2018-02-19
A. Palindromic Supersequence
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string A. Find a string B, where B is a palindrome and A is a subsequence of B.

A subsequence of a string is a string that can be derived from it by deleting some (not necessarily consecutive) characters without changing the order of the remaining characters. For example, "cotst" is a subsequence of "contest".

A palindrome is a string that reads the same forward or backward.

The length of string B should be at most 104. It is guaranteed that there always exists such string.

You do not need to find the shortest answer, the only restriction is that the length of string B should not exceed 104.

Input

First line contains a string A (1 ≤ |A| ≤ 103) consisting of lowercase Latin letters, where |A| is a length of A.

Output

Output single line containing B consisting of only lowercase Latin letters. You do not need to find the shortest answer, the only restriction is that the length of string B should not exceed 104. If there are many possible B, print any of them.

Examples
input
aba
output
aba
input
ab
output
aabaa
Note

In the first example, "aba" is a subsequence of "aba" which is a palindrome.

In the second example, "ab" is a subsequence of "aabaa" which is a palindrome.

感想:大水题 但是 1e3*2=2e3<1e4  一直以为大于后者 导致写了很久 算了 就算学了STL

code1

#include<string.h>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
#define MAX 0x3f3f3f3f
#define fi first
#define se second
#define Len 1e8+5
int main()
{
string s,a; //字符串用string 不是char
cin>>s;
cout<<s;
a.assign(s.rbegin(),s.rend());
cout<<a<<endl; //换行不换行是有区别的 }

code2

#include<string.h>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
#define MAX 0x3f3f3f3f
#define fi first
#define se second
#define Len 1e8+5
int main()
{
string s;
cin>>s;
cout<<s;
reverse(s.begin(),s.end()); //
cout<<s<<endl;
}

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