CodeForces 581A Vasya the Hipster(简单题)——Codeforces Beta Round #322 (Div. 2)

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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 test(s)
input
3 1
output
1 1
input
2 3
output
2 0
input
7 3
output
3 2

Note

In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.

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题意：有a双红袜子和b双蓝袜子，刚开始每天两只脚要穿不同颜色的袜子，即左脚穿红袜子，右脚穿蓝袜子，或者左脚穿蓝袜子，右脚穿红袜子，而且每天穿过的袜子都会扔掉，问有多少天可以穿不同颜色的袜子，接下来又有多少天可以穿袜子(即两只脚都有袜子穿)

解题思路：首先针对第一问，有多少天可以穿不同颜色的袜子，因为每天会消耗一只蓝袜子和一只红袜子，所以天数就是袜子少的那种颜色的袜子的数量，即min(a,b)

而接下来有多少天可以穿袜子，这个也简单，除去之前用掉的袜子，剩下的袜子除以2即为所求

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>#include<vector>#include<map>#include<set>#include<stdlib.h>#include<cmath>#include<string>#include<algorithm>#include<iostream>#define exp 1e-10using namespace std;const int N = 1001;const int inf = 1000000000;const int mod = 2009;int main(){    int a,b;    scanf("%d%d",&a,&b);    if(a<b)        swap(a,b);    printf("%d %d\n",b,(a-b)/2);    return 0;}

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