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can u guys tell me a counter-example where my program fails? i am not seeing it! and the problem looks so easy, i am fealling ashamed to ask about this one... (



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

float divider,number,temp,answer,temp2;


main()
{
int i;
number =1;
scanf("%f %f",&number,

try "25 5" and "5 5".

I had some problems with this one, too. Unfortunately, the judge thinks that if n<2 or m<2, then it's boring. Of course that's total crap, but the judge is always right...

<font size=-1>[ This Message was edited by: Stefan Pochmann on 2002-03-24 05:10 ]</font>

can u see any problem in that?


my input:

25 5
5 5
2 2
1 2
2 1
1 1
0 1
1 0
0 2
2 0
125 5
30 3
80 2
81 3
20000000000 2
125 125

my answer:
25 5 1
5 1
2 1
Boring!
Boring!
Boring!
Boring!
Boring!
Boring!
Boring!
125 25 5 1
Boring!
Boring!
81 27 9 3 1
Boring!
125 1

Looks "correct" (meaning the judge should like it).

well, the judge didnt like it, and the code still the same...

My code works fine with all the test cases. But I am getting "Output limit exceeded" when I submit it. Can any body help me with the reason????

ram: That means you print to much. Either you have too much debug output (remove it) or you screw the input (fix it), so that you for example process the last testcase again and again and again. Or you have an infinite loop that prints something (fix it). Or...

Thanks stefan,

It just got accepted. I used
"while(scanf("%lf %lf",&num,&div))"
instead of
"while((scanf("%lf %lf",&num,&div))==2)"

thanks for the reply.


<font size=-1>[ This Message was edited by: ram on 2002-03-27 01:43 ]</font>

/*@BEGIN_OF_SOURCE_CODE*/
#include<stdio.h>
#include<math.h>


int main()

{
long long n,div,x,y;

while(scanf("%lld%lld",&n,&div)==2){

if(div < 2 || n< 2) printf("Boring!\n");

else {
x = (log10(n)/log10(div));
y = pow(div,x);
if(y != n) printf("Boring!");

else
for( ;n>=1 ;n/=div)
printf("%lld ",n);

printf("\n");
}
}



return 0;
}
/*@END_OF_SOURCE_CODE*/

**********
Can anybody help me to find out the error? All input stated here works fine

program v10190;

var
n, a: double;

function isPow: boolean;
var
k: double;
begin
k := ln(n)/ln(a);

if abs(k-trunc(k)) < 1e-8 then
isPow := true
else
isPow := false;
end;

procedure solve;
begin
write(n:0:0);
while true do
begin
n := trunc(n/a);
write(' ', n:0:0);
if abs(n-1) < 1e-8 then
break;
end;
writeln;
end;

begin
{ TODO -oUser -cConsole Main : Insert code here }
while not eof do
begin
readln(n, a);
if (n < 2) or (a < 2) then
begin
writeln('Boring!');
continue;
end;
if isPow then
solve
else
writeln('Boring!');
end;
end.

Whenever I submitt 10190, it says "output limit exceeded". I don't understand where is the problem. My code is like this:

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

void main()
{
long m,n,result;

for(;(scanf("%ld %ld",&m,&n))==2;)
{
if((ceil(logl(m)/logl(n))) == (floor(logl(m)/logl(n))))
{
printf("%ld ",m);
for(;;)
{
m=m/n;
printf("%ld ",m);
if(m==1)
break;
}
}
else
{
printf("Boring!");
}
printf("\n");
}
}

HELP ME![/c]

Try with input:
0 2

your program prints 0 0 0 0 0 0....................

Good luck.

actually I'm just using a simple calculation which is

if (n%i==0)
n = n/m;


using only calculation like that I got accepted in 0.00 sec.

hint : don't think too hard on this one, it's not that hard to solve this

the judge is nuts.
1 1 should be accepted.. but it's Boring!

if (a < 2 || b < 2 || (a < b)) /* Boring! */