PAT A1136 A Delayed Palindrome （20 分）——回文，大整数

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:

Sample Output 1:

97152 + 25179 = 122331

122331 + 133221 = 255552

255552 is a palindromic number.

Sample Input 2:

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。

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