Codeforces863A-Quasi-palindrome

Quasi-palindrome

 

Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string.

String t is called a palindrome, if it reads the same from left to right and from right to left.

For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes.

You are given some integer number x. Check if it's a quasi-palindromic number.

Input

The first line contains one integer number x (1 ≤ x ≤ 109). This number is given without any leading zeroes.

Output

Print "YES" if number x is quasi-palindromic. Otherwise, print "NO" (without quotes).

Example
Input
131
Output
YES
Input
320
Output
NO
Input
2010200
Output
YES

题解:求一个数去掉后缀零后是否是回文串。直接模拟。
Code:
var
  n:longint;
function check(n:longint):boolean;
var s:string;i,j,l:longint;
begin
  str(n,s);
  l:=length(s);
  i:=1;j:=l;
  while (s[i]=s[j])and(i<=j) do
  begin
    inc(i);dec(j);
  end;
  if j-i<=0 then exit(true) else exit(false);
end;
begin
  readln(n);
  while n mod 10=0 do n:=n div 10;
  if check(n) then writeln('YES') else writeln('NO');
end.
### Codeforces Problem 976C Solution in Python For solving problem 976C on Codeforces using Python, efficiency becomes a critical factor due to strict time limits aimed at distinguishing between efficient and less efficient solutions[^1]. Given these constraints, it is advisable to focus on optimizing algorithms and choosing appropriate data structures. The provided code snippet offers insight into handling string manipulation problems efficiently by customizing comparison logic for sorting elements based on specific criteria[^2]. However, for addressing problem 976C specifically, which involves determining the winner ('A' or 'B') based on frequency counts within given inputs, one can adapt similar principles of optimization but tailored towards counting occurrences directly as shown below: ```python from collections import Counter def determine_winner(): for _ in range(int(input())): count_map = Counter(input().strip()) result = "A" if count_map['A'] > count_map['B'] else "B" print(result) determine_winner() ``` This approach leverages `Counter` from Python’s built-in `collections` module to quickly tally up instances of 'A' versus 'B'. By iterating over multiple test cases through a loop defined by user input, this method ensures that comparisons are made accurately while maintaining performance standards required under tight computational resources[^3]. To further enhance execution speed when working with Python, consider submitting codes via platforms like PyPy instead of traditional interpreters whenever possible since they offer better runtime efficiencies especially important during competitive programming contests where milliseconds matter significantly.
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