Codeforces 932A Palindromic Supersequence （超级水）

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本文介绍了一种简单有效的算法，用于根据给定的字符串构造一个包含该字符串作为子序列的回文串。通过将原字符串及其反转字符串连接起来，可以轻松地生成满足条件的回文串。

题目：

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

inputCopy
aba
output
aba
inputCopy
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.

题意：

有两个字符串a，b；已知a是b的子串并且b是一个回文串，现给出字符串a，让你求字符串b。

思路：

因为这里是让随便输出一个满足条件的b，所以直接把a正反输出两遍即可。

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>

using namespace std;
const int N=1e5+10;
char a[N];

int main()
{
    ios::sync_with_stdio(false);
    int n;
    while(cin>>a)
    {
        cout<<a;//正着输出一遍
        for(int i=strlen(a)-1;i>=0;i--)//逆过来再输出一遍
            cout<<a[i];
        cout<<endl;
    }
    return 0;
}