map容器

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1004

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

解法一：模拟

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct J//结构体的定义{    int a;    char b[20];} s[1005];int cmp(J x,J y)//自定义cmp函数，用来排序{    return x.a>y.a;}int main(){    int t;    while(~scanf("%d",&t)&&t)    {        int k=0;        for(int i=0;i<1005;i++)            s[i].a=0;        for(int i=0; i<t; i++)//两层for循环很关键        {            scanf("%s",s[k].b);            for(int j=0; j<k; j++)            {                if(!strcmp(s[k].b,s[j].b))                {                    s[j].a++;                    break;                }            }            s[k++].a++;        }        sort(s,s+t,cmp);        printf("%s\n",s[0].b);    }}解法二：map容器：
#include <iostream>#include <string>#include <iterator>#include <map>#include<stdio.h>using namespace std;map<string,int>mapp;char s[30];int main(){    int t;    while(~scanf("%d",&t)&&t)    {        mapp.clear();        int sum=0;       string sum1;        for(int i=0;i<t;i++)        {            scanf("%s",s);            mapp[s]++;        }        for(map<string,int>::iterator iter=mapp.begin();iter!=mapp.end();iter++)        {            if(iter->second>sum)            {                sum1=iter->first;                sum=iter->second;            }        }        cout<<sum1<<endl;    }}