OJ Board

Thanks for your smart hint. And I have to throw away my common sense and common interpretation on english words again.

After someone taught me, I've the following interpretation: G(n) represents the number of ways to form the n-bit gray code. For n-bit, you can fix 1st bit, 2nd bit, ..., n-th bit. After each bit fixed, you can have G(n-1) ways to generate the upper half, and G(n-1) ways to generate the lower half. I...

what does "dots" & "values" mean? i didn't find anywhere contains description of "dots" & "value". one piece of domino has two "values", say (0,1). there are "dots" to represent the "value", so on the domino there is one ...

I used DP to store a 3D table, storing in how many ways using M numbers, starting from I, and sum up to S. After constructing the table, can determine each individual number starting from left to right. I cannot think of a better way (likely due to limited ability/knowledge), though I believe there ...

Just this:

Input:
1
1 1
0 0 40

Output:
1

I wanna say something. first, the judge is wrong. actually i think this is an NP-complete problem, but Adil's algorithm works, which is polynomial running time. i first guess this is just an approximation algorithm, but, after the algorithm, i verified that it doesn't invite all the politicians! the...

I'm not sure about the speed either, as my program doesn't run fast. but, for IDA*, i suppose a stack is enough, not a heap. i used Manhattan distance heuristic too (quite standard?) i guess if you adjust/tune your scoring functions, say F() = number of steps + 80* estimated steps, it may terminate ...

yes, there's such a way. if you search the web.
if i remember correctly, i used the following:
let X = determine number of inversions + the row number of the empty square (numbered as row 1 to row 4)
if X is odd, no solution
if X is even, do the search

i forget exactly how i solve it. but think about prime number sequences. eg. for n=5 start from 1, 1+(i*j) , j=0,1,2,3,4 for i=0, then { 1,1,1,1,1 } for i=1, { 1,2,3,4,5 } for i=2, { 1,3,5,2,4 } for i=3, { 1,4,2,5,3 } for i=4, { 1,5,4,3,2 } you see that there's no rectangle. hope that helps. and wis...

i used recursion. though i guess there're better methods than mine. (i heard somebody use looping) my algorithm: go from the top-row, - for each one in top-row, you can switch or not switch for the rest of the rows, - the upper row would determine whether you can switch or not. ie. if the upper row ...

thanks to all of you. i think it's more than a year when i first tried to solve this problem.

i just wonder, how am i suppose to know the output format in such a way. how am i suppose to handle 0 and 180, if not your hints. or it's just a good proof that i'm not clever enough for ACM?

just then, I noticed that there's not a special correction program, does that imply side1 >= side2 in the output?

DM is called Fermat point, a search engine will tell you about it. one thing, there're two cases for Fermat point. [i'm surprised that Fermat is such a great mathematician] if a triangle cannot be formed, only circumcentre is -1. other values still can be obtained. i think you would know what should...

it can, please trust me.

OOXO
XXXX
XOOO

also can, do you think so?

our code are alike, but what my guess is,

maybe you try to return real number in s()

then change to integer when output
(int) (s(m2+1,n)-s(2,m1)+1)