Let the Balloon Rise

开始写技术博客了。

发发自己写的ACM的练习题。

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 53022Accepted Submission(s): 19135

Problem Description

Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges’ favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result.This year, they decide to leave this lovely job to you.

Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) — the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters.A test case with N = 0 terminates the input and this test case is not to be processed.

Output

For each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.

Sample Input

5

green

red

blue

red

red

3

pink

orange

pink0

Sample Output

red

pink

//JAVA CODE

import java.util.ArrayList;import java.util.Scanner;

public class Main{

public static void main(String args[]){　　Scanner scan = new Scanner(System.in);　　while(scan.hasNextInt()){　　　　int n = scan.nextInt();　　　　if(n == 0)　　　　　　break;　　　　ArrayList<String> color = new ArrayList<String>();　　　　ArrayList<Integer> count = new ArrayList<Integer>();　　　　String c = scan.next();　　　　color.add(c);　　　　count.add(1);　　　　String modeColor=c;　　　　Integer modeCount=1;　　　　while(–n > 0){　　　　　　String s = scan.next();　　　　　　if(color.contains(s)){　　　　　　　　count.set(color.indexOf(s),count.get(color.indexOf(s))+1);　　　　　　　　if(count.get(color.indexOf(s)) > modeCount){　　　　　　　　　　modeColor = s;　　　　　　　　　　modeCount = count.get(color.indexOf(s));　　　　　　　　}　　　　　　}　　　　　　else{　　　　　　　　color.add(s);　　　　　　　　count.add(1);　　　　　　}　　　　}　　　　System.out.println(modeColor);　　} }}

，网站空间，服务器空间，香港空间与一个赏心悦目的人错肩，真真实实的活着，也就够了。