Element, 17 characters plus 1 space
_'[_ 2:n;0>[n~+]]`
This is my first constructed language. It is designed to be very compact and human-readable. All of the instructions are one character long and perform a single function.
Element has two stacks and a hash as memory structures. The two stacks are called the main stack and the control stack. The main stack is where arithmetic, I/O, and hash manipulation occurs. The control stack is where logic operations occur, and this stack controls the while and for loops.
The basic idea behind Element is that there is a hash that stores numbers/strings, while the stack is used to perform calculations on these numbers. The results to these calculation can then assigned a certain place in the hash for future use. The different contents of the hash are called elements, so it is similar to an array but can have non-numerical names.
EDIT: You can find an interpreter for Element (written in Perl) here.
Here is the list of operators: In some of these examples, m and n represent numbers already on the stack.
text --pushes the string "text" onto the main stack
' --pops from main stack and pushes onto control stack
" --pops from control stack and pushes onto main stack
# --pops from main stack and destroys
[] --FOR statement (view the top number number from control stack and eval those many times)
{} --WHILE (loop until top number on control stack is 0)
( --pops from main stack, removes first character, pushes the remaining string onto stack, and pushes the removed character onto stack
) --pops from main stack, removes last character, pushes the remaining string onto stack, and pushes the removed character onto stack
~ --pops from main stack, pushes contents of the element with that name
+-*/%^ --pops two most recently named elements, adds/negates/multiplies/divides/modulates/exponentiates them, and places the result on the stack
mn; --pops m and n and assigns element n the value of m
mn@ --pops m and n and moves mth thing in stack to move to place n in stack
m$ --pops m and pushs size of m onto the stack
mn: --pops m and n and pushes m onto the stack n times
mn. --pops m and n and pushes m concatonated with n
m? --pops m and pushes 0 onto control stack if m is '0' or and empty string, else pushes 1
\ --escapes out of next character, so it isn't an operator and con be pushed onto the stack
><= --pops two numbers off of stack and tests, pushes 1 onto control stack if true and 0 if false
` --pops from main stack and prints
&| --pops two items from control stack, performs and/or respectively, and pushes result back onto control stack
! --pops a number off of control stack, pushes 1 if 0 or empty string, 0 otherwise
_ --inputs a word and pushes onto main stack
m, --pops m from main stack, coverts it to char and pushes, converts to num and pushes
Newlines and spaces separate different elements to be pushed onto the stack individually, but can pushed onto the stack using \
Here is a walkthrough of how the program works:
_'[ --take the first line of input, transfer it to the control stack, and start a for loop
_ 2: --take one more line of input, and duplicate it so that there are two copies
n; --take one copy and put into element n
0> --push a zero onto the stack, remove the zero and the other copy of the input, and compare. A 1 will be placed on the control stack if the input was greater than zero, a 0 otherwise.
[ --starts another for loop if the comparison was true. This loop will be repeated once if the comparison was true and no times if it was false, so it is the same as an IF statement.
n~ --pushes n onto the main stack, then pops it ans replaces it with the contents of n, which is the number stored earlier
+ --takes the number and adds it to the running total, which is contained as the last item on the stack
] --ends the inner for loop
] --ends the outer for loop
` --print the top item (also the only item) on the stack to output
T
(number of ...errr...numbers (?) ) is not an option...since the testcases involve extra data afterT
numbers...your code will fail at SPOJ. Everyone (3 answers below) seemed to have cleverly skipped the first integer. \$\endgroup\$