Oleg the client and Igor the analyst are good friends. However, sometimes they argue over little things. Recently, they started a new company, but they are having trouble finding a name for the company.

To settle this problem, they've decided to play a game. The company name will consist of n letters. Oleg and Igor each have a set of n letters (which might contain multiple copies of the same letter, the sets can be different). Initially, the company name is denoted by n question marks. Oleg and Igor takes turns to play the game, Oleg moves first. In each turn, a player can choose one of the letters c in his set and replace any of the question marks with c. Then, a copy of the letter c is removed from his set. The game ends when all the question marks has been replaced by some letter.

For example, suppose Oleg has the set of letters {i, o, i} and Igor has the set of letters {i, m, o}. One possible game is as follows :

Initially, the company name is ???.

Oleg replaces the second question mark with 'i'. The company name becomes ?i?. The set of letters Oleg have now is {i, o}.

Igor replaces the third question mark with 'o'. The company name becomes ?io. The set of letters Igor have now is {i, m}.

Finally, Oleg replaces the first question mark with 'o'. The company name becomes oio. The set of letters Oleg have now is {i}.

In the end, the company name is oio.

Oleg wants the company name to be as lexicographically small as possible while Igor wants the company name to be as lexicographically large as possible. What will be the company name if Oleg and Igor always play optimally?

A string s = s1s2...sm is called lexicographically smaller than a string t = t1t2...tm(where s ≠ t) if si < ti where i is the smallest index such that si ≠ ti. (so sj = tjfor all j < i)

Input

The first line of input contains a string s of length n (1 ≤ n ≤ 3·105). All characters of the string are lowercase English letters. This string denotes the set of letters Oleg has initially.

The second line of input contains a string t of length n. All characters of the string are lowercase English letters. This string denotes the set of letters Igor has initially.

Output

The output should contain a string of n lowercase English letters, denoting the company name if Oleg and Igor plays optimally.

Examples

Input
tinkoff
zscoder
Output
fzfsirk
Input
xxxxxx
xxxxxx
Output
xxxxxx
Input
ioi
imo
Output
ioi

题意:A,B都有N个字符,A和B轮流贡献一个字符,A想组成的字典序越小,B想组成的字典序越大。

思路:A肯定是贡献出最小的(N+1)/2个字符,B贡献出最大的N/2个字符。 错误思路是A从小到大贡献字符,B从大到小贡献字符,但是会出现错误,

比如A=cbxy,B=aaaa,错误解=baca,正解=abac。正确的贪心策略应该是:对A,如果最小值小于B的最大值,那么把最小值填到前面,否则把最大值填到最后。B同理。

#include<bits/stdc++.h>
using namespace std;
const int maxn=;
char a[maxn],b[maxn],ans[maxn];
int main()
{
int N,L,R,h1,t1,h2,t2;
scanf("%s%s",a+,b+); N=strlen(a+);
sort(a+,a+N+); sort(b+,b+N+);
L=; R=N; h1=; t1=(N+)/; h2=N; t2=N-N/+;
while(L<=R){
if(a[h1]<b[h2]) ans[L++]=a[h1++];
else ans[R--]=a[t1--];
if(L>R) break;
if(b[h2]>a[h1]) ans[L++]=b[h2--];
else ans[R--]=b[t2++];
}
for(int i=;i<=N;i++) printf("%c",ans[i]);
return ;
}

CodeForces - 794C:Naming Company(博弈&简单贪心)的更多相关文章

  1. [刷题]Codeforces 794C - Naming Company

    http://codeforces.com/contest/794/problem/C Description Oleg the client and Igor the analyst are goo ...

  2. CF 628C --- Bear and String Distance --- 简单贪心

    CF 628C 题目大意:给定一个长度为n(n < 10^5)的只含小写字母的字符串,以及一个数d,定义字符的dis--dis(ch1, ch2)为两个字符之差, 两个串的dis为各个位置上字符 ...

