[Solved] 1225 – Palindromic Numbers (II) of LightOJ

[Solved] 1225 – Palindromic Numbers (II) of LightOJ

Main LightOJ Problem is here

A palindromic number or numeral palindrome is a
‘symmetrical’ number like 16461, that remains the same when its digits are
reversed. In this problem you will be given an integer, you have to say whether
the number is a palindromic number or not.

Input

Input starts with an integer T (≤ 20000),
denoting the number of test cases.

Each case starts with a line containing an integer n (0 ≤
n < 109).

Output

For each case, print the case number and ‘Yes’ if n
is palindromic, otherwise print ‘No’.

Sample Input

Output for Sample Input

5

1

21

16161

523125

0

Case 1: Yes

Case 2: No

Case 3: Yes

Case 4: No

Case 5: Yes

 

Full Source Code Using C Plus Plus

Code
  1. #include<iostream>
  2. #include<cstring>
  3. #include<stdio.h>
  4. using namespace std;
  5. int main()
  6. {
  7.     int t,i,j,k,flag=1;
  8.     char n[20];
  9.     cin>>t;
  10.     for(i=1;i<=t;i++)
  11.     {
  12.         flag=1;
  13.         scanf(“%s”,n);
  14.         for(j=0,k=strlen(n)–1;k>=j;j++)
  15.         {
  16.             if(n[j]!=n[k–j])
  17.             {
  18.                 flag=0;
  19.                 break;
  20.             }
  21.         }
  22.         if(flag==1) cout<<“Case “<<i<<“: Yes\n“;
  23.         else cout<<“Case “<<i<<“: No\n“;
  24.     }
  25.     return 0;
  26. }
  27.  
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