# Sum of positive integers. [closed]

## 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)

• 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
• why is this question tagged python.. Are we interested in python solution only? – Aman ZeeK Verma Feb 24 '11 at 13:56
• I don't think questions from contest sites should be posted here. – fR0DDY Feb 24 '11 at 14:01
• 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
• 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

# 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=

• (+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

p eval [*$<].join.gsub(/\A\d+|-\d+|\n/, '+0')  Call like ruby scriptname file_with_ints . • I can't read much Ruby, but does that even read the number of test cases? – Joey Feb 25 '11 at 9:15 • No it doesn't... – st0le Feb 25 '11 at 9:31 • @st0le: Just noticed that apparently no solution currently solves the task. – Joey Feb 25 '11 at 23:49 # Ruby, 52 t=gets.to_i;s=0;t.times{i=gets.to_i;s+=i if i>0};p s  • I don't see where you are printing the value of s. – Dogbert Feb 27 '11 at 18:27 • yeah I forgot – david4dev Feb 27 '11 at 19:21 # Haskell, 58 Properly operates on only t integers. Haven't run it against Spoj because I just don't care to register there. f (x:l) = take x l main = interact$ show . sum . f . map (max 0.read) . lines

• What are "t integers"? – wchargin Feb 10 '15 at 20:25

## 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.

• 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 '14 at 11:34
• Doesn't compiles with gcc 4.8.1 error: initializer element is not computable at load time x="%d" – manav m-n Jul 21 '14 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.

• 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*)))))

• 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

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))

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.

• 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; }


• 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 '14 at 12:18

# 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

• 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).

Mathematica: 18 16

Boole[#>0]&/@x.x

• 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 '14 at 23:08

## PowerShell, 44

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


## Q,12

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


sample usage

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


:0j&1-\&:0*+\:03*j$.@  with a little inspiration by seeing marinus answer, I've also managed 24 characters. but I've got a completely different approach. # PYTHON 2.x, 50 chars r=input print sum(i for i in (r() for j in range(r())) if i>0)  # C, 70 72 characters s;main(i,c){for(;c--;i>0?s+=i:0)scanf("%d",s?&i:&c);printf("%d",s-1);}  The results on the SPOJ site definitely seem unreal - I have no idea how to get this down to 63. However, 68 characters is reachable on some compilers by abusing undefined behaviour. The following works on x86 Linux with 32-bit gcc, on which all arguments are passed on the stack. s;main(i,c){for(;c--;i>0?s+=i:0)scanf("%d",&i+!s);printf("%d",s-1);}  ## excel, 27 =SUM(INDIRECT("A2:A"&1+A1))  count t in A1, rest of data a2 and down ## Clojure, 108 (let [[n & m] (->> *in* java.io.BufferedReader. line-seq (map read-string))] (->> m (take n) (filter pos?) (apply +) println))  I really wish I could avoid the java.io.BufferedReader. part, since it costs 24 chars itself. But AFAIK there's no facility to read lines from STDIN without it. # Perl, 20 I know it is old and trivial, but the Perl answer can be still improved: #!perl -p$.<2or$\+=$_*!/-/}{

• This is awesome! But what does }{ mean/do? – daniero Jul 19 '15 at 13:40

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 ?

• 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;


### 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]))

• 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;
if i>0) yield i)sum)


## 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);
}

• 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 --- ex-moderator kitten 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

## Ruby, 42

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


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

# 45 chars in python

c=0
j=input
for i in j()*:
b=j()
c+=b*(b>0)
print c
`
• 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