HDU-1029 Ignatius and the Princess IV

题目：
“OK, you are not too bad, em… But you can never pass the next test.” feng5166 says.

“I will tell you an odd number N, and then N integers. There will be a special integer among them, you have to tell me which integer is the special one after I tell you all the integers.” feng5166 says.

“But what is the characteristic of the special integer?” Ignatius asks.

“The integer will appear at least (N+1)/2 times. If you can’t find the right integer, I will kill the Princess, and you will be my dinner, too. Hahahaha…” feng5166 says.

Can you find the special integer for Ignatius?

Input
The input contains several test cases. Each test case contains two lines. The first line consists of an odd integer N(1<=N<=999999) which indicate the number of the integers feng5166 will tell our hero. The second line contains the N integers. The input is terminated by the end of file.

Output
For each test case, you have to output only one line which contains the special number you have found.

Sample Input
5
1 3 2 3 3
11
1 1 1 1 1 5 5 5 5 5 5
7
1 1 1 1 1 1 1

Sample Output
3
5
1
题目是英文的，在此我就不过多赘述了···
刚开始我是用桶排序来写的，然后一直在死循环···自己半天也找不出来···如果有哪位好心人能看出错误，麻烦您跟我讲一下···实在是自己也搞不明白
错误代码在这：

#include
int a[1000000],b[1000000];
int main(){
 int N;
 while(~scanf("%d",&N)){
  for(int i = 0;i < N; i++){
   scanf("%d",&a[i]);
   b[a[i]]++;
     }
     for(int i = 0;i < 10000000; i++){
      if(b[i] >= (N+1)/2)
       printf("%d",i);
  }
  printf("\n");
 }
 return 0;
}

最后我换了一种方法，先用sort将输入的数字从小到大排序，然后记录下每个数字出现的次数，大于（N+1）/2就将此数字输出，然后过了

#include
#include
using namespace std;
int a[1000000];
int main(){
	int N;
	while(~scanf("%d",&N)){
		for(int i = 1;i <= N; i++){
			scanf("%d",&a[i]);
	    }
	   sort(a,a+N);
	   int flag = 1;
	   for(int i = 1;i <= N; i++){
			if(a[i+1] == a[i])
				flag++;
			else{
				if(flag >= (N+1)/2)
					printf("%d",a[i]);
				flag = 1;
			} 
	   }
		printf("\n");
	}
	return 0;
}