HDOJ 1004 Let the Balloon Rise

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
  
  

   想法很简单，建立一个结构体数组存放颜色和出现次数，
  
  

   如果新加入的颜色之前出现过，那么只要把次数加1，
  
  

   如果没出现过，push进去既可以，num=1。
  
  

   然后扫描一次，选出最大的
  

#include <iostream> #include <algorithm> #include <string> #include <vector> using namespace std; struct data { string color; int num; }; int main() { int n; while(cin>>n&&n) // n个测试数据 n=0时退出 { string scolor; data temp; vector<data>::iterator i; vector<data> lis; while(n--) { cin>>scolor; for(i=lis.begin();i!=lis.end();i++) //如果当前向量中已经存在，那么数值+1即可 { if((*i).color==scolor) { (*i).num++; break; } } if(i==lis.end()) //如果不存在，则加到队列中 { temp.color=scolor; temp.num=1; lis.push_back(temp); } } int max=0; string max_color; for(i=lis.begin();i!=lis.end();i++) { if((*i).num>max) { max=(*i).num; max_color=(*i).color; } } cout<<max_color<<endl; } }