HDU 1004 Let the Balloon Rise

最新推荐文章于 2021-06-24 22:53:24 发布

原创最新推荐文章于 2021-06-24 22:53:24 发布 · 433 阅读

CC 4.0 BY-SA版权

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 73297 Accepted Submission(s): 27379

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

学习map容器的使用。map容器的基础题，没什么好说的。代码如下：

#include <iostream>
#include<cstdio>
#include<map>
#include<string>
using namespace std;
int n,maxn;
string maxc;
//char* a;
string a;
map<string, int> bollon;
int main()
{
while(scanf("%d",&n)&&n!=0)
{
int i;
bollon.clear();//多次输入，使用前注意清空bollon
for(i=0;i<n;i++)
{
//scanf("%s",a);
cin>>a;
bollon[a]++;
}
maxn=0;
map<string ,int>::iterator ite;
for(ite=bollon.begin();ite!=bollon.end();ite++)
if(ite->second>maxn)
{
maxn=ite->second;
maxc=ite->first;
}
cout<<maxc<<endl;
}
return 0;
}

附录：关于map容器的一些知识：

C++中map容器提供一个键值对容器，map与multimap差别仅仅在于multiple允许一个键对应多个值。

一、map的说明
  1   头文件
  #include   <map>

  2   定义
  map<string,   int>   my_Map;
  或者是typedef     map<string,   int>   MY_MAP;
  MY_MAP   my_Map;

  3   插入数据
  (1)   my_Map["a"]   =   1;
  (2)   my_Map.insert(map<string,   int>::value_type("b",2));
  (3)   my_Map.insert(pair<string,int>("c",3));
  (4)   my_Map.insert(make_pair<string,int>("d",4));

  4   查找数据和修改数据
  (1)   int   i   =   my_Map["a"];
            my_Map["a"]   =   i;
  (2)   MY_MAP::iterator   my_Itr;
            my_Itr.find("b");
            int   j   =   my_Itr->second;
            my_Itr->second   =   j;
  不过注意，键本身是不能被修改的，除非删除。

  5   删除数据
  (1)   my_Map.erase(my_Itr);
  (2)   my_Map.erase("c");
  还是注意，第一种情况在迭代期间是不能被删除的，道理和foreach时不能删除元素一样。

  6   迭代数据
  for   (my_Itr=my_Map.begin();   my_Itr!=my_Map.end();   ++my_Itr)   {}

  7   其它方法
  my_Map.size()               返回元素数目
  my_Map.empty()       判断是否为空
  my_Map.clear()           清空所有元素
  可以直接进行赋值和比较：=,   >,   >=,   <,   <=,   !=   等等

  更高级的应用查帮助去吧，^_^;