#include <stdio.h>
#include <math.h>
int main()
{
	int n,l=0;
	int sb[9999][9] = { 0 };
	scanf("%d", &n);
	for (int i = 1; i <=n; i++)
	{
		for (int j = 1; j <=3; j++)
		{
			scanf("%d", &sb[i][j]);
		}

	}
	for (int i = 1; i <=n; i++)
	{
		for (int j = i+1; j <=n; j++)
		{
			if ((abs((sb[i][1] - sb[j][1]) <= 5))&&(abs((sb[i][2] - sb[j][2]) <= 5)) &&(abs((sb[i][3]- sb[j][3])<= 5))&&(abs(((sb[i][3]+sb[i][1]+sb[i][2])-(sb[j][3]+sb[j][1]+sb[j][2]))<=10)))
			{
				l++;
			}
		}
	}
	printf("%d", l);
	return 0;
}

完全按照答案第一个条的思路来打的代码，思路一模一样，细节也感觉没毛病，能过样例，但是全wa呜呜呜