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 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 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=0;
    for(int i=s1.length()-1;i>=0;i--){
        int tmp=s2[i]-'0'+s1[i]-'0'+carry;
        res+=(tmp%10+'0');
        carry=tmp/10;
    }
    if(carry!=0)res += (carry+'0');
    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 0;
      }
      for(i=0;i<10;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 0;
          }
          s1=s3;
      }
      printf("Not found in 10 iterations.
");
}

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。