+1
Just use Dirichlet's principle.

1)
1
1 1
5 5
0 0
2 2
0 10

=>
OK
3

2)
2
1 1
2 2
-1 5
0 0
0 1
100 0
0 100
54 45
67 76
99 99

=>
OK
2 4

3)
2
1 2
3 4
5 6
7 8
9 10
11 12
-13 -14
-15 -16
-17 -18
-19 -20

=>
OK
1 2

В задаче, кстати, нифига не понятен случай n=1,т.к. понятие УВАЖЕНИЯ введено для ПАР. Но в тестах подразумевается, что человек сам себя уважает(хотя пары для него нет).

Another hint: imagine if you got n points and you could cheat and make them all respect each other by stretching or compressing the coordinate grid (while keeping the points static); what would be the simplest way to transform the grid? In this problem you can't cheat like that, but on the other hand you only need to have n/5 points work together.

The checker program is fixed. All submits were rejudged.

#include <cstdio>
int n;
int main()
{
    scanf("%d", &n);
    puts("OK");
    for(int i = 0; i < n; i++)
    {
        printf("1 ");
    }
    return 0;
}