Editorial

Algorithm Analysis

Read the input number as a string into a character array s. Let len be the length of the string s, decreased by 1 (the index of the cell in the character array where the last digit of the number is located). The string is a palindrome if $s [i] = s [l e n - i]$ for all $0 \leq i < l e n - i$ .

Consider the number 4570754 of length 7. Then len = 7 – 1 = 6. The following equalities hold:

s[0] = s[6 - 0] = s[6] = ‘4’;
s[1] = s[6 - 1] = s[5] = ‘5’;
s[2] = s[6 - 2] = s[4] = ‘7’;

Algorithm Implementation

Read the input number into a character array s.

char s[50];

Read the input number as a string. Set len equal to the index of the last digit.

gets(s);
len = strlen(s) - 1; flag = 0;

Set flag = 0, which means that the number is a palindrome. If the equality $s [i] = s [l e n - i]$ is violated for some i, then the number is not a palindrome, set flag = 1.

for(i = 0; i < len - i; i++)
    if (s[i] != s[len - i]) flag = 1;

Depending on the value of flag, print the result.

if (flag == 0) printf("Yes\n"); else printf("No\n");

Implementation – Function IsPalindrome

#include <stdio.h>
#include <string.h>

char s[50];
int flag;

int IsPalindrome(char *s)
{
    int i = 0, j = strlen(s) - 1;
    while(i < j)
    {
        if (s[i] != s[j]) return 0;
        i++; j--;
    }
    return 1;
}

int main(void)
{
    gets(s);
    flag = IsPalindrome(s);
    if (flag == 1) printf("Yes\n"); else printf("No\n");
    return 0;
}

Implementation – STL reverse

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

string p, q;

int main(void)
{
    cin >> p;
    q = p; reverse(q.begin(),q.end());
    if (p == q) printf("Yes\n"); else printf("No\n");
    return 0;
}

Java Implementation

import java.util.*;

public class Main
{
    public static void main(String[] args)
    {
        Scanner con = new Scanner(System.in);
        char[] s = con.nextLine().toCharArray();
        int flag = 0, len = s.length - 1;
        for(int i = 0; i < len - i; i++)
            if (s[i] != s[len - i]) flag = 1;
        
        if (flag == 0)
            System.out.println("Yes");
        else
            System.out.println("No");    
        con.close();
    }
}

Java Implementation – Function IsPalindrome

import java.util.*;

public class Main
{
    public static int IsPalindrome(String s)
    {
        int i = 0, j = s.length() - 1;
        while(i < j)
        {
            if (s.charAt(i) != s.charAt(j)) return 0;
            i++; j--;
        }
        return 1;
    }

    public static void main(String[] args)
    {
        Scanner con = new Scanner(System.in);
        String s = con.next();
        
        int flag = IsPalindrome(s);
        
        if (flag == 1)
            System.out.println("Yes");
        else
            System.out.println("No");
        con.close();
    }
}

Python Implementation

a = input()
if a == a[::-1]:
    print("Yes")
else:
    print("No")