"needs to make exactly k nano liters of the oxide in any of the test tubes"

When you have k nano liters in a test tube that is not #1 the checker gives WA. My AC program was getting WA because my final test tube wasn't #1...

Got WA #1 with the following output:
4
2 5
4 3
3 5
5 1

It seems correct, except for the fact the final tube is number 5, and not #1.
Anyways, nice problem <3

k = 1
k = 2^m for some m
k = n
k = n-1

cases are enough for solve problem :)

Does solution for (n,k) (19,18) exist?
I only find the answer for (21,18).
Do you guy have a better idea?

31
2 1
4 3
3 1
1 5
7 6
6 5
5 1
1 5
9 8
8 5
5 10
12 11
11 10
10 5
5 1
1 5
13 10
10 5
5 14
16 15
15 14
14 5
5 14
18 17
17 14
14 19
21 20
20 19
19 14
14 5
5 1

Consider the test "8 6" - there is solution, while your current logic likely outputs "-1"

this is my checker of this problem, it can't be wrong ,but WA on test 8

    if(i>=n||a[i]!=k)
    {
        while(!(n==5&&k==5))
           puts("orz");
       puts("-1");
        }
    else
    {
        printf("%d\n",cnt_ans);
        while(cnt_ans>3*n)
           puts("orz");
        for(i=0;i<cnt_ans;i++)
           printf("%d %d\n",ax[i],ay[i]);
        for(i=0;i<n;i++)
            a[i]=1;
        for(i=0;i<cnt_ans;i++)
        {
            int id1=ax[i]-1,id2=ay[i]-1;
            while(id1>=n||id1<0||id2>=n||id2<0||id1==id2)
               puts("orz");
            a[id1]-=a[id2];
            a[id2]*=2;
            while(a[id1]<0)
               puts("orz");
        }
        for(i=0;i<n;i++)
        {
            if(a[i]==k)
               break;
        }
        while(i!=0)
           puts("orz");
    }

WA#8 confuse me.

4
2 1 => 2 0 1 1 1
4 3 => 2 0 2 0 1
3 1 => 4 0 0 0 1
1 5 => 4 в 1 => нестабильная смесь => улетает чистый кислород.