OJ Board

Thank you very much

I don't know if it's proper to post the troubles here, but i lost all my C codes for the ACM problems. (because my hard disk broke) :( Is it possible to contact the online-judge to recover my codes with their back-up? I hope they can mail my codes back to me. Please tell me where I should post or wh...

[quote="Per"]Ah OK, then I see why you need it. I just used Ford-Fulkerson, but augment paths using Bellman-Ford instead of DFS/BFS (not sure what Successive Shortest Path Algorithm is, but it sounds like something similar). It wasn't fast, but it worked. :)[/quote] I also use Ford-Fulkerson impleme...

Since the problen says that "Each game starts with an integer m, 6 <= m <= 18, indicating the moves have been made" We can found a solution with worse case in 2^(18-m) steps. (each edge is chosen by A or B) ps: I got AC in 1.468 sec. You can also try a DP solution :lol: My friend use a DP solution t...

TO Julien Cornebise:

Try to use link-list to store edges instead of adj-matrix.
Because this prob have 10000 edges (500*20) fewer than 500*500.
I use this method to get AC. (but very slow)

Or you can optimize Dijkstra with heap (ElogV)
can get AC within 2 secs

Best regards

TO LIMON : I didn't say 9 is ana.prime. I say : "If n is a ana.prime ,
the digits of n can only contain 1, 3, 7, 9. "

TO Observer: Thanks for your data

my approach is : 1: Get all input and sort them. (using qsort + strcmp) 2: store all data (since ASCII range is 0-255) in array[MAXSIZE][300] 3: [c]for(i = 0; i < input; i++) for(j = i+1; j < input; j++) if(check(data , data[j]) == YES) output(input , input[j]);[/c] best regards!

We can know that from n > 10, if n is a ana.prime , the digits of n can only contain 1, 3, 7, 9. (since even number & 5 obviously cant be ana.prime) But, does anyone know why numbers bigger than 1000 with digits of 1, 3, 7, 9 can't be ana. prime? (or just by this program result?) I try to find it , ...

Try to solve this by DP.
Because test data must contain complex numbers.
(I originally used solve a+b, a*b individually)

After using DP, it still contains a prob....a large exp.
So u must find the iteration to simplify prob.
(eg: a+b = 0, a*b = 1. a= i^3, b=-i^3)




good luck

10153EN wrote:I think for this problem, the sample output of the problem is just to give us the format of the output only

10153EN is right.
The sample output is just a "FORMAT",not part of the anwser.
I just got AC by using DFS.

Give it a try

My programs response: :o I don't understand the problem. Can anyone help me? :o Input: :-? 10 2 3 5 4 3 2 9 6 10 10 6 2 7 3 4 7 4 5 3 2 4 47 6 11 2 7 3 4 8 5 10 7 8 3 7 4 2 3 9 10 2 5 3 31 9 5 8 2 1 7 7 11 3 6 5 4 9 10 5 2 1 5 3 9 7 16 3 11 2 7 3 4 8 5 10 7 8 3 7 4 2 3 9 10 2 5 3 25 1 7 7 7 7 8 8 8...

Try this method: only same states are allowed, also create biggest figure of good constant, it is: x[.(x)+][{E|e}(x)+] where (x)+ means one or more digit [x] means optional part of expression {x|y} means one and exactly one of specified in brackets values I've got AC by using Dominik's method :lol:...

I've got AC by using [c] double n, factor; factor = floor( sqrt(n) ); if(n - factor*factor < 0.00000001) printf("yes\n"); else printf("no\n");[/c] also got AC by using [c] long long n, factor; factor = (long long) sqrt(n); if(n == factor*factor) printf("yes\n"); esle printf("no\n");[/c] But I think ...

gvcormac wrote:http://plg.uwaterloo.ca/~acm00/S98-1/test_data/ (problem B)

Thx. According this test data~ I found my mistake and got AC

Only need to worry when s+p != y , Puff should +1 age before Spot.
(Using trial & error , I finally found that )

following code might help: [c] typedef struct { int index; char num[20]; } array; array a[100000]; [/c] good luck :) -sohel I am afraid this structure array's size still not big enough... I just got AC by changing my arraysize output[10000][20] --> output[10000][55] So the judge input must have som...