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Problem:

Given a set of integers, find the sum of all positive integers in it.

Input:

  • t – number of test cases [t < 1000]
  • On each of next t lines, an integer N [-1000 ≤ N ≤ 1000]

Output

The program should output the sum of all the positive integers.

Check your code in online judge

Score

Score is equal to size of source code of your program except symbols with ASCII code ≤ 32.

Here is the best score list: Python Best Scores (Best score is 29)

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9  
There are a lot more codegolf challenges at spoj.pl/SHORTEN. Don't see the point of duplicating them here though. –  hallvabo Feb 24 '11 at 12:22
3  
why is this question tagged python.. Are we interested in python solution only? –  Aman ZeeK Verma Feb 24 '11 at 13:56
18  
I don't think questions from contest sites should be posted here. –  fR0DDY Feb 24 '11 at 14:01
2  
I already did this one on SPOJ. A while ago they promoted all the Python2.6 answers to Python3 even though some of them wont run under Python3 and would be longer in Python3 - eg have to use int(input()) instead of input() and print(x) instead of print x. So I don't take SPOJ very seriously anymore. I'm tied with Tim Peters and that's good enough for me :) –  gnibbler Feb 24 '11 at 21:22
4  
I just want to point out, that skipping the T (number of ...errr...numbers (?) ) is not an option...since the testcases involve extra data after T numbers...your code will fail at SPOJ. Everyone (3 answers below) seemed to have cleverly skipped the first integer. –  st0le Feb 25 '11 at 9:36

41 Answers 41

Whitespace, 0

I couldn't resist. S= space, T= tab, N = newline, all have ASCII codes <= 32.

SSSSNSSSSNTTSSSSSTNTNTTNSSSNSSSSTNTTTNTSTNSSSSTNSSSSTNTTTSSSSTNTSSTTTSSSSSTSNTNTTSSSSTSNTTTNTTSNSSSSNSSSSTSNTTTSSSSNTTTTSSSTTSNSNSNNSSTNSSSSNTTTTNSTSSSSTSTSNTNSSNNN

Base64 encoded for easy copy & paste.

ICAgIAogICAgCgkJICAgICAJCgkKCQkKICAgCiAgICAJCgkJCQoJIAkKICAgIAkKICAgIAkKCQkJ
ICAgIAkKCSAgCQkJICAgICAJIAoJCgkJICAgIAkgCgkJCQoJCSAKICAgIAogICAgCSAKCQkJICAg
IAoJCQkJICAgCQkgCiAKIAoKICAJCiAgICAKCQkJCQogCSAgICAJIAkgCgkKICAKCgo=
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3  
(+1) Nice program! A tiny "FWIW": 9 characters can be removed due to 9 instances of an unneeded S in the binary coding of a number. These are all in push-number-to-stack instructions of the form SSSS...N, where the 4th S codes a superfluous leading 0. (Of course this has no effect on the score.) –  r.e.s. Mar 29 '12 at 12:18

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
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5  
An entry like this would be much improved with a pointer to a working environment. –  dmckee Mar 26 '12 at 4:09
3  
I don't think you understand what "human-readable" means. –  WChargin Feb 25 at 2:44

Perl, 31

<>;$i+=$_*($_>0)while<>;print$i
share|improve this answer
    
Wouldn't using say make this just a bit shorter? It'd tie with the 29 character best. –  GigaWatt Apr 23 '12 at 15:52
    
No, because say isn’t built-in and (at least) requires a command-line switch which would count towards the character count. –  Timwi Apr 23 '12 at 18:19
    
It can be shortened to 29 bytes by using $\ instead of $i: <>;$\+=$_*($_>0)while<>;print –  Heiko Oberdiek Mar 5 at 9:59

Ruby, 52

t=gets.to_i;s=0;t.times{i=gets.to_i;s+=i if i>0};p s
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I don't see where you are printing the value of s. –  Dogbert Feb 27 '11 at 18:27
    
yeah I forgot –  david4dev Feb 27 '11 at 19:21

Haskell, 58

Properly operates on only t integers. Haven't run it against Spoj because I just don't care to register there.

f (x:l) = take x l
main = interact $ show . sum . f . map (max 0.read) . lines
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Ruby 1.9.2, 37

p eval [*$<].join.gsub(/\A\d+|-\d+|\n/, '+0')

Call like ruby scriptname file_with_ints .

