Kickstart Practice Round 2017---A

Problem

The Constitution of a certain country states that the leader is the person with the name containing the greatest number of different alphabet letters. (The country uses the uppercase English alphabet from A through Z.) For example, the name GOOGLE has four different alphabet letters: E, G, L, and O. The name APAC CODE JAM has eight different letters. If the country only consists of these 2 persons, APAC CODE JAM would be the leader.

If there is a tie, the person whose name comes earliest in alphabetical order is the leader.

Given a list of names of the citizens of the country, can you determine who the leader is?

Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case starts with a line with an interger N, the number of people in the country. Then N lines follow. The i-th line represents the name of the i-th person. Each name contains at most 20 characters and contains at least one alphabet letter.

Output

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the name of the leader.

Limits

1 ≤ T ≤ 100.
1 ≤ N ≤ 100.

Small dataset

Each name consists of at most 20 characters and only consists of the uppercase English letters A through Z.

Large dataset

Each name consists of at most 20 characters and only consists of the uppercase English letters A through Z and ' '(space).
All names start and end with alphabet letters.

Sample

Input	Output
2 3 ADAM BOB JOHNSON 2 A AB C DEF	Case #1: JOHNSON Case #2: A AB C

一开始的错误解法：

 

思路：将每个名字的字母放到一个数组中，统计不同字母的个数，每个案例中个数最多的就是Leader，没有考虑到如果遇到相同字母数量的情况。incorrrect

 

package kickstart2017;
import java.io.*;
public class CountryLeader {

    public static void main(String[] args) {
        File outfile = new File("D://Code//Java//workspace//kickstart2017//src//kickstart2017//outputforsmall.txt");    //创建输出文件对象
        try {
            FileWriter out =new FileWriter(outfile);    //创建FileWriter对象
            BufferedWriter bufw = new BufferedWriter(out);  //创建BufferedWriter类对象
            FileReader fr = new FileReader("D://Code//Java//workspace//kickstart2017//src//kickstart2017//A-small-practice.in");
            BufferedReader bufr = new BufferedReader(fr);     
            String cases = null;      
            //读取第一行信息得到case的值,为字符变量
            cases = bufr.readLine();
            int numofcases = Integer.parseInt(cases);    //字符串转变成int常量
            for(int j = 1;j < numofcases+1; j++){        //对每一个案列分别进行处理
                String N = null;                
                N = bufr.readLine();
                int numofnames = Integer.parseInt(N);
                int numofcharacter[] = new int[numofnames];    //数组存放每个名字的字母个数
                String nameofarrays[] = new String[numofnames];//将所有名字放入到一个字符串数组中去
                for(int k = 0;k < numofnames; k++){   //对每个名字即每行进行处理                    
                    int ch[] = new int[26];   //数组存放26个字母的出现次数
                    String names = bufr.readLine();   //读取一行，得到名字中包含所有的字母
                    nameofarrays[k] = names;
                    for(int m = 0;m < names.length();m++){
                        char c = names.charAt(m);  //依次取出每个字母
                        int index = c-'A';         //
                        ch[index] = ch[index] + 1;// 对应字母出现则存储字母的数组加1                        
                    }
                    int numofalp = 0;             //求出每个数组中不为0的元素的个数即为不同字母的个数            
                    for(int n = 0;n < 26; n++){
                        if(ch[n]>= 1){
                           numofalp++;
                        }
                    }
                    numofcharacter[k] = numofalp;  //将每个名字包含的字母数存储到数组中
                }
                int maxzhi = 0;
                int maxzhiindex = 0;
                for(int p = 0;p < numofnames;p++ ){    //求取每个案列中的最大值和其对应的名字
                    maxzhi = numofcharacter[0];
                    if(numofcharacter[p] > maxzhi){
                        maxzhi = numofcharacter[p];
                        maxzhiindex = p;
                    }
                }
                
                System.out.println("Case #"+ j +":" +" " + nameofarrays[maxzhiindex]);
                bufw.write("Case #"+ j +":" +" " + nameofarrays[maxzhiindex]);
                bufw.newLine();
            }
                        
            bufr.close();          
            fr.close();              //将FileReader流关闭
            bufw.close();
            out.close();             // 将输出流关闭
            
        } catch (FileNotFoundException e) {
            
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        
    }
    
    
}