Let the Balloon Rise

Problem Description

Contest time again! How excited it is to see balloonsfloating around. But to tell you a secret, the judges' favorite time isguessing the most popular problem. When the contest is over, they will countthe 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 casestarts with a number N (0 < N <= 1000) -- the total number of balloonsdistributed. The next N lines contain one color each. The color of a balloon isa string of up to 15 lower-case letters.
A test case with N = 0 terminates the input and this test case is not to beprocessed.

Output

For each case, print the color of balloon for the mostpopular problem on a single line. It is guaranteed that there is a uniquesolution for each test case.

Sample Input

5
green
red
blue
red
red
3
pink
orange
pink
0

Sample Output

red
pink

#include <stdio.h> 
#include <string.h>
#define N 1001

int main(){
    int number;
    char temp[16];    //temp存放输入的颜色
    char color[N][16];  //color所有不同的颜色
    int amount[N];   //amount存放所有不同颜色的数量
    int i;
    int j;
    int k;   //k为不同颜色的数量
    int max;
    int flag;

    while(1){
        scanf("%d",&number);

        if(number==0)
            break;

        for(i=0;i<number;i++)
            amount[i]=0;

        k=0; 
        for(i=0;i<number;i++){
            scanf("%s",temp);

            if(i==0){
                strcpy(color[0],temp);
                k++;
                continue;
            }

            for(j=0;j<k;j++){
                if(strcmp(temp,color[j])==0){
                    amount[j]++;
                    break;
                }
            }

            if(j==k){
                strcpy(color[k],temp);    
                k++;
            }
        }

        max=amount[0];
        flag=0;
        for(i=0;i<k;i++){
            if(amount[i]>max){
                max=amount[i];
                flag=i;
            }
        }

        printf("%s
",color[flag]);
    }
    
    return 0;
}