PAT甲级2017秋季1136 A Delayed Palindrome (20分)-蒲公英云

PAT甲级2017秋季1136 A Delayed Palindrome (20分)

1136 A Delayed Palindrome (20分)
在这里插入图片描述

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:

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 <iostream>
#include <algorithm>
using namespace std;
string rev(string s) { 
    reverse(s.begin(), s.end());
    return s;
}
string add(string s1, string s2) { 
    string s = s1;
    int carry = 0;
    for (int i = s1.size() - 1; i >= 0; i--) { 
        s[i] = (s1[i] - '0' + s2[i] - '0' + carry) % 10 + '0';
        carry = (s1[i] - '0' + s2[i] - '0' + carry) / 10;
    }
    if (carry > 0) s = "1" + s;
    return s;
}
int main() { 
    string s, sum;
    int n = 10;
    cin >> s;
    if (s == rev(s)) { 
        cout << s << " is a palindromic number.\n";
        return 0;
    }
    while (n--) { 
        sum = add(s, rev(s));
        cout << s << " + " << rev(s) << " = " << sum << endl;
        if (sum == rev(sum)) { 
            cout << sum << " is a palindromic number.\n";
            return 0;
        }
        s = sum;
    }
    cout << "Not found in 10 iterations.\n";
    return 0;
}
#include <bits/stdc++.h>
using namespace std;
bool ifPalin(string a){ 
    for(int i=0;i<a.length()/2;i++){ 
        if(a[i]!=a[a.length()-i-1])return false;
    }
    return true;
}
string add(string a, string b) { 
    string c = a;
    int carry = 0;
    for (int i = a.size() - 1; i >= 0; i--) { 
        c[i] = (a[i] - '0' + b[i] - '0' + carry) % 10 + '0';
        carry = (a[i] - '0' + b[i] - '0' + carry) / 10;
    }
    if (carry > 0) c = "1" + c;
    return c;
}
int main(){ 
  string a;
  int k=0;
  cin>>a;
  while(!ifPalin(a)) { 
      string b("");
      for(int i=0;i<a.length();i++){ 
          b.push_back(a[a.length()-1-i]);
      }
     // printf("%s + %s = %s\n",a.c_str(),b.c_str(),add(a,b).c_str());
      cout<<a<<" + "<<b<<" = "<<add(a,b)<<endl;
      a=add(a,b);
      k++;
      if(k==10)break;
  }
  if(k<=10 && ifPalin(a)) cout<<a<<" is a palindromic number.";
  else cout<<"Not found in 10 iterations.";
  return 0;
}