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2^(n-1) could be larger than ans. Its very large for 10^6. The worst case.

So possible solution is to take 2^something with loop, taking modulo 10^9+7 in every iteration.

What if someone says that this round is harder than the previous one? Would everyone quit giving it? That depends on us the participants how the round is. And this website is a source of giving us the idea in which topic you lag behind. So lets find and give every round a full shot as if its only created for you. Lets enjoy the process!

On surajanand218ranking in a contest, 6 years ago
0

+1 and +2 symbolises the number of times you attempted the given problem wrong but finally got it correct.

As the time increases the penalty increases and is same as the time passed so the person who solves fast will get more points.

More information at this link. This is how increase and decrease of rating calculated.

The A2OJ website is on renovation. It is in testing mode and hence the website is down.