most simple with bash..using formula 11*f(n+6)+11*f(n+6)%10
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algoge
simple with AWK
-
PRAMENDRA RATHI
NIT ALLAHABAD
2011/6/16 • » νιρυℓ « •
> Binet's Formula.
> ( keep shortening :P )
>
>
> On Thu, Jun 16, 2011 at 5:57 PM, kartik sachan wrote:
>
>> @vipul what algo u have applied for o(1)??
>>
>> --
>> You received this
Binet's Formula.
( keep shortening :P )
On Thu, Jun 16, 2011 at 5:57 PM, kartik sachan wrote:
> @vipul what algo u have applied for o(1)??
>
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@vipul what algo u have applied for o(1)??
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hint : think it as matrix raised to some power ( then u can compute that in
log(n) ) or other way around is to find cycle length . :-)
On Thu, Jun 16, 2011 at 3:56 PM, saurabh singh wrote:
> Very sorry i did used memoization..Ya not possible with o(2^n)
> solution...Apologies once again
>
>
Very sorry i did used memoization..Ya not possible with o(2^n)
solution...Apologies once again
On Thu, Jun 16, 2011 at 3:21 PM, saurabh singh wrote:
> Well dont know...My pure recursive solution in python got AC in 0.07s(93
> bytes).No DP involved.And I don't think python is faster than c?
>
Well dont know...My pure recursive solution in python got AC in 0.07s(93
bytes).No DP involved.And I don't think python is faster than c?
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My AC solution is O(1).
On Thu, Jun 16, 2011 at 2:29 PM, PRAMENDRA RATHi rathi
wrote:
> without DP it will be TLE..
> -
> PRAMENDRA RATHI
> NIT ALLAHABAD
>
>
>
>
> On Thu, Jun 16, 2011 at 9:04 AM, saurabh singh wrote:
>
>>
>> t,m;f(n){return n<=2?1:f(n-1)+f
without DP it will be TLE..
-
PRAMENDRA RATHI
NIT ALLAHABAD
On Thu, Jun 16, 2011 at 9:04 AM, saurabh singh wrote:
>
> t,m;f(n){return n<=2?1:f(n-1)+f(n-2);}
>
>
> main(){scanf("%d",&t);while(t--)scanf("%d",&m)&&printf("%d\n",f(m+11)-f(m+1)+f(6+m)%10));}
got it repetition in cycles of 60
On Thu, Jun 16, 2011 at 12:35 PM, vaibhav agarwal <
vibhu.bitspil...@gmail.com> wrote:
> shuldn't it be f(246+m)%10 last one
>
>
> On Thu, Jun 16, 2011 at 9:04 AM, saurabh singh wrote:
>
>>
>> t,m;f(n){return n<=2?1:f(n-1)+f(n-2);}
>>
>>
>> main(){scanf("%d",&t);
ACTUALLY FIBBONACCI SERIERS REAPEAT AFTER 60 TERM SO 246%60 WILLBE 6 SO WE
ONLY HAVE TO FIND M+6 TERM
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shuldn't it be f(246+m)%10 last one
On Thu, Jun 16, 2011 at 9:04 AM, saurabh singh wrote:
>
> t,m;f(n){return n<=2?1:f(n-1)+f(n-2);}
>
>
> main(){scanf("%d",&t);while(t--)scanf("%d",&m)&&printf("%d\n",f(m+11)-f(m+1)+f(6+m)%10));}
> My best attempt with the c code..132 bytes still
> Now g
t,m;f(n){return n<=2?1:f(n-1)+f(n-2);}
main(){scanf("%d",&t);while(t--)scanf("%d",&m)&&printf("%d\n",f(m+11)-f(m+1)+f(6+m)%10));}
My best attempt with the c code..132 bytes still
Now gonna try perlIt definitely requires exceptional skills to bring it
down to 111 bytes but have this gut
problem:::
https://www.spoj.pl/problems/FIBSUM/
can anyone suggest idea to reduce my code to 111 byte..its currebt size is
about 174
int a[50]={0},t,m;
f(int n) { return a[n]=a[n]?a[n]:n<=2?1:f(n-1)+f(n-2); }
main() {
scanf("%d",&t);
while(t--)
{ scanf("%d",&m) ;
printf("%d\n",(f(m+11)-f(m+1)+f
You need to keep generating Fibonacci numbers until you meet the
condition.Check for even valued term by using TERM%2==0 and sum
up.Fibonacci series grows exponentially so n wont be very high.Take care
that it doesn't overflow integer range.
On Mon, Dec 20, 2010 at 8:36 PM, Shalini Sah <
shalini
Each new term in the Fibonacci sequence is generated by adding the previous
two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
Find the sum of all the even-valued terms in the sequence which do not
exceed four million. I'm just a beginner..plz
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