(ispali(s1)){
cout<<s1<<" is a palindromic number."<<endl;
return ;
}
for(i=;i<;i++){
s2=s1;
reverse(s2.begin(),s2.end());
s3=add(s1,s2);
cout<<s1<<" + "<<s2<<" = "<<s3<<endl;
if(ispali(s3)){
cout<<s3<<" is a palindromic number."<<endl;
return ;
}
s1=s3;
}
printf("Not found in 10 iterations.\n");
}
注意点：判断回文用string还是方便，string只能string+char，不能char+string，所以大数相加还是要最后反转一下。