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Results tagged with Search options user 31347

This challenge is intended to be solved by using, manipulating, accepting as input, or outputting numeric values.

3
votes
R, 113 101 bytes Not quite as elegant as @plannapus's answer. Go the All Blacks :) n=scan();for(C in 0:n)for(T in 0:n)for(P in 0:n)if(C*7+T*5+P*3==n)cat(P,'K,',C+T,'T,',C,'C\n',sep='') Test run …
answered Sep 19 '15 by MickyT
1
vote
converted to an odd number sequence. y is incremented on each recursion until it exceeds x. The first sequence which sums to the target will be returned. …
answered Jul 6 '17 by MickyT
5
votes
and outputs a vector of integers. Essentially this uses integer division to round the number down, adds 1 and multiples it by five. Anything divisible by 10 has 1 taken away. If n = 1 then it …
answered Sep 21 '15 by MickyT
6
votes
Cubix, 18 32 bytes I think I'll have to spend sometime on this later and see if I can compress it a bit. But for the moment here it is. Turns out I was thinking about this totally the wrong way. Now …
answered Feb 21 '17 by MickyT
0
votes
R, 33 bytes Implemented as an unnamed function function(x)rle(x%%10^(0:99))$v[2] This applies a mod of 10^0 through 10^99. rle is used to reduce the results down so that the second item is always …
answered Feb 23 '17 by MickyT
5
votes
Cubix, 19 bytes ;ww.1I!@s%Ow;)Sow.$ Try it online! ; w w . 1 I ! @ s % O w ; ) S o w . $ . . . . . Watch It Run A fairly straight forward implementation. 1 push 1 to the stack …
answered Jun 14 '17 by MickyT
1
vote
R, 77 72 bytes F=gmp::fibnum;i=0;d=n=scan();while(n)if(grepl(F(d),F(i<-i+1)))n=n-1;F(i) This makes use of the gmp library for the Fibonacci number. Fairly straight foward implementation of the …
answered Jun 15 '17 by MickyT
2
votes
R, 85 80 76 Uses the Hare Quota method. Removed a couple after seeing the spec that W will sum to 1 function(a,b){s=floor(d<-b*a);s[o]=s[o<-rev(order(d%%1))[0:(a-sum(s))]]+1;s} Test run > (funct …
answered Feb 16 '15 by MickyT
1
vote
Cubix, 22 bytes This will output the sequence indefinitely. The general idea is that it has a base number which 6 - 1 is added to. For each add the result is output multiplied by 10, which is …
answered Feb 18 by MickyT
2
votes
R, 86 bytes I thought there was already an answer (or two) in R for this question, but I must have been mistaken or they had the same issues that I had with R not doing signed ints. That issue too …
answered Nov 28 '17 by MickyT
2
votes
R, 28 bytes -((n=scan())%%3-1)*(n%/%3+1) Looks like this is a variation of most of the answers here. Zero based. n=scan() # get input from STDIN ( )%%3-1 # …
answered Jun 2 '16 by MickyT
1
vote
Cubix, 30 bytes 1..-.w>?^I3%?;,)O@...o-'.<u;;; Try it here. You will need to replace the current code with the above and enter an input number. This wraps onto a cube with an edge length of 3 … % ? ; , ) O @ . . . o - ' . < u ; ; ; . . . . . . . . . . . . . . . . . . . . . . . . Explanation: I 3 % ? Take a number from input, push a literal 3, mod on TOS and do a check. The …
answered Jun 2 '16 by MickyT
5
votes
Cubix, 23 24 25 bytes I1Wq/)s.;0..s;p-?\.+O@u 0 for truthy and nothing 0 for falsey. Brutes forces by incrementing counter, adding to cumulative sum and comparing to input. Now to try and fit it …
answered May 22 '17 by MickyT
7
votes
R, 86 83 Was working through this and then realised that I had essentially come up with the same solution as Optimizer and others I suspect. Anyway here it is as a function that takes a vector f=fu …
answered Jan 15 '15 by MickyT
4
votes
COUNT, thanks @nutki SELECT COUNT(R)FROM(SELECT ROW_NUMBER()OVER(ORDER BY V DESC)R,V FROM I)A WHERE R<=V SQLFiddle example Essentially it numbers the rows on a descending sort of the values. Then it returns the maximum row number where the row number is greater than equal to the value. …
answered Dec 22 '14 by MickyT

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