B. Game of Credit Cards
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and decided to continue their competitions in peaceful game of Credit Cards.

Rules of this game are simple: each player bring his favourite n-digit credit card. Then both players name the digits written on their cards one by one. If two digits are not equal, then the player, whose digit is smaller gets a flick (knock in the forehead usually made with a forefinger) from the other player. For example, if n = 3, Sherlock's card is 123 and Moriarty's card has number 321, first Sherlock names 1and Moriarty names 3 so Sherlock gets a flick. Then they both digit 2 so no one gets a flick. Finally, Sherlock names 3, while Moriarty names 1 and gets a flick.

Of course, Sherlock will play honestly naming digits one by one in the order they are given, while Moriary, as a true villain, plans to cheat. He is going to name his digits in some other order (however, he is not going to change the overall number of occurences of each digit). For example, in case above Moriarty could name 1, 2, 3 and get no flicks at all, or he can name 2, 3 and 1 to give Sherlock two flicks.

Your goal is to find out the minimum possible number of flicks Moriarty will get (no one likes flicks) and the maximum possible number of flicks Sherlock can get from Moriarty. Note, that these two goals are different and the optimal result may be obtained by using different strategies.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the number of digits in the cards Sherlock and Moriarty are going to use.

The second line contains n digits — Sherlock's credit card number.

The third line contains n digits — Moriarty's credit card number.

Output

First print the minimum possible number of flicks Moriarty will get. Then print the maximum possible number of flicks that Sherlock can get from Moriarty.

Examples
input
3
123
321
output
0
2
input
2
88
00
output
2
0
Note

First sample is elaborated in the problem statement. In the second sample, there is no way Moriarty can avoid getting two flicks.

/*
贪心:田忌赛马
*/ #include<bits/stdc++.h>
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
char s[];
char m[];
int a[];
int b[];
int n;
int main(){
// freopen("in.txt","r",stdin);
scanf("%d",&n);
scanf("%s",s);
for(int i=;i<n;i++){
a[i]=s[i]-'';//大王的马
}
scanf("%s",m);
for(int i=;i<n;i++){
b[i]=m[i]-'';//田忌的马
}
// for(int i=0;i<n;i++){
// cout<<a[i]<<" ";
// }cout<<endl;
// for(int i=0;i<n;i++){
// cout<<b[i]<<" ";
// }cout<<endl; sort(a,a+n);
sort(b,b+n);
int j;
j=n-;
for(int i=n-;i>=;i--){
if(a[i]<=b[j])j--;
}
cout<<j+<<endl;
j=;
for(int i=;i<n;i++){
if(b[i]>a[j])j++;
}
cout<<j<<endl;
return ;
}

code force 401B. Game of Credit Cards的更多相关文章

  1. Codeforces 777B Game of Credit Cards

    B. Game of Credit Cards time limit per test:2 seconds memory limit per test:256 megabytes input:stan ...

  2. Codeforces777B Game of Credit Cards 2017-05-04 17:19 29人阅读 评论(0) 收藏

    B. Game of Credit Cards time limit per test 2 seconds memory limit per test 256 megabytes input stan ...

  3. Game of Credit Cards(贪心+思维)

    After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle betwe ...

  4. Game of Credit Cards

    After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle betwe ...

  5. Codeforces 777B:Game of Credit Cards(贪心)

    After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle betwe ...

  6. pycharm debug后会出现 step over /step into/step into my code /force step into /step out 分别表示

    1.debug,全部打印 2.打断点debug,出现单步调试等按钮,只运行断点前 3.setup over 调试一行代码 4.setup out 运行断点后面所有代码 5.debug窗口显示调试按钮 ...

  7. code forces 1173 C. Nauuo and Cards

    本文链接:https://www.cnblogs.com/blowhail/p/10990833.html Nauuo and Cards 原题链接:http://codeforces.com/con ...

  8. 【Code Force】Round #589 (Div. 2) D、Complete Tripartite

    题目链接 大致题意 把一个图分成三块,要求任意两块之间是完全图,块内部没有连线 分析 首先根据块内没有连线可以直接分成两块 假定点1是属于块1的,那么所有与点1连接的点,都不属于块1:反之则是块1的 ...

  9. code force 424 A - Office Keys

    There are n people and k keys on a straight line. Every person wants to get to the office which is l ...

随机推荐

  1. 【框架学习与探究之定时器--Quartz.Net 】

    声明 本文欢迎转载,原文地址:http://www.cnblogs.com/DjlNet/p/7572174.html 前言 这里相信大部分玩家之前现在都应该有过使用定时器的时候或者需求,例如什么定时 ...

  2. Azure SQL Database (25) Azure SQL Database创建只读用户

    <Windows Azure Platform 系列文章目录> 本文将介绍如何在Azure SQL Database创建只读用户. 请先按照笔者之前的文章:Azure SQL Databa ...

  3. 初学者一些常用的SQL语句(一)

    一.数据库的创建create database 数据库名create database bbb二.表的创建 ***[]:可选项*** null:空值 not null 不为空***只有字符型能指定长度 ...

  4. Opengl4.5 中文手册—F

    索引 A      B    C      D     E     F     G H      I     J      K     L     M     N O      P    Q      ...

  5. MySQL高级查询(一)

    修改表 修改表名 语法: ALTER  TABLE<旧表名> RENAME  [TO] <新表名>; 添加字段 语法: ALTER  TABLE 表名 ADD 字段名 数据类型 ...

  6. poj3070矩阵快速幂

    Fibonacci Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7752   Accepted: 5501 Descrip ...

  7. uva11806

    [题意] n行m列网格放k个石子.有多少种方法?要求第一行,第一列,最后一行,最后一列必须有石子. [题解] 利用容斥原理.可以转到求"第一行.第一列.最后一行.最后一列没有石子" ...

  8. 上传本地项目到githup仓库

    1.在网上下载Git,然后安装 点击下一步 2.默认选择,下一步 3.选择使用命令行环境,下一步 4.后续步骤默认选择,点击下一步,等待安装完成 5.在githup上面新建一个仓库存放项目代码,具体方 ...

  9. Java并发(一、概述)

    离上次写博客又隔了很久,心中有愧.在我不断使用Java的过程中,几乎都是拿来就用,就Java并发这块我还没有系统的梳理过,趁着国庆有空余时间,把它梳理一遍.以下部分内容参考相关书籍,以作学习之用,特此 ...

  10. validators配置要点及No result defined for action报错解决方案

    在做JavaEE SSH项目时,接触到validators验证. 需要了解validators配置,或者遇到No result defined for action 这个错误时,可查阅本文得到有效解决 ...