A. Vasya the Hipster

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/581/problem/A

Description

One day Vasya the Hipster decided to count how many socks he had. It turned out that he had a red socks and b blue socks.

According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.

Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.

Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.

Can you help him?

Input

The single line of the input contains two positive integers a and b (1 ≤ a, b ≤ 100) — the number of red and blue socks that Vasya's got.

Output

Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.

Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.

Sample Input

3 1

Sample Output

1 1

HINT

题意

一个人有a个红袜子,b个蓝袜子,问他最多一只脚穿红一只脚穿蓝能穿多少天

并且之后还能穿多少天,他不想洗袜子,穿完就会扔啦

题解:

水题啦

代码:

//qscqesze
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 1205000
#define mod 1000000007
#define eps 1e-9
#define e exp(1.0)
#define PI acos(-1)
#define lowbit(x) (x)&(-x)
const double EP = 1E- ;
int Num;
//const int inf=0x7fffffff;
const ll inf=;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************* int main()
{
int a=read(),b=read();
cout<<min(a,b)<<" "<<abs(a-b)/<<endl;
}

Codeforces Round #322 (Div. 2) A. Vasya the Hipster 水题的更多相关文章

  1. Codeforces Round #281 (Div. 2) B. Vasya and Wrestling 水题

    B. Vasya and Wrestling 题目连接: http://codeforces.com/contest/493/problem/B Description Vasya has becom ...

  2. Codeforces Round #297 (Div. 2)A. Vitaliy and Pie 水题

    Codeforces Round #297 (Div. 2)A. Vitaliy and Pie Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: xxx  ...

  3. Codeforces Round #290 (Div. 2) A. Fox And Snake 水题

    A. Fox And Snake 题目连接: http://codeforces.com/contest/510/problem/A Description Fox Ciel starts to le ...

  4. Codeforces Round #373 (Div. 2) B. Anatoly and Cockroaches 水题

    B. Anatoly and Cockroaches 题目连接: http://codeforces.com/contest/719/problem/B Description Anatoly liv ...

  5. Codeforces Round #368 (Div. 2) A. Brain's Photos 水题

    A. Brain's Photos 题目连接: http://www.codeforces.com/contest/707/problem/A Description Small, but very ...

  6. Codeforces Round #359 (Div. 2) A. Free Ice Cream 水题

    A. Free Ice Cream 题目连接: http://www.codeforces.com/contest/686/problem/A Description After their adve ...

  7. Codeforces Round #355 (Div. 2) A. Vanya and Fence 水题

    A. Vanya and Fence 题目连接: http://www.codeforces.com/contest/677/problem/A Description Vanya and his f ...

  8. Codeforces Round #281 (Div. 2) D. Vasya and Chess 水

    D. Vasya and Chess time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  9. Codeforces Round #384 (Div. 2) A. Vladik and flights 水题

    A. Vladik and flights 题目链接 http://codeforces.com/contest/743/problem/A 题面 Vladik is a competitive pr ...

随机推荐

  1. Android开发之动画(转)

    activity跳转的过渡效果,很漂亮,很全 注意,切换方法overridePendingTransition只能在startActivity和finish方法之后调用. 第一个参数为第一个Activ ...

  2. Servlet个人总结

    netstat -an ——查看端口占用情况 netstat -an ——查看是谁占用了哪个端口 端口被占用之后可以关闭端口占用程序或者在conf/server.xml修改本身使用端口 javac - ...

  3. 在SQLite中使用索引优化查询速度

    在进行多个表联合查询的时候,使用索引可以显著的提高速度,刚才用SQLite做了一下测试. 建立三个表: create table t1 (id integer primary key,num inte ...

  4. SDOI2008Cave 洞穴勘测

    无限膜拜CLJ大牛…… 不会动态树的弱弱在CLJ的帮助下AC了此题 我想到了并查集(人人都会想到的吧……囧),但不知道应该如何处理destroy操作…… 其实 make操作的实质就是:把x节点到其所在 ...

  5. 【聚类算法】谱聚类(Spectral Clustering)

    目录: 1.问题描述 2.问题转化 3.划分准则 4.总结 1.问题描述 谱聚类(Spectral Clustering, SC)是一种基于图论的聚类方法——将带权无向图划分为两个或两个以上的最优子图 ...

  6. Mysql slave 状态之Seconds_Behind_Master

    在MySQL的主从环境中,我们可以通过在slave上执行show slave status来查看slave的一些状态信息,其中有一个比较重要的参数Seconds_Behind_Master.那么你是否 ...

  7. CVTE公司面经

    1.先是网上测评,通过后通知你参加一面. 2.关于一面:一共进行了10分钟左右,三四个人一起面,没有问什么技术. 一共问了3个问题:a.你为什么选择我们公司的这个职位.我答的大概意思是本科研究生期间, ...

  8. HDU 5536 Chip Factory 字典树+贪心

    给你n个数,a1....an,求(ai+aj)^ak最大的值,i不等于j不等于k 思路:先建字典树,暴力i,j每次删除他们,然后贪心找k,再恢复i,j,每次和答案取较大的,就是答案,有关异或的貌似很多 ...

  9. 自动化测试(二):QTP验证点

    1 程序自带验证点 自带验证点:图形界面insert  ->  checkpoint Standard Checkpoint 标准验证:用于检查测试对象的属性 Text Checkpoint 文 ...

  10. PHP 进行蜘蛛访问日志统计

    $useragent = addslashes(strtolower($_SERVER['HTTP_USER_AGENT'])); if (strpos($useragent, 'googlebot' ...