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 resultif (count == COUNT)y[k++] = tmp;}// printfor (int i = 0; i < k; i++)printf(" %d", y[i]);return 0;
}