《ACM程序设计》书中题目 O-15 color of balloons

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

   

   这道题的基本题意为每组输入n个数据为气球颜色，要求输出每组数据中出现次数最多的气球颜色。

   基本思路为输入气球颜色，然后储存在vector数组中，构建一个map数组用气球颜色作下标，数组值表示这种颜色气球的个数，最后找出个数最多的气球颜色。

源代码如下：

#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,i,max;
map<string,int>m;
vector<string>a;
string b,M;
while(cin>>n&&n!=0)
{ for(i=0;i<n;++i)
  { cin>>b;
    a.push_back(b);
    m[b]=0;
       }
 
 for(i=0;i<n;++i)
  if(m.find(a[i])!=m.end())++m[a[i]];
 max=m[a[0]];
 M=a[0];
 for(i=1;i<n;++i)
  if(m[a[i]]>max)
  { max=m[a[i]];
    M=a[i];
  }
 cout<<M<<endl;
 m.clear();
 a.clear();
}
} 

   这道题的难点在于如何将各种颜色的气球进行分类，并将其对应的个数求出来，然后再注意一下每次循环初始化个数组就ok了。