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Posted: Thu Dec 16, 2004 3:59 pm
by sumankar
Sum[floor(n/p^i)] i = 1...k, such that p^(k+1) > n
suman
Posted: Sat Dec 18, 2004 11:37 am
by sumankar
I have umpteen WA's here and I am really really
pissed off with this
problem.My prog o/p matches all i/o given here.It seems I am missing some
stupid little trick.
Any ideas, fellows?
Suman.
Posted: Mon Dec 20, 2004 8:18 am
by sumankar
HI all
Any ideas!!
Posted: Sat Dec 25, 2004 6:46 pm
by neno_uci
Hi Suman:
Here is my algo, i hope you find it useful...
1. Find all the primes from 1..5000.
2. For each prime x that divides m find such a1 that m % x^a1 == 0 and a2 that n! % x^a2 == 0, if a1 > a2 then n! is not divided by m else store the minimal a2 / a1 and that is the result.
I Hope you solve that!!!
Posted: Fri Jan 21, 2005 5:16 am
by minskcity
Posted: Fri Jan 21, 2005 3:11 pm
by misof
(correction: apparently you meant the
greatest power of 13)
No need to be angry, just read the problem statement more carefully. The problem description is correct, as is your (corrected) second sentence. Let's see some quotes:
problem description wrote:Given a number n you have to determine the largest power of m, not necessarily prime, that divides n!.
This is the sentence you are referring to. Basically, this sentence tells you what to do.
output description wrote:The result is either an integer if m divides n! or a line "Impossible to divide" (without the quotes).
This is the
output description. After you have solved the problem, this section tells you how to
format your output. In my humble opinion, the case "m doesn't divide n!" is exactly the case when the greatest power of m dividing n! is 0. So, this section tells you: instead of zero, output a message. What's wrong with that?
Posted: Fri Jan 21, 2005 5:06 pm
by minskcity
misof wrote:
output description wrote:The result is either an integer if m divides n! or a line "Impossible to divide" (without the quotes).
This is the
output description. After you have solved the problem, this section tells you how to
format your output. In my humble opinion, the case "m doesn't divide n!" is exactly the case when the greatest power of m dividing n! is 0. So, this section tells you: instead of zero, output a message. What's wrong with that?
For many problems from Valladolid the second option (such as "no solution") never needs to be printed. How do I know that it's not the case here? It is quite clear to me that division is always possible according to problem statement...
Posted: Wed Feb 16, 2005 11:23 pm
by Sedefcho
Of course there're cases when you print
"Impossible to divide"
For example what is the highest degree of M=23 which
divides N ! = 5 !
Of course it is 0 so you print "Impossible to divide"
Posted: Thu Feb 17, 2005 11:05 pm
by minskcity
Why exactly am I supposed to print "Impossible to divide" when I can divide by 23^0? Can you give me the sentence in problem statement that says so? My understanding of "impossible to divide" is impossible to divide and not "impossible to divide for the exception of M^0". I see nothing that makes zero special in the problem statement. And if I have to print "impossible to divide" when answer is 0, why should not I print "impossible to divide" when answer is 1, 2, 3....????
Posted: Fri Feb 18, 2005 12:44 am
by Adrian Kuegel
I agree that the problem statement could be clearer. But here is what is meant:
The powers of m are integers values m^p with p a integer >0 (in this problem). So, if there is no such power of m which divides n, you have to print "Impossible to divide".
Posted: Wed May 25, 2005 1:07 pm
by Dominik Michniewski
I have a problem with this problem
My algorithm is:
1. generate primes in range [2...5000]
2. for each M,N pair
- factorize M using prime table from step 1
- factorize N! using prime table from step 1
- output (power of max prime in M in N!)/(power of max prime in M) if it's greater than 0 or appropriate message
Is this wrong ?
I check all I/O posted, but in all cases I got correct answer. Could anyone help me ?
Best regards
DM
Posted: Wed May 25, 2005 1:26 pm
by dumb dan
Dominik Michniewski wrote:My algorithm is:
1. generate primes in range [2...5000]
2. for each M,N pair
- factorize M using prime table from step 1
- factorize N! using prime table from step 1
- output (power of max prime in M in N!)/(power of max prime in M) if it's greater than 0 or appropriate message
Is this wrong ?
Only consideringing the max prime factor in M is not enough. Consider cases such as:
Posted: Wed May 25, 2005 1:54 pm
by Dominik Michniewski
Thanks.
Case 96 96 was exactly what I missed
Best regards
DM
10780 WA plz help!!!!!
Posted: Mon Sep 12, 2005 7:04 am
by georgemouse
I tried all the input posted in former topics and all he output is right.
BUT I still got WA!!!!!Why?
Can someone tell me what's wrong in my code?
Thanks!!!!!!
Posted: Mon Sep 12, 2005 7:16 am
by georgemouse
The way I used is to find the largest prime factor in M (the largest prime factor is K)
then find the largest power of M that divides N:
if K^a divides N!,
then I'll find the largest 'b' that k^b divides M.
At last, I'll output a/b (if it's 0 , output "Impossible to divide").
example:
if N=11,M=8=2^3
8's largest prime factor is 2
11/2->5
5/2->2
2/2->1
a=5+2+1=8
b=3
8/3=2
output 2