hdu 5123 who is the best?（Bestcoder Round #20)

who is the best?

                                                                Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                 Total Submission(s): 128    Accepted Submission(s): 87

Problem Description

There are N people want to choose the best person. Each person select the best person  ai , .John wants to know that who received the most number of votes.

 

Input

The first line contains a single integer  T(1≤T≤50) ,indicating the number of test cases.
Each test case begins with an integer  N(1≤N≤100) ,indicating the number of person.
Next N lines contains an integer  ai(1≤ai≤N) .

 

Output

For each case, output an integer means who is the best person. If there are multiple answers, print the minimum index.

 

Sample Input

   
   
   
   

    
    
    
    2
10
1
2
3
4
5
6
7
8
9
10
5
3
3
3
3
3
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
3
   
   
   
   

 

    官方题解：

我们对于每个 ai 都进行计数，即 b[ai]++ ，如此之后，我们可以一个循环语句 i=1−>n 来寻找最大的 bi ,注意此时应是 bi>MAX ，而不是 bi≤MAX ，这样才能保证出现的是编号最小的。另外需要注意的是每次做完后数组应当清0，否则会影响下次的答案，由于各种非确定性因素我在小数据就已经把没清0的程序卡死了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int a[110];
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        memset(a,0,sizeof(a));
        int max=0;
        for(int i=0;i<n;i++)
        {
            int x;
            scanf("%d",&x);
            a[x]++;
            if(a[x]>max)
            max=a[x];
        }
        int ans;
        for(int i=0;i<=n;i++)
        if(a[i]==max)
        {
            ans=i;
            break;
        }
        printf("%d\n",ans);
    }
    return 0;
}