Let the Balloon Rise（STL的map）_let the balloon rise map算法-优快云博客

本文链接：https://blog.youkuaiyun.com/Enjoying_Science/article/details/40950549

本文介绍了一个简单的编程问题，即统计编程比赛中每种颜色气球的数量并找出最受欢迎的问题颜色。使用C++实现，涉及输入输出流、字符串处理及映射数据结构。

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Link：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2104

ZOJ Problem Set - 2104

Let the Balloon Rise

Time Limit: 2 Seconds Memory Limit: 65536 KB

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: Zhejiang Provincial Programming Contest 2004

#include<iostream>
#include<algorithm>
#include<map>
#include<cstdio>
#include<cstring>
#include<sstream>
using namespace std;
map<string,int>ball;
map<string,int>::iterator loc,maxpos;
int main()
{
	int n,max;
	string color;
	while(~scanf("%d",&n)&&n)
	{
	//	if(n==0)
	//	break;
	    ball.clear();//注意每次都要清空，否则会wrong answer！！！ 
		while(n--)
		{
			cin>>color;
			ball[color]++;
		}
	    max=0;
	for(loc=ball.begin();loc!=ball.end();loc++)
	{
		if(max<loc->second)
		{
			maxpos=loc;
			max=(*loc).second;
		}
	}
	//cout<<(*maxpos).first<<endl;
	cout<<maxpos->first<<endl;//注意“.”与“->”的用法差别！！！否则会报错。迭代器本身就是指针地址，可直接用“->”
	// 或加“*”取其值后用点“.”访问对应的数据。 
   }
   return 0;
}