CodeForces 688B Lovely Palindromes（偶数位回文数）

最新推荐文章于 2022-09-19 08:00:09 发布

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最新推荐文章于 2022-09-19 08:00:09 发布

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本文介绍了一个算法问题——如何找到第n个具有偶数位数的回文数，并提供了一种简单直接的解决方案。回文数是指从前往后读和从后往前读都一样的数。

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http://codeforces.com/problemset/problem/688/B

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.

题意：

找第 n 个偶数位的回文数。

思路：

一看表，就懂了、自己之前还按照 int 来敲代码的，一看不对劲！

Code：

#include<stdio.h>
#include<cstring>
const int MYDD=1103+2e5;

char n[MYDD];
char ans[MYDD];

int main() {
	while(scanf("%s",n)!=EOF) {
		int lenn=strlen(n);
		strcpy(ans,n);//拷贝一份到前面 
		int v=0;
		for(int j=lenn-1; j>=0; j--) {
			ans[lenn+v]=n[j];//逆序赋值 
			v++;
		}
		printf("%s\n",ans);
	}
	return 0;
}