Codeforces Round #322 (Div. 2) A. Vasya the Hipster 水题

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-10 ;
int Num;
//const int inf=0x7fffffff;
const ll inf=999999999;
inline ll read()
{ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}
//*************************************************************************************int main()
{int a=read(),b=read();cout<<min(a,b)<<" "<<abs(a-b)/2<<endl;
}

转载于:https://www.cnblogs.com/qscqesze/p/4845054.html