  3. Codeforces 437C The Child and Toy(贪心)

    题目连接:Codeforces 437C  The Child and Toy 贪心,每条绳子都是须要割断的,那就先割断最大值相应的那部分周围的绳子. #include <iostream> ...

  4. Codeforces 828B Black Square(简单题)

    Codeforces 828B Black Square(简单题) Description Polycarp has a checkered sheet of paper of size n × m. ...

  5. Uva 11729 Commando War (简单贪心)

    Uva 11729  Commando War (简单贪心) There is a war and it doesn't look very promising for your country. N ...

  6. Codeforces Round #546 (Div. 2) D 贪心 + 思维

    https://codeforces.com/contest/1136/problem/D 贪心 + 思维 题意 你面前有一个队列,加上你有n个人(n<=3e5),有m(m<=个交换法则, ...

  7. CDOJ 1502 string(简单贪心)

    题目大意:原题链接 相邻两个字母如果不同,则可以结合为前一个字母,如ac可结合为a.现给定一个字符串,问结合后最短可以剩下多少个字符串 解体思路:简单贪心 一开始读题时,就联想到之前做过的一道题,从后 ...

  8. ACM_发工资(简单贪心)

    发工资咯: Time Limit: 2000/1000ms (Java/Others) Problem Description: 作为广财大的老师,最盼望的日子就是每月的8号了,因为这一天是发工资的日 ...

  9. ACM_Ruin of Titanic(简单贪心)

    Ruin of Titanic Time Limit: 2000/1000ms (Java/Others) Problem Description: 看完Titanic后,小G做了一个梦.梦见当泰坦尼 ...

随机推荐

  1. Java泛型详解(转)

    文章转自  importNew:Java 泛型详解 引言 泛型是Java中一个非常重要的知识点,在Java集合类框架中泛型被广泛应用.本文我们将从零开始来看一下Java泛型的设计,将会涉及到通配符处理 ...

  2. hadoop06---多线程

    .1.1. 实现线程的两种方式 1.继承Thread的方式 见代码MyThreadWithExtends 2.声明实现 Runnable 接口的方式 见代码MyThreadWithImpliment ...

  3. 如何计算android设备的屏幕物理尺寸

    https://segmentfault.com/q/1010000002794179 通过android提供的接口,我们可以获取到如下的信息:int densityDpi = mContext.ge ...

  4. I.MX6Q(TQIMX6Q/TQE9)学习笔记——内核启动与文件系统挂载

    经过前面的移植,u-boot已经有能力启动内核了,本文主要来看下如何通过之前移植的u-boot来启动内核.如果按照前面的文章完成了LTIB 的编译,那么,Linux的内核应该就会出现rpm/BUILD ...

  5. jQuery消息提示框插件Tipso

    在线演示 本地下载

  6. 谈Swift中的访问控制

    访问控制(Access Control) 访问控制可以限定其他源文件或模块中的代码对你的代码的访问级别.这个特性可以让我们隐藏代码的一些实现细节,并且可以指定一些代码和访问和使用的优先接口. 你可以明 ...

  7. Virtual Container Hosts(VCHs) 介绍

    In vSphere Integrated Containers, you deploy virtual container hosts (VCHs) that serve as Docker API ...

  8. mac 查看C++及各种环境的命令

    MacBook-Air:$ which g++/usr/bin/g++MacBook-Air:$ archi386MacBook-Air:$ g++ --versionConfigured with: ...

  9. JMeter学习(七)聚合报告之 90% Line 正确理解

    90% Line 参数正确的含义: 虽然,我的上面理解有一定的道理,显然它是错误的.那看看JMeter 官网是怎么说的? 90% Line - 90% of the samples took no m ...

  10. 简学Python第五章__模块介绍,常用内置模块

    Python第五章__模块介绍,常用内置模块 欢迎加入Linux_Python学习群  群号:478616847 目录: 模块与导入介绍 包的介绍 time &datetime模块 rando ...