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I can't read much Ruby, but does that even read the number of test cases? –  Joey Feb 25 '11 at 9:15
    
No it doesn't... –  st0le Feb 25 '11 at 9:31
    
@st0le: Just noticed that apparently no solution currently solves the task. –  Joey Feb 25 '11 at 23:49

code in C 89 characters


x="%d";  main(b,a,t)  {  
  for(scanf(x,&t);t;t--)
    {  scanf(x,&a); a>0?b+=a:a; }  printf(x,b-1);
       return 0; }

I tried a lot to reduce my code less than 63 bytes, but i can reduce it only to 89 bytes. Please help me to reduce it to 63 bytes or even less.

share|improve this answer
    
1) I have counted 90 characters. 2) return 0; is not necessary, the for cycle may be contracted to for(scanf(x,&t);t--;scanf(x,&a),a>0?b+=a:a); == which make that 78 characters... –  V-X Mar 5 at 11:34
    
Doesn't compiles with gcc 4.8.1 error: initializer element is not computable at load time x="%d" –  Manav Jul 21 at 8:49

Perl, 33

<>;while(<>){$i+=$_ if$_>0}print$i

Although the space is necessary, so it seems odd not to count it. Oh well, the rules is the rules.

Hmm. I could probably get away with using a variable name which doesn't count towards the total either. The thing is, I'm not sure how I'd paste the code then.

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Just show them as $^A-$^Z, but beware that many of that variables have special meanings. –  ninjalj Feb 24 '11 at 20:58

Clojure, 71

(reduce + (filter pos? (map #(Integer/parseInt %) (next (line-seq *in*)))))
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This doesn't produce any output, and fails because *in* is not a java.io.BufferedReader, as required by line-seq. –  John Cromartie Jul 11 '13 at 15:24
    
Also it ignores the number of lines input t. –  John Cromartie Jul 11 '13 at 15:39

Haskell, 51

main = interact $ show . f . lines
f (x:l) = foldl (+) 0 $ map read l

(extra spaces for clarity, since they don't count)

Haskell is ... interesting, since you tend to get programs with a significant number of necessary spaces.

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2  
You forgot a filter (>0). –  FUZxxl Apr 10 '11 at 14:58

C,88


x="%d";  main(b,a,t)  {  
for(scanf(x,&t);t--;)  
{  scanf(x,&a); a>0?b+=a:0; }  printf(x,b-1);
return 0; }

After another big effort, code is one character less, please help me to reduce it more.

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6  
just edit the original answer next time –  ratchet freak Jul 3 '11 at 19:18
    
remove the (return 0;) and ({} for for) –  l0n3sh4rk Apr 21 '12 at 19:39
    
b,x="%d";main(a,t){for(scanf(x,&t);t--&&scanf(x,&a);)b+=(a>0)*a;printf(x,b);} <- 77 bytes –  walpen Jun 9 '12 at 19:59
    
@walpen: they used the fact, that their "argc" like parameter was set to 1, your b is uninitialized... –  V-X Mar 5 at 12:18

In memoriam Dennis M. Ritchie

unix 57¹72:

n=$(head -n1 i); echo $(($(head -n $((n+1)) i | tail -n $n | grep -v "-" | tr '\n' '+')0))

assuming i is the file, containing the ints.

¹) was wrong, included the number of lines, and added 1 line too less.

echo $(($(cat i | head -n $(head -n1 i) | grep -v "-" | tr '\n' '+')0))

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Befunge-98 (24)

(Make sure you use an interpreter that can read negative numbers (seems to be a somewhat common bug, but RcFunge works))

<;-1\+*`0:&\_\#;.@;:;#&0 

Perl (25)

(Perl allows control characters in variable names, I named my variable ^B (ASCII 2) so that it does not count towards the goal.)

<>;$^B+=$_*!/-/for<>;print$^B

(Normal variant (27 chars)):

<>;$B+=$_*!/-/for<>;print$B
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i first ignored your perl answer when i saw the variable naming and completely missed the excellent one below it –  ardnew Jul 9 '13 at 22:55

APL (10)

+/{0⌈⎕}¨⍳⎕

Explanation:

  • ⍳⎕: read a line, gives a list [1..N] for user's input N
  • ¨: for each element in this list... (i.e. do N times)
  • 0⌈⎕: read a line, return the maximum of 0 and the entered N
  • We now have a list with all positive Ns the user entered, and 0s where the user entered something negative.
  • +/ gives the sum of this list.
  • The result is output by default (because we're not doing anything else with it).
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Mathematica: 18 16

Boole[#>0]&/@x.x
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Nice function, but how does this handle the specified newline-separated input? How does it not incorporate the number of test cases parameter t as part of the sum? How does it sum only up to the number of test cases specified, even if more are given? –  Jonathan Van Matre Mar 5 at 23:08

PowerShell, 44

($i=$input|%{+$_})[1..$i[0]]-gt0-join'+'|iex
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Q,12

{0+/x(&)x>0}

sample usage

q){0+/x(&)x>0} 1 -1 2 3 -1
6
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befunge, 3524

:0`j&1-\&:0`*+\:0`3*j$.@

with a little inspiration by seeing marinus answer, I've also managed 24 characters. but I've got a completely different approach.

