HDU_1004 Let the Balloon Rise

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 58567    Accepted Submission(s): 21505

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

 

Author

WU, Jiazhi

 

Source

ZJCPC2004

 

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JGShining

CODE:

#include<stdio.h>
#include<string.h>
int main()
{
char a[10001][16];
int i,j,num,max,t,flag;
while(scanf("%d",&num)!=EOF && num)
{
getchar();
for(i=0;i<num;i++)
gets(a[i]);
max=1;
flag=0;
for(i=0;i<num;i++)
{
t=1;
if(a[i][0]=='\0')continue;     //判断字符串是否为空
for(j=i+1;j<num;j++)
{
if(a[j][0]=='\0')continue;    //判断字符串是否为空
if(strcmp(a[i],a[j])==0)
{
t++;                   //记录颜色相同的气球数目
a[j][0]='\0';          //将比较过后的气球颜色变为空，以免重复计数
}
}
if(t>max)                      //找出数目最多的气球颜色
{
max=t;                 
flag=i;                    //记录数目最多的气球颜色的下标
}
}
puts(a[flag]);                     //输出最多数目的气球颜色
}
return 0;
}