687D: Dividing Kingdom II

Codeforces Round #360 Editorial [+ Challenges!]

B. Lovely Palindromes
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.

Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.

Now Pari asks you to write a program that gets a huge integer n from the input and tells what is the n-th even-length positive palindrome number?
Input

The only line of the input contains a single integer n (1 ≤ n ≤ 10100 000).
Output

Print the n-th even-length palindrome number.
Examples
Input

1

Output

11

Input

10

Output

1001

Note

The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001.

Hint

Try to characterize even-length palindrome numbers.

Solution

For simplifications, in the following solution we define lovely integer as an even-length positive palindrome number.

An even-length positive integer is lovely if and only if the first half of its digits is equal to the reverse of the second half.

So if a and b are two different 2k-digit lovely numbers, then the first k digits of a and b differ in at least one position.

So a is smaller than b if and only if the first half of a is smaller than the the first half of b.

Another useful fact: The first half of a a lovely number can be any arbitrary positive integer.

Using the above facts, it's easy to find the first half of the n-th lovely number — it exactly equals to integer n. When we know the first half of a lovely number, we can concatenate it with its reverse to restore the lovely integer. To sum up, the answer can be made by concatenating n and it's reverse together.

The complexity of this solution is O(log n).

C++ code

//     . .. ... .... ..... be name khoda ..... .... ... .. .     \\

#include <bits/stdc++.h>
using namespace std;

inline int in() { int x; scanf("%d", &x); return x; }
const long long N = 2002;

int main()
{
	string s;
	cin >> s;
	cout << s;
	reverse(s.begin(), s.end());
	cout << s << endl;
}

Python code

s = raw_input()
print s + ''.join(reversed(s))

原文链接： Codeforces Round #360 Editorial [+ Challenges!] - Codeforces