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
5greenredblueredred3pinkorangepink0
Sample Output
redpink
输入n个气球颜色，输出颜色最多的气球的颜色。第一遍用结构体做的，，想到还可以用map做，就用map做了，但还是出现了一些状况···map用的不熟练。map很强大！代码如下：
#include<cstdio>#include<map>#include<iostream>#include<string>using namespace std;int main(){int n;map<string,int>balloon;string maxcolor,color;while(scanf("%d",&n)==1&&n){balloon.clear();int maxn=0;for(int i=0;i<n;i++){cin>>color;balloon[color]++;}map<string,int>::iterator it=balloon.begin();for(;it!=balloon.end();it++)if(it->second>maxn){maxn=it->second;maxcolor=it->first;}cout<<maxcolor<<endl;}return 0;}
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