#433 - B. Maxim Buys an Apartment

待我称王封你为后i 2022-06-08 04:25 169阅读 0赞

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题目连接：[http://codeforces.com/contest/854/problem/B][http_codeforces.com_contest_854_problem_B]

## 题目描述 ##

### Description ###

Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.

Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn’t know their indices yet.

Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.

### Input ###

The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).

### Output ###

Print the minimum possible and the maximum possible number of apartments good for Maxim.

### Sample Input ###

6 3

### Sample Output ###

1 3

## 解题思路 ##

题意：一个人买房子，喜欢旁边有人住，n为房子个数（呈一排），m为已经居住的人，但是不知道哪个房子已经住人了，输出挨着人的最大和最小可以居住的位子的个数，旁边一定有人住。  
比如※有人住，×没人住  
8 2 的情况有4个位子  
×※××※×  
除去0的情况，即n == m , m == 0 , 的情况；  
剩下的由规律得min(n-m, m<<1)

## AC代码 ##

#include<iostream>
    #include<algorithm>
    
    using namespace std;
    
    int main() {
        int n, m;
        while(cin >> n >> m) {
            if(m == 0 || n == m) {
                cout << "0 0" << endl; 
                continue;
            }
            cout << "1 " << min(n-m, m<<1) << endl;
        }
        return 0;
    }

[http_codeforces.com_contest_854_problem_B]: http://codeforces.com/contest/854/problem/B