HDU 1004

New~ 本学期最后一次入选"ACM校队"的机会——12月份月赛(赶快报名吧~)

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 62684    Accepted Submission(s): 23105

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

 

   这是一道 支付串相关 的题 主要用到 strcmp  函数进行比较 返回值为1 时 说明两个字符串相同；

  算法思想：

               1、 用什么数据结构 ，用a[1000][15] 二位数组来存放字符串 这个很关键

               2.、思想主要就是 查找 用第一个开始查找有没有和他相同的 有的话 计数加1 并且要注意的是 查找到相同的数组记得删除 即可；

               3、其实一套主要的是 max 的赋值方法 开始 给Max 赋值一个较小的值 如果计数 的数大于 max 将其值付给 max  并且记录位置 然后继续往下走

源代码：

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int main(){    char a[1000][15];    int n, po;    int i, j;    while(cin>>n)    {        po = n;        int maxm = 1;        if(n == 0)        {            return 0;        }        for(i = 0; i < n; i++)        {            scanf("%s", a[i]);        }        for(i = 0; i < n; i++)        {           int coun = 1;        //开始时没注意 这个计数要在 for里面进行计数的           if(a[i][0] == '\0') continue;           for(j = i+1; j < n; j++)           {               if(strcmp(a[i], a[j]) == 0)               {                  coun++;                  a[j][0] = '\0';               }           }           if(maxm <= coun)           {               maxm = coun;               po = i;           }        }        puts(a[po]);    }    return 0;}