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 ...

[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 ...

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 ...

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 ...

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 ...

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 ...

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 ...