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.
InputInput 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.
This year, they decide to leave this lovely job to you.
A test case with N = 0 terminates the input and this test case is not to be processed.
OutputFor 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 0Sample Output
red pink
#include<stdio.h>
#include<string.h>
int main()
{
int n;
int i,j=0;
char store[1000][20];//储存字符串，未初始化即为0
char t[20];
int max;
int count[1000]={0};//用来计第几个字符串出现的次数，并且用0初始化只能进行一次
int flag=0;
while(scanf("%d",&n)，n!=0)
{
memset(store, 0, sizeof store);//清零操作
memset(count, 0, sizeof count);
for(i=0;i<n;i++)//i只负责计数
{
scanf("%s",t);
for(j=0;j<n;j++)//j从0开始，可以保证用count将所有重复的字符串进行计数
{
flag=0;//标记是否重复
if(strcmp(t,store[j])==0)//与原有字符串进行比较
{
count[j]+=1;//如果原来存在，给相应地方加一
flag=1;
break;
}
}
if(flag==0)//如果没有录入，进行录入
strcpy(store[i],t);
}
max=0;
for(j=0;j<n;j++)//选出值最大下标
{
if(count[max]<count[j])
max=j;
}
printf("%s\n",store[max]);
scanf("%d",&n);
}
return 0;
}