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;
}

ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined) A的更多相关文章

  1. Codeforces 932 A.Palindromic Supersequence (ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined))

    占坑,明天写,想把D补出来一起写.2/20/2018 11:17:00 PM ----------------------------------------------------------我是分 ...

  2. ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined)

    靠这把上了蓝 A. Palindromic Supersequence time limit per test 2 seconds memory limit per test 256 megabyte ...

  3. Codeforces 932 C.Permutation Cycle-数学 (ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined))

    C. Permutation Cycle   time limit per test 2 seconds memory limit per test 256 megabytes input stand ...

  4. Codeforces 932 B.Recursive Queries-前缀和 (ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined))

    B. Recursive Queries   time limit per test 2 seconds memory limit per test 256 megabytes input stand ...

  5. 【ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined) D】Tree

    [链接] 我是链接,点我呀:) [题意] 让你在树上找一个序列. 这个序列中a[1]=R 然后a[2],a[3]..a[d]它们满足a[2]是a[1]的祖先,a[3]是a[2]的祖先... 且w[a[ ...

  6. 【ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined) C】 Permutation Cycle

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] p[i] = p[p[i]]一直进行下去 在1..n的排列下肯定会回到原位置的. 即最后会形成若干个环. g[i]显然等于那个环的大 ...

  7. 【ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined) B】Recursive Queries

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 写个记忆化搜索. 接近O(n)的复杂度吧 [代码] #include <bits/stdc++.h> using nam ...

  8. 【ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined) A】 Palindromic Supersequence

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 字符串倒着加到原串右边就好 [代码] #include <bits/stdc++.h> using namespace ...

  9. ICM Technex 2017 and Codeforces Round #400 (Div. 1 + Div. 2, combined) A map B贪心 C思路前缀

    A. A Serial Killer time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

随机推荐

  1. redhat7.5在H3C机器上黑屏无显

    现象:H3C机器上,PXE安装/ISO安装系统,多用户模式启动,过内核启动界面后,屏幕黑屏无显,但是可以通过SSH登陆系统,服务正常 环境:redhat7.5/H3C R4900G3/Purely平台 ...

  2. 让bat以管理员权限运行

    有的电脑是非管理员登录,运行程序时,需要提示是否运行运行.解决方法如下: @ echo off % % ver|find "5.">nul&&goto :Ad ...

  3. WebSocket.之.基础入门-建立连接

    WebSocket.之.基础入门-建立连接 1. 使用开发工具(STS.Eclipse等)创建web项目.如下图所示,啥东西都没有.一个新的web项目. 2. 创建java类.index.jsp页面. ...

  4. 219. 存在重复元素 II

    给定一个整数数组和一个整数 k,判断数组中是否存在两个不同的索引 i 和 j,使得 nums [i] = nums [j],并且 i 和 j 的差的绝对值最大为 k. 示例 1: 输入: nums = ...

  5. hbase-java-api002(flush)

    package api; import java.io.IOException; import org.apache.hadoop.conf.Configuration; import org.apa ...

  6. [4]传奇3服务器源码分析一 SelGate

    1. 2 留存 服务端下载地址: 点击这里

  7. django 网站的搭建(2)

    这里使用nginx+uwsgi的方法来搭建生产环境 1,pip3.5  install uwsgi 下载uwsgi ,这里就不做测试了,一般不会出错 2,将django与uwsgi连接在一起 毕竟ru ...

  8. python:基于tkinter的定时关机程

    本人使用python3 from tkinter import* import os from PIL import Image, ImageTk root=Tk() a=Label(root,tex ...

  9. Python全栈-异常处理

    一.异常 1.异常的定义 异常是错误发生的信号,程序一旦出错就会抛出错误信息,如果不及时处理就会程序就会随之停止运行 异常有三部分组成: 1)异常类型 2)异常追踪 3)异常的值 2.异常的分类 1) ...

  10. 找不到System.Web.Optimization命名空间

    找不到System.Web.Optimization命名空间,无法完成BundleConfig.cs内容的添加. 解决方法如下:打开程序包管理控制台,在控制台中输入:Install-PackageMi ...