PAT1136:A Delayed Palindrome

Posted 2020-10-17 0kk470

1136. A Delayed Palindrome (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Consider a positive integer N written in standard notation with k+1 digits a_i as a_k...a₁a₀ 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.

思路
  和 Pat1024 一样的题目，字符串处理问题。
代码

#include<iostream>
#include<algorithm>
using namespace std;

bool isPalindrome(const string& s)
{
  for(int i = 0,j = s.size() - 1;i <= j;i++,j--)
  {
      if(s[i] != s[j])
        return false;
  }
  return true;
}

string add(const string& a,const string& b)
{
    string tmp;
    int len = a.size();
    int carry = 0;
    for(int i = len - 1;i >= 0;i--)
    {
        int cur = (a[i] - \'0\') + (b[i] - \'0\') + carry;
        carry = cur / 10;
        cur = cur % 10;
        tmp += (to_string(cur));
    }
    reverse(tmp.begin(),tmp.end());
    if(carry > 0)
        tmp.insert(0,"1");
    return tmp;
}

int main()
{
   string a;
   while(cin >> a)
   {
      int cnt = 0;
      while(!isPalindrome(a) && ++cnt <= 10)
      {
          string b = a;
          reverse(b.begin(),b.end());
          string c = add(a,b);
          cout << a << " + " << b << " = " << c << endl;
          a = c;
      }
      if(cnt == 11)
        cout << "Not found in 10 iterations." << endl;
      else
        cout << a << " is a palindromic number." << endl;
   }
}