Codeforces Round #360 (Div. 2) B. Lovely Palindromes

最新推荐文章于 2021-02-15 21:42:14 发布

夜幕下的ACM之路

最新推荐文章于 2021-02-15 21:42:14 发布

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本文介绍了一种高效的方法来找到指定位置的偶数位回文数。通过将输入数字的一半直接拼接其反转后的部分，可以快速生成对应的回文数。文章提供了完整的代码实现。

B. Lovely Palindromes
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

题目让求最长的回文串，其实就是把原串输出一遍后，再倒着输出一遍。
下面是AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char a[1000005];
int main()
{
    while(~scanf("%s",a))
    {
        int len=strlen(a);
        for(int i=0;i<len;i++)
        {
            printf("%c",a[i]);
        }
        for(int i=len-1;i>=0;i--)
        {
            printf("%c",a[i]);
        }
        printf("\n");
    }
    return 0;
}