HDOJ-1004 Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)	Memory Limit: 65536/32768 K (Java/Others)

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
5
green
red
blue
red
red
3
pink
orange
pink
0
Sample Output
red
pink

统计出现次数最多的字符串然后输出，用冒泡的方法将相同字符串出现的次数统计到一个数组，然后记住出现次数最多位置，将这个位置的字符串输出。

//AC: 15MS  1692K
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main(){
    int N;
    while(scanf("%d",&N)!=EOF){
        char color[1005][20];
        int max=0,t=0,num[1005];
        if(N==0)
            break;
        for(int i=0;i<N;i++){
            scanf("%s",color[i]);
        }
        memset(num,0,sizeof(num));
        for(int i=1;i<N;i++)
            for(int j=0;j<i-1;j++)
                if(strcmp(color[i],color[j])==0)
                    num[i]++;
        for(int i=1;i<N;i++){
            if(max<num[i]){
                max=num[i];
                t=i;
            }
        }
        printf("%s\n",color[t]);
    }
    return 0;
}

开始还准备用map< string,int >来写，结果自己完全没理解map，满脑子骚操作。