Let the Balloon Rise

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

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Total Submission(s): 86334    Accepted Submission(s): 32613

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

Author

WU, Jiazhi

Source

​​ZJCPC2004​​

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解析：这里我主要用到两个函数：

             strcmp（s1，s2）判断字符串s1与s2是否相等：

                  1. s1==s2，其值为 0

                  2. s1<s2，  其值 <0

                  3. s1>s2，  其值>0

             strcpy(s1，s2)，将字符串s2赋值给字符串s1。

代码：

#include<cstdio>
#include<cstring>
#define maxn 1000
using namespace std;

char s[maxn+10][20];
int sum[maxn+10];

int main()
{
  freopen("1.in","r",stdin);
  int n,i,j,k;
  while(scanf("%d",&n),n)
    {
      for(sum[0]=0,i=1;i<=n;i++)
        {
          scanf("%s",s[0]);
          for(i=1;i<=sum[0];i++)
            if(strcmp(s[0],s[i])==0)sum[i]++;
          if(i>sum[0])
        {
          sum[++sum[0]]=1;
          strcpy(s[sum[0]],s[0]);
        }  
        }
      for(k=1,i=2;i<=sum[0];i++)
      if(sum[i]>sum[k])k=i;
      printf("%s\n",s[k]);      
    }
  return 0;
}