int dfs(int x){
	if(ans[x] != -1) return ans[x];
	return ans[x] = min(1 + dfs((2 * x) % md), 1 + dfs((1 + x) % md));
}

You have test input

4
19 32764 10240 49

Why don't we try it?

1. x = 19, produces two inner calls (dfs(38) and dfs(20))
2. Assume, that first call would be first (it is not standardized btw)
3. dfs(38) produces another two inner calls (dfs(76) and dfs(39))
4. And so on... dfs(76) produces dfs(152) and dfs(77)
5. ...
6. And now you on x = 16384. This will produce dfs(0) and dfs(16385)
7. dfs(0) indeed returns 0, but dfs(16385)? Moving on
8. dfs(16385) produces dfs(2) and dfs(16386)
9. dfs(2) produces dfs(4) and dfs(3)
10. ....
11. And you at some point will come to x = 16384
12. Repeat infinitely steps 6-11.

See problem? If not, then your inner dfs calls never terminates.