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
pink
0
 
Sample Output
red
pink

思路

1.逐个读入(n)
2.对比第I个是否已经读入过,若有,则指示该类型增1,若无增添加该类型,若该类型的个数过半,则该类型为最多的。
 
最差的情况是,n个都互不相同,n(n-1+ n-2 + n-3 。。。。。+1),0.5*n^3
最好的情况是 n

还好AC了。

1

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Text; namespace hdoj {
class Programweb {
private static StringBuilder sb = new StringBuilder();
static void Main(string[] args) {
// //DateTime d1 = DateTime.Now;//local
// var lines = File.ReadLines("input2.txt").ToList();//local
int num = ; int lineIndex = ;
int lineNumIndex = -;
//获取行数
while (true) {
string s1 = "";
List<string> keys = new List<string>();
List<int> values = new List<int>();
bool keyExist = false;
int index = ;
// int max = 0;
if (int.TryParse(Console.ReadLine(), out num)) {//web
// lineNumIndex = lineNumIndex + num + 1;//local 指示数字行所在的行数=之前数字行所在的位置+上次的行数+数字本身的站1行
// if (int.TryParse(lines[lineNumIndex], out num)) {//local
for (int i = ; i < num; i++) {
s1 = Console.ReadLine();//web
// s1 = lines[i + lineNumIndex+1]; //local
// handOneLine(s1, i,num);
for (int j = ; j < keys.Count; j++) {
if (keys[j] == s1) {
keyExist = true;
index = j;
break;
} else {
keyExist = false;
}
}
//是否存在不同处理
if (keyExist) {
values[index]++;
if (values[index] > num / ) {
// max = index;
break;
}
} else {
keys.Add(s1);
values.Add();
} } if (num == ) {
break;
} //一次结束
if (values.Count != ) {
int max = ;
int maxIndex = ;
for (int i = ; i < values.Count; i++) {
if (values[i]>max) {
maxIndex = i;
max = values[i];
}
}
sb.AppendLine(keys[maxIndex]);
}
keys.Clear();
values.Clear(); //
// sb.AppendLine(string.Format("Case {0}:", i));
//sb.AppendLine(); } } Console.Write(sb.ToString()); //DateTime d2 = DateTime.Now;
//TimeSpan ts1 = d2 - d1;
//Console.WriteLine("Milliseconds:" + ts1.Milliseconds);
// Console.Read();//local
} } }

2

HDOJ 1004 Let the Balloon Rise的更多相关文章

  1. HDOJ 1004 Let the Balloon Rise (字符串+stl)

    题目: Problem Description Contest time again! How excited it is to see balloons floating around. But t ...

  2. hdu 1004 Let the Balloon Rise(字典树)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1004 Let the Balloon Rise Time Limit: 2000/1000 MS (J ...

  3. HDU 1004 Let the Balloon Rise(map的使用)

    传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1004 Let the Balloon Rise Time Limit: 2000/1000 MS (J ...

  4. HDU 1004 Let the Balloon Rise【STL<map>】

    Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Oth ...

  5. hdu 1004 Let the Balloon Rise

    Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Oth ...

  6. HDU 1004 Let the Balloon Rise map

    Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Oth ...

  7. hdu 1004 Let the Balloon Rise strcmp、map、trie树

    Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Oth ...

  8. 杭电1004 Let the Balloon Rise

    Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Oth ...

  9. hduoj#1004 -Let the Balloon Rise [链表解法]

    原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1004 Problem Description Contest time again! How exci ...

随机推荐

  1. 四则运算appNABCD模型

    团队: 郭志豪:http://www.cnblogs.com/gzh13692021053/ 杨子健:http://www.cnblogs.com/yzj666/ 刘森松:http://www.cnb ...

  2. 【oracle】oracle表结构导出到Word

    因为需要写数据库文档,所以需要把数据库里边的表结构在word中用表格列出来,之前一直用powerdesigner,感觉有些麻烦,后来在网上找到了一段sql语句,经测试完全符合我的需求,不敢独享,语句如 ...

  3. 从零开始山寨Caffe·零:必先利其器

    工作环境 巧妇有了米炊 众所周知,Caffe是在Linux下写的,所以长久以来,大家都认为跑Caffe,先装Linux. niuzhiheng大神发起了caffe-windows项目(解决了一些编译. ...

  4. 踩坑事件:不能对基于文本的临时表使用sql insert语句

    先来描述一下问题: 如果你是从基于文本的数据源来创建DataFrame的,当你将DataFrame注册为临时表后,如果对这个临时表进行insert into 操作,会抛出异常的. 问题答案参见:htt ...

  5. POJ 2251 题解

    Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 27520   Accepted: 10776 ...

  6. 基于FS4412的DS18B20温度采集编程实现(1-时序分析)

    作者:秦老师,华清远见嵌入式学院讲师. 一.DS18B20简介 DS18B20是常用的数字温度计.DS18B20数字温度计提供9至12位(可配置)温度读数,表明该设备的温度. 信息通过单总线接口被发送 ...

  7. Velocity 语法(转)

    一.基本语法 1."#"用来标识Velocity的脚本语句,包括#set.#if .#else.#end.#foreach.#end.#iinclude.#parse.#macro ...

  8. About_类与对象03

    php中的static: 1:属于静态变量:: 2:是全部类的属性: 3:调用静态变量要用::(两个冒号). eg:1 <html> <head> <title>s ...

  9. Android 之 ListView的学习

    ListView 是一个控件,一个在垂直滚动的列表中显示条目的一个控件,这些条目的内容来自于一个ListAdapter .EditText Button TextView ImageView Chec ...

  10. BSBuDeJie_02

    一 左边的类别数据 1 模型 和 字典中的数据对应 /* id */ @property (nonatomic, assign) NSInteger *id; /* 总数 */ @property ( ...