Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 87703    Accepted Submission(s): 33157


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
 
//std:map<int, string> personnel;这样就定义了一个用int作为索引,并拥有相关联的指向string的指针.
 1 #include<stdio.h>
 2 #include<string>
 3 #include<map>
 4 #include<iostream>
 5 using namespace std;
 6 int main()
 7 {
 8     int n;
 9     map<string,int>color; 
10     string str,temp;
11     while(~scanf("%d",&n),n)
12     {
13         int max=0,i;
14         for(i=0;i<n;i++)
15         {
16             cin >> str;
17             color[str]++;
18             if(color[str]>max)
19             { temp=str; max=color[str];}
20             
21         }
22         cout << temp << endl;
23     }
24     return 0;
25 }

 

posted on 2015-06-23 00:13  cleverbiger  阅读(163)  评论(0编辑  收藏  举报