Which algorithm?? Help

You can try to binary search the test case.

Put somewhere in your program the following

if (clock() >= 0.9 * CLOCK_PER_SEC) {
  assert(K <= BINARY_SEARCH_MIDDLE);
}

You need to include <cassert> and <ctime> for these to work.

Now  (double)clock() / CLOCK_PER_SEC gives you program execution time in seconds.
clock() >= 0.9 * CLOCK_PER_SEC   checks that your program runs more than 0.9 seconds, in other words, approaching TL of 1s.  This 'if' allows to execute code that will probably work only on 41th test case.

assert( boolean expression ); exits with non-zero value if condition is not met.

You could also do
if (K <= BINARY_SEARCH_MIDDLE) {
   exit(your favourite non-zero number);
}

So, you can start with assert(K <= 100000 / 2) and so on. If you program fails on 41th test with RE instead of TL you change the bin search value.

Yep, graph is very sparse. Almost always (without several trivial cases) you can consider only range [0..2*k]. To avoid any kinds of checking simply get n = max(10, 2*k). For each number in [1..n] try to find cycle with previous ones. Use e.g. disjoint sets for simplicity. Note, that solution always exists!

Interesting question...

Here are the correct answers:
1 3
2 5
3 5
4 7
5 6
6 8
7 10
8 12
9 11
10 11
11 8
12 15
13 14
14 17
15 20
16 23
17 18
18 20
19 15
20 24
21 24
22 18
23 21
24 32
25 19
26 27
27 23
28 31
29 13
30 35
31 22
32 39
33 29
34 35
35 40
36 44
37 30
38 33
39 42
40 44
41 31
42 45
43 32
44 39
45 50
46 43
47 44
48 60
49 27
50 39

Administration does.

Just emulate the professor's operations using some precalculation...

my algorithm is that:
if a conect to b,and b conect to c,so a%n==c%n;
that means c=a+n;and the first cycle will happen at max(b+n,a+n);
if a conect to a+n, ans the first cycle will happen at a+2*n;

so there is no way to output -1,
Am i right?
I can't even past the 3th testcase.