OJ Board

By the way what's the output for this: This input was inspired by the fact sqrt(2) = 1.41, so the answer should be .
However, most of the AC code returns

@yiuyuho: you are right, the fact that L*(B^n-factor) is divisible by (B*factor-1) does not gurantee, that x -- the corresponding quotient -- is n digits long.
But my solution gets AC. The problem is erroneous, if somebody has a really correct solution, please post here your comments.

Hey, I didn't check your code, but the thing is: there can be edges (i,j) such that either i < 0 or j < 0.
The answer is, of course, "NO".

Input: Output: 0.00000

The only tricky case I can think of is a rhombus: output is the same. And I got it AC with upper envelope + lower envelope also.

My algo is: value is 0, if the i-th cell is special, otherwise it is given in the input; sm = sm[i-1] + value , thus the weight of the segment [i,j] is calculated as follows: (1ULL << sm[j]-sm[i-1]); pref = pref[i-1] + (is_special ? 1 : 0); Now suppose (j,k) be such a state that we have optimally di...

I solved it with multinomial coefficients -- i.e. without any DP except for a set to store some duplicating bitmasks. Got AC in 1.560. I now wonder how those fellows solve it in 0.000? We just enumerate all the bitmasks, and if the corresponding sum of digits is divisible by 5, we form all permutati...

It was I who was lying when I said that (a,b) uniquely defines the configuration. Obviously it does not.

I haven't solved the problem yet. Still, I have a few remarks: i-th car is definitely lying if a[i]+1+b[i] > n; every pair (a[i],b[i]) _uniquely_ defines the configuration of the cars Now, if a particular car -- say, j-th -- is definitely lying in the above sense, and i-th car's data is plausible, t...

1. Do Avenues run from South to North and Streets from West to East?
To put it simply, Avenues are vertical lines and Streets are horizontals, right?
2. "Riding" is a large rectangle we have obtained by S horizontal and A vertical cuts?

I would rather say that

Reversing the sequence and finding the length of its longest common subsequence with the original sequence. Th? answer is n-LCS. Now, LCS problem is still O(n^2). There is also a O(rlog(n)) algo, where r is the number of pairs (i,j) such that ai = bj...

All I can think of is a lazy DP: the minimum cost of making a palindrome of s[i:j] is stored in a map if the cost is <= K...
Of course this is still O(n^2), but I thought that K's being small will make up for this deficiency... Any ideas?

Now I see how that color stunt is helpful in dfs. In the sample input above one edge is given twice, so it is easy to mistake it for a cycle during dfs, if one doesn't use colors.

I would like to strongly emphasize that UVa264 and UVa880 are not one and the same: for instance, for input you would have and respectively.