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/Others) Total Submission(s): 123800    Accepted Submission(s): 48826

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

Author

WU, Jiazhi

Source

ZJCPC2004

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1004

分析：STL大法好啊，本题要求最大的气球数，可以用到STL标准库中的map<string,int>来保存，当map的first插入一个string的时候，map的second++，就可以统计出具体的string的个数了，然后遍历找到最大值就可以了！

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     map<string,int>ballon;
 6     int n;
 7     string str;
 8     while(cin>>n&&n)
 9     {
10         ballon.clear();
11         while(n--)
12         {
13             cin>>str;
14             ballon[str]++;
15         }
16         int maxn=0;
17         string maxcolor;
18         map<string,int>::iterator it;
19         for(it=ballon.begin();it!=ballon.end();it++)
20         {
21             if(it->second>maxn)
22             {
23                 maxn=(*it).second;
24                 maxcolor=(*it).first;
25             }
26         }
27         cout<<maxcolor<<endl;
28     }
29     return 0;
30 }