HNU Missing Pages

For this problem we assume the classic approach is used for folding paper to make a booklet that has a number of pages that is a multiple of four. As an example, a newspaper with 12 pages would be made of three sheets of paper (see figure below). One sheet
would have pages 1 and 12 printed on one side, and pages 2 and 11 printed on the other. Another piece of paper would have pages 3 and 10 printed on one side and 4 and 9 printed on the other. The third sheet would have pages 5, 6, 7, and 8.

When one numbered page is taken from the newspaper, the question is what other pages disappear.

Input:  Each test case will be described with two integers
N and P, on a line, where 4 ≤ N ≤ 1000 is a multiple of four that designates the length of the newspaper in terms of numbered pages, and
1 ≤ P ≤ N is a page that has been taken. The end of the input is designated by a line containing only the value 0.

Output:  For each case, output, in increasing order, the page numbers for the other three pages that will be missing.

12 2
12 9
8 3
0

1 11 12
3 4 10
4 5 6

#include<stdio.h>
int main()
{
    int N,p,m,tp,p1,p2,p3;
    while(scanf("%d",&N)>0&&N)
    {
        scanf("%d",&p);
        m=N/4;
        if(2*m>=p)
        {
            tp=(p+1)/2;
            int k1=1,k2=N;
            for(int i=2;i<=tp;i++)
            k1+=2,k2-=2;
            if(k1!=p)p1=k1;else p1=k1+1;
            p2=k2-1; p3=k2;
        }
        else
        {
            tp=(p-m*2+1)/2;
            int k1=m*2,k2=m*2+1;
            for(int i=2;i<=tp;i++)
            k1-=2,k2+=2;
            if(k2!=p)p3=k2;else p3=k2+1;
            p1=k1-1;p2=k1;
        }
        printf("%d %d %d\n",p1,p2,p3);
    }
}