HDU 1004 Let the Balloon Rise （map使用）

分析：题意简单，就用map的知识即可

#include <stdio.h>#include <iostream>#include <string.h>#include <map>using namespace std;int main()          {int i,a,b,k;int N;char num[10];    while (~scanf("%d",&N))    {map<string,int> mp;    while (N--)    {scanf("%s",num);mp[num]++;    }map<string,int>::iterator iter;int max=-1;for (iter=mp.begin();iter != mp.end();iter++){if (iter->second > max){max=iter->second;}}for (iter=mp.begin();iter != mp.end();iter++){if (max==iter->second){cout<<iter->first<<endl;break;}}    }return 0;}

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 87130    Accepted Submission(s): 32907

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

 

Author

WU, Jiazhi

 

Source

ZJCPC2004

2015-10-17

#include <iostream>#include <string>#include <map>using namespace std;int main(){int t;int num;int max;    string s;string smax;while(cin>>t,t){ max = 0;  map<string , int> m; while(t--) {  cin>>s; m[s]++; if(m[s] > max) { max = m[s];     smax = s; } } cout<<smax<<endl; }return 0;}