Search found 33 matches

by popel
Tue Dec 31, 2002 6:56 am
Forum: Other words
Topic: Problems Solved
Replies: 2
Views: 1051

...go to judge status page..
They have said that they are rebuilding their database..
by popel
Mon Dec 30, 2002 6:07 pm
Forum: Other words
Topic: Happy New Year to everyone!!!
Replies: 6
Views: 1445

Hi Everyone,
Happy new year.

....but dawn shows the day.So the judge system should be recovered
before 1st January-unless it will say that the judge system will be
damaged for 2003 year long. :wink: :(
by popel
Sun Dec 22, 2002 7:23 pm
Forum: Volume 100 (10000-10099)
Topic: 10018 - Reverse and Add
Replies: 169
Views: 21980

Once again, use long long... 4000000000 should yield 1 4000000004 *. At first Thank you. Because your answer pointed my my mistake *. Is 4000000004 a long long ? Unsigned long is enough here. So one should not use long long.....(its slower) *. So silly I am ! I couldn't find the normal case :oops: ...
by popel
Sun Dec 22, 2002 12:24 pm
Forum: Volume 100 (10000-10099)
Topic: 10018 - Reverse and Add
Replies: 169
Views: 21980

Try this testcase: Input: 2 2 99 Output: 1 4 6 79497 .....Checking problems once accepted is a boring task. ...But what about me? My code yields the same output you presented. Here is my one: [c] #include<stdio.h> unsigned reverse(unsigned n); unsigned reverse(unsigned n){ unsigned len,temp,j,k,rev...
by popel
Mon Nov 18, 2002 11:09 am
Forum: Volume 3 (300-399)
Topic: 371 - Ackermann Functions
Replies: 196
Views: 30311

This problem requires temporary values stored. The optimization by Mahbub is very well. But to solve it in 0.2 ~ 0.3 temporary values should be saved in an array. The interval of judge inputs are not more than 400000. So, a good way to solve this problem is having an array of 400000 elements. If one...
by popel
Thu Aug 01, 2002 3:51 pm
Forum: Volume 5 (500-599)
Topic: 583 - Prime Factors
Replies: 171
Views: 38470

:-? :-? :-? :-? :-? :-? :-? :-? :-? :-? :-? :-? :-? :-? :-? My solution should give you compile error. I don't know why but there may be some error when I was editing it in the poll editor. See,I have corrected it. (There was a declaration double a;it should be int a) Send it, And see,It will get ac...
by popel
Wed Jul 31, 2002 4:37 pm
Forum: Volume 5 (500-599)
Topic: 583 - Prime Factors
Replies: 171
Views: 38470

Better you can compare with my code. Here's my one: [cpp] #include<stdio.h> #include<math.h> int prime[5000]; int cp(); int cp(){ int a,t; int b,c=3,p; prime[0]=2;prime[1]=3;prime[2]=5; for(a=7;a<=47000;a+=2){ p=0; t=(int)sqrt((double)a); for(b=2;(b<=t)&&(p!=1);b++) {if(a%b==0)p=1;} if(p!=1){prime[c...
by popel
Sat Jul 27, 2002 12:32 pm
Forum: Volume 5 (500-599)
Topic: 583 - Prime Factors
Replies: 171
Views: 38470

For Simplicity, Let, if P i is a Prime and x is a Natural Number: x = P 1. P 2. P 3.... ... ... P n Suppose, P n > sqrt(x) Then , P 1. P 2. P 3.... ... ... P n-1 <sqrt(x) So a Number x CANNOT have more than ONE prime factor greater than sqrt(x) Can you smell something new ? 8)
by popel
Fri Jul 26, 2002 1:37 pm
Forum: Volume 5 (500-599)
Topic: 583 - Prime Factors
Replies: 171
Views: 38470

You should know that what to do if you want to find the prime factors of a number efficiently.Trial for primes less that sqrt(x) and think a while.
by popel
Thu Jul 25, 2002 10:14 am
Forum: Volume 103 (10300-10399)
Topic: 10311 - Goldbach and Euler
Replies: 98
Views: 23796

I think there are many processes to do it not exceeding TL. One process is by first enlisting primes<1e4 by sieve process. Because you need only to check with the primes less than sqrt(x) to find whether x is a prime. My program needed 424 KB memory to solve the problem in 19 secs.But surely there a...
by popel
Thu Jul 25, 2002 10:11 am
Forum: Volume 1 (100-199)
Topic: 107 - The Cat in the Hat
Replies: 278
Views: 22720

By Solving the Equation in Bi-Section Method.
(like Binary Search)
by popel
Sun Jul 21, 2002 12:29 pm
Forum: Volume 103 (10300-10399)
Topic: 10302 - Summation of Polynomials
Replies: 29
Views: 15422

Formula

1^3 + 2^3 + 3^3 + ... ... ... +n^3

Sn=(n*(n+1)/2)^2

:wink:
by popel
Mon Jun 10, 2002 3:21 pm
Forum: Volume 4 (400-499)
Topic: 478 - Points in Figures: Rectangles, Circles, Triangles
Replies: 44
Views: 11024

Re: Hi!

cyfra wrote: I got your program Accepted without any changes....
Thank you to show me the right path. As I became sure
that the program is correct.
...And as the others my problem was also line breaking in mailer..

8)
by popel
Fri May 03, 2002 9:21 am
Forum: Volume 4 (400-499)
Topic: 478 - Points in Figures: Rectangles, Circles, Triangles
Replies: 44
Views: 11024

478:Points in Figures:Rectangles,Circles and Triangles

Hello Everyone,
Is there any tricks in this problem ?
I think my algorithm ok,so what causes it WA?
I have checks for point circles,triangles and rectangles.
Can anyone give me some critical test inputs ?
and check the code below please.....
[c]
:roll:
...code has been removed...
:o
[/c]
by popel
Sat Apr 27, 2002 10:25 am
Forum: Volume 1 (100-199)
Topic: 107 - The Cat in the Hat
Replies: 278
Views: 22720

Your program is not efficient. To realize this give inputs 483736625 481890304 The output should be 615441 1931252289. In this problem,your main task is to solve the equation : log(x+1)/log(x)=log(i)/log(w) in an efficient process. as i and w is given the right part is a constant (=k) x must be>1 an...

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