HDOJ-1004(map)

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 103000    Accepted Submission(s): 39524

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

 

先上题意：

首先输入数据个数n，以下n行输入n个字符串表示气球颜色，输出出现次数最多的颜色。

本题也是做过好久的老题了，上次我们探讨的是用字符串比较函数strcmp()来比较两字符串是否相等从而判断出现次数（见这里）。这次我们用map映照容器来重新做一下这个题目。

用map容器相对来说就要简单一些，每当输入相应的颜色，与之相对应的数值就+1，最后再输出最大数值最大的颜色就好。

附AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;

int main(){
    map<string,int> Bollon;//定义map容器 
    string Color,Maxcolor;
    int n;
    while(~scanf("%d",&n)&&n!=0){
        Bollon.clear();//每次处理下一组数据前要清空容器 
        while(n--){//输入气球颜色，相应颜色气球个数加一 
            cin>>Color;
            Bollon[Color]++;
        }
        int max=0;
        map<string,int>::iterator it;//定义前向迭代器，向前遍历容器 
        for(it=Bollon.begin();it!=Bollon.end();it++){
            if(it->second>max){//不知为何换成it.second就编译不过 
                max=it->second;
                Maxcolor=it->first;//找出个数最多的气球颜色 
            }
        }
        cout<<Maxcolor<<endl;
    }
    return 0;
}