找出那个数字出现3次的数字

找出那个数字出现3次的数字，简单的C++

#include <stdio.h>
#define COUNT  3
#define  ARRAY_SIZE(_) (sizeof (_) / sizeof (*_))


int _tmain(int argc, _TCHAR* argv[])
{
    // 找出那个数字出现3次的数字
    // x=[1 2 3 1 2 5 1 2]
    int x[] = {1, 2, 3, 1, 2, 5, 1, 2};
    int y[] = {0, 0, 0, 0 ,0, 0, 0, 0};

    int n = ARRAY_SIZE(x);
    int k = 0;

    // second 
    for (int i = 0; i < n; i++)
    {
        int count = 0;
        int flag = 0;
        int tmp = x[i];
        // check this value have exit ?
        for (int j = 0; j < k; j++)
        {
            if (!(y[j] ^ tmp))
            {
                flag = 1;
                break;
            }
        }

        if (flag)
            continue;

        // count 
        for (int j = i/* 0 */; j < n; j++)
        {
            if (!(tmp ^ x[j]) && (count++ > COUNT))
                break;
        }

        // save result
        if (count == COUNT)
            y[k++] = tmp;
    }

    // print
    for (int i = 0; i < k; i++)
        printf(" %d", y[i]);


    return 0;
}