When we speak about arithmetic progression (or arithmetic sequence) we mean a series of numbers with a special property - each value is followed by the other, greater by predefined amount (step).
I.e. difference of (K+1)-th and K-th values is a constant. Here are examples of sequences
1 2 3 4 5 6 7 ...
4 6 8 10 12 14 16...
10 13 16 19 22 25 28...
Since so, arithmetic sequence is completely defined by the first member (A) and the increment value - step size - (B). First few members could be expressed as
A + (A + B) + (A + 2B) + (A + 3B) + ...
Here we could of course write loops and solve the problem that way, but it gets a lot simpler if we factor out our solution a bit, and get some method to the madnes,
since we have A, B ad N as shown up we can reduce this to A*N + (1+2+...+b-1) * b the only not so obvious part here is the parenthesized part but it turns out it's rather easy to get in oforth, since we have #seq the part then becomes A N * B 1 - seq sum B * + in proper reverse notation, we just need to get this to deal with the stack, and it doesn't need much stack juggling so that means we get to not use locals
- Code: Select all
: getLn { System.Console askln }
: getNumber { getLn asInteger }
: pprint { apply(#.) printcr }
: getInput { [] getNumber #[ getLn + ] times }
: parseLine { words map(#asInteger) }
: 3list>st { dup first swap dup second swap third } // [a,b,n] -- a b n
: arithseq { tuck 1 - seq sum * tor * + } // a * n + (1 + 2 + ... [b-1] )b
: solveline { parseLine 3list>st arithseq }
getInput map(#solveline) pprint