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. 一个tomcat下,两个系统的jar包可以相互引用。

    将道路挖占管理系统(rems)从交通设备设施系统(tms)中剥离出去以后,在本地调试的时候是在同一个Tomcat下启动的,上传文件成功. 然后部署到西安以后,分成两个tomcat以后,发现rems上传 ...

  2. why big data

    很多人都知道大数据很火,就业很好,薪资很高,想往大数据方向发展.但该学哪些技术,学习路线是什么样的呢?用不用参加大数据培训呢?如果自己很迷茫,为了这些原因想往大数据方向发展,也可以,那么大讲台老师就想 ...

  3. python 将word另存为txt

      import os import os.path from win32com import client as wc c=[] rootdir=["d:/77"] #以该路径为 ...

  4. ubuntu安装mysql,redis,python-mysqldb

    sudo apt-get install mysql-server sudo apt-get install redis-server sudo apt-get install python-redi ...

  5. 05 enumerate index使用

    # enumerate 自动生成一列, 默认0开始,每次自增+1li = ["电脑","鼠标垫","U盘","游艇"]f ...

  6. c# 共享事件处理程序

    使用同一个方法来处理多个Button实例的Click事件. 1.全选所有的Button,在事件添加中的Click点击事件中添加处理函数. 2.假如一个label控件用于显示按钮按下输出文本 3.处理函 ...

  7. 创建一个简单的WCF程序

    1.创建WCF服务库 打开VS2010,选择文件→新建→项目菜单项,在打开的新建项目对话框中,依次选择Visual C#→WCF→WCF服务库,然后输入项目名称(Name),存放位置(Location ...

  8. bzoj4445 小凸想跑步

    题目链接 半平面交,注意直线方向!!! 对于凸包上任意一条边$LINE(p_i,p_{i+1})$都有$S_{\Delta{p_i} {p_{i + 1}}p} < S_{\Delta{p_0} ...

  9. linux 查找locate find

    1.locate locate指令和find找寻档案的功能类似,但locate是透过update程序将硬盘中的所有档案和目录资料先建立一个索引数据库,在 执行loacte时直接找该索引,查询速度会较快 ...

  10. js 显示刚刚上传的图片 (onchange事件)

    <table> <tr width="100"> <td>上传商场图片:</td> <td> <input typ ...