OJ Board

Hi...
I can tell you that moving the cube on the board may be simulated. For the algo, I think a BFS is the solution

In this problem we are talking about simple graph which does not have same endpoints for more than one edge, and also does not have edges with the same endpoint.

what does it mean?
Is the graph directed?

Yeeeess...I've solved it. And lastly got AC. 4 ms and 420 k memory.

Some explanations:
- a cell can be repeated more than once
- it is preferably to use a BFS (Bradth First Search) in Labirint: that means you must search in all four directions simultaneously to get to final destination in small ...

I think that a cell can be repeated. So the input:

1

3 2
1 1
2 2


may give an answer like that:

6


otherwise "No solution".
But if I do a DF traversal then I'll get TLE.

Here is my DFS code:

#include <stdio.h>
#include <memory.h>


/* die faces */
#define UP 0
#define DOWN 1
#define LEFT ...

How can I solve this problem?
I've tried DFS first, then BFS, but WA. It seems a cell can be visited more than once. So it would go for TLE, in't it?

Oh... I've forgot
Delete the line
printf("\n")
before while(i<n) from your source code and you'll get a clean AC.

check the link:
http://acm.uva.es/problemset/minput.html

Thank you very much, tan_Yui.

Man man man

What happened? Are you new to acm.uva.es problemset?
Some key explanations:
- this problem requires multiple input
- revise your code, see that csum variable is not initialized after first calculation process
- the line
"printf("%.2f",a1); " must be transformed into
"printf("%.2f%s ...

Probable you must not put a blank line after last case e.g printf("\n\n").
You can write in this way:

while(i < n)
{
/* ...... */
if(i == n)
printf("\n");
else
printf("\n\n");
}

Hye... I am very disappointed. I submitted this problem 20 times and I've got WA for all. Is there some treaky inputs?
Please, people, help me with this easy problem.
Look a little at my code:

#include <stdio.h>
#include <ctype.h>

int m, n; /* num of line & columns */
char grid[60][60];
char viz ...

Lastly I've solved this problem. I've got TLE because of square() function which was evaluated every step in the for() loop. Also the above code would give WA even if we improve it.

Problem solution is easy as PI, because it's pure mathematic:
For a number n = p1^a1 * p2^a2 * . . . * pn^an, where ...

Hy!
I used the clasical algorithm to solve this problem, that means Euler phi function and I still get TLE. I don't know why..
please help me understand why...
Here goes my code:

#include <stdio.h>
#include <math.h>

int main(void)
{
int n, i, phi;
while(scanf("%d", &n) == 1)
{
if(n == 0 ...