• HDOJ(HDU) 2192 MagicBuilding(用Java的Map做了下)


    Problem Description
    As the increase of population, the living space for people is becoming smaller and smaller. In MagicStar the problem is much worse. Dr. Mathematica is trying to save land by clustering buildings and then we call the set of buildings MagicBuilding. Now we can treat the buildings as a square of size d, and the height doesn’t matter. Buildings of d1,d2,d3….dn can be clustered into one MagicBuilding if they satisfy di != dj(i != j).
    Given a series of buildings size , you need to calculate the minimal numbers of MagicBuildings that can be made. Note that one building can also be considered as a MagicBuilding.
    Suppose there are five buildings : 1, 2, 2, 3, 3. We make three MagicBuildings (1,3), (2,3), (2) .And we can also make two MagicBuilding :(1,2,3), (2,3). There is at least two MagicBuildings obviously.

    Input
    The first line of the input is a single number t, indicating the number of test cases.
    Each test case starts by n (1≤n≤10^4) in a line indicating the number of buildings. Next n positive numbers (less than 2^31) will be the size of the buildings.

    Output
    For each test case , output a number perline, meaning the minimal number of the MagicBuilding that can be made.

    Sample Input
    2
    1
    2
    5
    1 2 2 3 3

    Sample Output
    1
    2

    其实说了这么多,就是找出现次数最多的那个数。

    练习了下Map的使用。也可以不用Map的。

    import java.util.Map;
    import java.util.Scanner;
    import java.util.TreeMap;
    
    public class Main{
    
        public static void main(String[] args) {
            Scanner sc =new Scanner(System.in);
            int t =sc.nextInt();
            while(t-->0){
                Map<Integer,Integer> map = new TreeMap<Integer, Integer>();
                int n=sc.nextInt();
                int m[] = new int[n];
                for(int i=0;i<n;i++){
                    m[i] = sc.nextInt();
                    if(map.get(m[i])==null){
                        map.put(m[i], 1);
                    }else{
                        map.put(m[i], map.get(m[i])+1);
                    }
                }
                int max=0;
                for(int i=0;i<n;i++){
                    if(map.get(m[i])>max){
                        max=map.get(m[i]);
                    }
                }
                System.out.println(max);
            }
        }
    }
    
  • 相关阅读:
    docker参数--restart=always的作用
    docker参数expose使用
    Linux主机添加路由和端口转发
    docker自动开启端口转发功能
    【Tips】【UE】总结自己常用的UltraEdit使用技巧
    浅谈I2C总线
    I2C总线简介(很经典)
    ECN
    视频编码未来简史
    爬虫与反爬虫
  • 原文地址:https://www.cnblogs.com/webmen/p/5739216.html
Copyright © 2020-2023  润新知