# 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. Feb 24 '11 at 12:22
• why is this question tagged python.. Are we interested in python solution only? Feb 24 '11 at 13:56
• I don't think questions from contest sites should be posted here. 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 :) 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. 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.) 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 [*$<].join.gsub(/\A\d+|-\d+|\n/, '+0')  Call like ruby scriptname file_with_ints . • I can't read much Ruby, but does that even read the number of test cases? – Joey Feb 25 '11 at 9:15 • No it doesn't... Feb 25 '11 at 9:31 • @st0le: Just noticed that apparently no solution currently solves the task. – Joey Feb 25 '11 at 23:49 # Ruby, 52 t=gets.to_i;s=0;t.times{i=gets.to_i;s+=i if i>0};p s  • I don't see where you are printing the value of s. Feb 27 '11 at 18:27 • yeah I forgot Feb 27 '11 at 19:21 # Haskell, 58 Properly operates on only t integers. Haven't run it against Spoj because I just don't care to register there. f (x:l) = take x l main = interact$ show . sum . f . map (max 0.read) . lines

• What are "t integers"? 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" 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. 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. Jul 11 '13 at 15:24
• Also it ignores the number of lines input t. 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))

### 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.

• You forgot a filter (>0). 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.

• just edit the original answer next time Jul 3 '11 at 19:18
• remove the (return 0;) and ({} for for) 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 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 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? Mar 5 '14 at 23:08

## PowerShell, 44

($i=$input|%{+$_})[1..$i[0]]-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? 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. 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. Jul 3 '13 at 11:37

### scala 55 54:

println ((for (r <- 1 to readInt;
i=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. 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. :) 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. 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


(based on david4dev's answer)

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