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PYTHON 2.x, 50 chars

r=input
print sum(i for i in (r() for j in range(r())) if i>0)
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C, 70 72 characters

s;main(i,c){for(;c--;i>0?s+=i:0)scanf("%d",s?&i:&c);printf("%d",s-1);}

The results on the SPOJ site definitely seem unreal - I have no idea how to get this down to 63.

However, 68 characters is reachable on some compilers by abusing undefined behaviour. The following works on x86 Linux with 32-bit gcc, on which all arguments are passed on the stack.

s;main(i,c){for(;c--;i>0?s+=i:0)scanf("%d",&i+!s);printf("%d",s-1);}
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C++:

#include<iostream>
using namespace std;
int main()
{
    int c,n,s=0;cin>>c;
    while(c--)
    {
        cin>>n;s+=n*(n>0);
    }
cout<<s;return 0;
}

115 characters long. Need to optimize it to 90. Any suggestions ?

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2  
Just the standard tricks: The return is unnecessary in standard C++ or C99, there is an implicit return 0 in main. By making the variables global you can drop the =0 initialization. Finally, for(;;) is the same number of characters as while() but you get two extra places to put an expression in. –  han Mar 25 '12 at 11:10
    
This is already old, but also, writing std:: before cin and cout and getting rid of the using namespace std; can save 5 more characters. –  Morwenn Jul 3 '13 at 11:35

PHP, 71

<?for($s=0,$t=fgets(STDIN)+0;$t--;$s+=($n=fgets(STDIN))>0?$n:0);echo$s;
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Python: (92 characters)

t = int(raw_input())
n = [int(raw_input()) for i in range(t)]
print(sum([n[i] for i in range(t) if n[i]>0]))
share|improve this answer
    
Using a=raw_input and r=range and using a() and r() later can save quite a few characters. –  Morwenn Jul 3 '13 at 11:37

scala 55 54:

println ((for (r <- 1 to readInt;
  i=readInt;
    if i>0) yield i)sum)
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C

void main()
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
     {
     if(i>0)
     sum=sum+i;
     }
printf("sum of positive numbers is %d",sum);
}
share|improve this answer
1  
Welcome to CodeGolf.SE! If you look at the other answer, you'll see that they have formatted code and a minimal header noting the implementation language; on more complicated challenges many also have notes on the implementation and any limits or surprises in the code. Without some of this, you answer is unlikely to be well received. –  dmckee Oct 12 '11 at 15:28
    
I counted the chars, added the indentation to make code-layout work, and removed the decoration of the output. Oh - now I have to count again. :) –  user unknown Oct 13 '11 at 14:35
    
Added language name. There's a lot of room for reductions here - sum can be reduced to s, the output string can just be "%d", etc. –  Gareth Oct 31 '12 at 9:17

excel, 27

=SUM(INDIRECT("A2:A"&1+A1))

count t in A1, rest of data a2 and down

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Ruby, 42

s=0
gets.to_i.times { 
    i=gets.to_i
    s+=i if i>0
}
p s

(based on david4dev's answer)

Best score for Ruby on spoj: http://www.spoj.com/ranks/SIZECON/lang=RUBY

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45 chars in python

c=0
j=input
for i in j()*[0]:
    b=j()
    c+=b*(b>0)
print c
share|improve this answer
1  
How you counted that? It gives me 54 characters. –  manatwork Jul 2 '13 at 7:02
    
@manatwork, this question has non-standard scoring rules which don't count whitespace. –  Peter Taylor Jul 2 '13 at 11:00
    
Oops, sorry. I missed that. Thanks, @PeterTaylor. –  manatwork Jul 2 '13 at 11:39

Perl, 30

so close to that python best of 29, but had to add +1 for the -p switch

$s.=$_}{$s=~s/\s\d+/$_+=$&/eg
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Python 3, score: 63 61

r=0
q=input
w=int
for i in[1]*w(q()):
 t=w(q())
 r+=t+abs(t)
print(r/2)

I don't really know python, so I'm having a considerably difficult time making this any smaller.

Python, score: 49 48

r=0
q=input
for i in[1]*q():
 t=q()
 r+=t+abs(t)
print r/2

Same as above, but python 2 doesn't need the casting to int, apparently.

share|improve this answer
    
So do I. I'm not entirely sure where 49 came from. –  Cereal Jul 9 '13 at 19:34
    
my mistake -- apparently characters with ascii value < 33 are discounted. –  boothby Jul 9 '13 at 19:49
    
if whitespace etc doesn't count, then it's shorter to put "compound" statements like r,q,w=0,input,int as single statements on separate lines; r=0 q=input w=int (where space is newline) –  daniero Jul 10 '13 at 11:40
    
Thanks, edited that in. –  Cereal Jul 10 '13 at 12:41
    
The reason for casting to int in Python 3 is input is not the same function : see this SO answer –  plg Feb 2 at 11:04

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