why are strings 1, 1001,1001001 ,1001001001, 1001001001001, k*001*
where k is the previous string all non primes. i saw an article from
dijkstra but couldn't understand other than when sum of string is 3.
can someone explain to me why
link:
yes, we can use a min heap of size k to keep track of the top k matching
words.
On 3/28/07, Dhruva Sagar [EMAIL PROTECTED] wrote:
OK! I get it now.
In that case what we can do is instead of sorting the result, what we can
do is simply while performing a linear search for the words (that are
well...its simple as given in the article..
note that the numbers are all decimal numbers (not binary..in case
misled by just 0 and 1 in the number) therefore
if you have have k copies of 001 and k%3 == 0, then the sum of the
digits is divisible by 3 hence the number is divisible by 3 - The
In case you are wondering about the part of the proof for k%30, it
goes like this, the given number can be written as 1 + 1000 + 100
+ 10 +..and so on
now if for a given k, if we choose the string of k 1s and try to find
the modulus with each of the terms in the above expression,
for
Oooh...I almost forgot to add this..
notice the relation between the proof for part (ii) and the discrete
logarithm problem. The proof is no mere coincidence. This is the reason
behind using primes in encryption schemes that rely on the hardness of the
discrete logarithm problem. This will ensure
Hi, Kevin,
you can try soundex. compute each word's soundex, and store these
soundex, Listword in a HashMap. for a giving word, compute it's
soundex, and just find the soundex in the hashmap. It would cost O(1).
good luck.
On Mar 29, 9:49 am, Kevin [EMAIL PROTECTED] wrote:
This is from an
hi friends i m trying to solve this problem
http://acm.uva.es/p/v111/11151.html but getting wrong answer..
my algorithm is i m trying to enumarate all the n*(n+1)/2 substrings of
string ,where n is string length...
and checking for palindrom ..
suppose i have input ADAM .then all
