tag:blogger.com,1999:blog-9099779.post6818043418799613689..comments2020-07-22T23:36:44.643+02:00Comments on StalkR's Blog: smpCTF challenge #1 write-upStalkRhttp://www.blogger.com/profile/15113480981262771031noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-9099779.post-88651533253263751632010-07-15T22:09:26.286+02:002010-07-15T22:09:26.286+02:00Btw thanks for the walk though on # 5 it drove me ...Btw thanks for the walk though on # 5 it drove me nuts. also my comment above I wrote that in python, formatting was removed when I posted it.f8lerrornoreply@blogger.comtag:blogger.com,1999:blog-9099779.post-39644317343090731002010-07-15T22:07:31.577+02:002010-07-15T22:07:31.577+02:00This is all I did...
s = 0
p = 0
prv = 1
while s&...This is all I did...<br /><br />s = 0<br />p = 0<br />prv = 1<br />while s<51997:<br /> s = s + 1<br /> p = p + 1<br /> a = (s*p+prv+62)<br /> print a<br /> prv = af8lerrornoreply@blogger.comtag:blogger.com,1999:blog-9099779.post-89561128615828204432010-07-15T09:47:44.552+02:002010-07-15T09:47:44.552+02:00I didn't bother to do the maths but you're...I didn't bother to do the maths but you're right. Very cool solve! :)<br /><br />I'd love a challenge where the "easy" way to solve the problem takes too long and the answer expires, so that you are forced to find a smarter way, possibly doing some maths or just smart algorithmic.<br /><br />Thanks for your input!StalkRhttps://www.blogger.com/profile/15113480981262771031noreply@blogger.comtag:blogger.com,1999:blog-9099779.post-67109449130399168612010-07-14T16:17:24.580+02:002010-07-14T16:17:24.580+02:00When I saw this challenge, it reminded me of Serie...When I saw this challenge, it reminded me of Series (Math).<br /><br />S[n] = (n^2) + S[n-1] + R , S[0] = 1<br />can be solved to<br />S[n] = n(n+1)(2n+1)/6 + 1 + nR<br /><br />I just solved it ;). When playing, I also coded similar to you but using C.sleepyanoreply@blogger.com