In this 294C - Шаасс и лампочки I do not understand clearly what to do next. I have figured out that if there are n off lights between two on lights , number of ways to turn them on is 2^(n-1) . And for consecutive off lights having only one adjacent on light can only be turned on in one way. In the third sample test case lights from 5 to 7 can be turned on in 2^2 ways . And lights from 1 to 3 can be turned on in 1 way. Also, lights from 9 to 11 can be turned on in 1 way. So answer will be X * 1 * 4 * 1 . Now how to find this "X" ??