880  Cantor Fractions
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880  Cantor Fractions
I solve many such problems(for example 264),but this one I'm getting TLE(on Pascal).Somebody tell me what to do or may be I must write it on C++.
Thanks.
Thanks.
someone who like to solve informatic problems.
http://acm.uva.es/cgibin/OnlineJudge?AuthorInfo:29650
http://acm.uva.es/cgibin/OnlineJudge?AuthorInfo:29650

 Experienced poster
 Posts: 151
 Joined: Tue Nov 16, 2004 7:23 pm
 Location: Norway
 Contact:
Hi all!!!
I used the following algorithm to solve this problem:
1. Find the maximux number n for which n(n+1)/2 is closest to the input value(x).
2. s = (n * (n + 1)) / 2;
if (x == s)
cout << 1;
else
cout << n + 2  x + s;
cout << '/';
if (x  s)
cout << x  s << endl;
else
cout << n << endl;
I have tried a lot of test cases and all are fine..., where is my mistake???
Thanx in advance,
Yandry.
Btw i used unsigned long long to store the number...
I used the following algorithm to solve this problem:
1. Find the maximux number n for which n(n+1)/2 is closest to the input value(x).
2. s = (n * (n + 1)) / 2;
if (x == s)
cout << 1;
else
cout << n + 2  x + s;
cout << '/';
if (x  s)
cout << x  s << endl;
else
cout << n << endl;
I have tried a lot of test cases and all are fine..., where is my mistake???
Thanx in advance,
Yandry.
Btw i used unsigned long long to store the number...

 Experienced poster
 Posts: 183
 Joined: Thu Nov 11, 2004 12:35 pm
 Location: AIUB, Bangladesh
Hi, I got TLE, can anyone say what's wrong ?
Code: Select all
#include<stdio.h>
void main()
{
unsigned long long n,x,p;
while(scanf("%llu",&n)==1)
{
x=1;
while((p=x*(x+1)/2)<n)
{
if(p>=n)
break;
else
x++;
}
printf("%llu/%llu\n",1+(pn),x(pn));
}
}
Jalal : AIUB SPARKS
Well, I haven't attempted the problem, but if the input value can really be up to 2^64, then it is quite normal you get TLE since you would run the while loop more than 2^32 times. Use a quicker algo to get the square root of 2n, either use sqrt and double, or use the babylonian algorithmCodeMaker wrote:Hi, I got TLE, can anyone say what's wrong ?Code: Select all
#include<stdio.h> void main() { unsigned long long n,x,p; while(scanf("%llu",&n)==1) { x=1; while((p=x*(x+1)/2)<n) { if(p>=n) break; else x++; } printf("%llu/%llu\n",1+(pn),x(pn)); } }
http://www.math.jhu.edu/courses/107/arc ... ode16.html.
Hmm... is it so hard to figure out an O(1) solution? I suggest anyone who get TLE to spend some time on that

A quotation:

A quotation:
gvcormac wrote:I suggest that it is your responsibility to explain why your program should work than to invite others to find errors.
7th Contest of Newbies
Date: December 31st, 2011 (Saturday)
Time: 12:00  16:00 (UTC)
URL: http://uva.onlinejudge.org
Date: December 31st, 2011 (Saturday)
Time: 12:00  16:00 (UTC)
URL: http://uva.onlinejudge.org
CodeMaker,
here is your code.
Note that you set X to 1 before starting your while cycle.
x=1;
That is the problem. If we call the value 1 SEED for X then
your problem is that your SEED is too low, you can use a better
value for it , thus ensuting you have in your while just a couple
of iterations.
A better SEED is this one:
x = Math.sqrt(2*N)1
converted to the appropriate INT type you use
( I think you said it is long long or something like that ).
This way you just avoid a huge amount of looping in your
WHILE.
Make the seed x = Math.sqrt(2*N)3 , if you want, you won't
have TLE in both cases.
here is your code.
Code: Select all
#include<stdio.h>
void main()
{
unsigned long long n,x,p;
while(scanf("%llu",&n)==1)
{
x=1;
while((p=x*(x+1)/2)<n)
{
if(p>=n)
break;
else
x++;
}
printf("%llu/%llu\n",1+(pn),x(pn));
}
}
x=1;
That is the problem. If we call the value 1 SEED for X then
your problem is that your SEED is too low, you can use a better
value for it , thus ensuting you have in your while just a couple
of iterations.
A better SEED is this one:
x = Math.sqrt(2*N)1
converted to the appropriate INT type you use
( I think you said it is long long or something like that ).
This way you just avoid a huge amount of looping in your
WHILE.
Make the seed x = Math.sqrt(2*N)3 , if you want, you won't
have TLE in both cases.
 dovier_antonio
 New poster
 Posts: 47
 Joined: Fri Feb 18, 2005 5:00 am
 Location: Havana, Cuba
Hi Sedefcho !!!
source code deleted...
Last edited by dovier_antonio on Fri Feb 03, 2012 9:23 am, edited 1 time in total.
 dovier_antonio
 New poster
 Posts: 47
 Joined: Fri Feb 18, 2005 5:00 am
 Location: Havana, Cuba
Hi Sedefcho !!!
source code deleted...
Last edited by dovier_antonio on Fri Feb 03, 2012 9:24 am, edited 1 time in total.

 Learning poster
 Posts: 70
 Joined: Sat Feb 05, 2005 9:38 am
 Location: Gurukul
Let Reply
Hi Sumanker,
Very late replay !!! . I dont know wheither u read or not. Well, long long is enough for this problem. Thankx.
Very late replay !!! . I dont know wheither u read or not. Well, long long is enough for this problem. Thankx.
Some Love Stories Live Forever ....
880 Cantor Fractions...
Isnt this the same as 264.... except for the printing thing and the order of printing. I am getting TLE for the program.
The time required for :
10000000
999999
999998
287234
100000
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
44
45
is
real 0m0.010s
user 0m0.000s
sys 0m0.010s
The time required for :
10000000
999999
999998
287234
100000
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
44
45
is
real 0m0.010s
user 0m0.000s
sys 0m0.010s
The input limit is higher here...Think about 64 bit int....
And better to find a formula.....
And better to find a formula.....
Ami ekhono shopno dekhi...
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 Experienced poster
 Posts: 120
 Joined: Sat Nov 01, 2003 6:16 am
 Location: Dhaka (EWU)
Hi
I think the input limit is not more than (2^31  1).
i have derive a short cut form of this problem.
this is my function.
I think this will be help for you.
Regards
MAP
I think the input limit is not more than (2^31  1).
i have derive a short cut form of this problem.
this is my function.
Code: Select all
function(int N) {
int res;
res = ceil(((1 + sqrt(1 + 4.0 * 2.0 * N))/(double)2.0));
int ans = ((((res * res) + res) /(double) 2.0));
printf("%d/%d\n", (ans  N) + 1,res);
}
Regards
MAP