rating 1900+ for Div2-only competitions. I thought yyc registration would be miserable if I didn't play, it turns out that if you don't open the question, it doesn't seem to be participating. .
It was too late and I didn’t want to fight, but I’d better have a fight with Sinogi bosses, and practice before apio. .
A: The sb question doesn't say much, it's very pitiful. After handing in a WA, I was told that the special sentence should output YES. Bad review.
B: 4*n grid, two people go from (1,1) to (4,n) and one from (1,n) to (4,1), put obstacles in the middle to make the number of shortest paths for the two people the same . Boundaries cannot be placed. It is guaranteed that n is an odd number, and the number of obstacles cannot be placed without the boundary.
It is so obvious that it is obvious that the left and right symmetrical placement is over. If there is no NO, it will also output YES, which is a bad review. The title is too ghostly, bad review.
C: I didn't understand the question, bad review. Holding Sinogi's thigh, orz.
D: A sequence of 5000, |ai|<=1e8. For a string, there is an ans indicating the minimum number of groups that can be divided so that the product of each group is a perfect square. A single one can be considered, and the grouping can be divided at will without being connected, but it must be grouped. Ask k=1~n, how many substrings have ans=k.
Obviously a*b is legal, b*c is legal, then a*c must also be legal. Then directly divide into groups, and enumerate the interval to see how many different groups there are. The specific method is that each point records where las is the last occurrence of the same group. Pay attention to the case where ai=0 is specially judged, at least the average person is because of this fst. I also fst, very sad to say.
E: A tree, the weight of the point numbered i is 2^i. Given k, take a connected block of size nk such that the sum of weights is the largest.
Such weights are greedy at first glance, and you can give up as many points as you want in order to choose a point with a large label. Build a tree with n as the root, adding points from large to small. Each time the distance from a point to the selected connected block (multiplied processing will do) plus the existing one not greater than nk can be selected. Violence is marked and selected. No time to write. .
The question is very watery, but the answer is so bad, I can't be a mouth contestant anymore, not to mention most of the questions can't be AC. Get the title of the meeting, and you won't regret it.
emmm, rating-- is very spiritual.