Sum of positive integers. (Spoj - SIZECON)

Question

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)

Answer 1

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=

Answer 2

Perl, 31

<>;$i+=$_*($_>0)while<>;print$i

Answer 3

Ruby, 52

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

Answer 4

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

Answer 5

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.

Answer 6

Ruby 1.9.2, 37

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

Call like ruby scriptname file_with_ints .

Answer 7

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

Answer 8

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.

Answer 9

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.

Answer 10

Clojure, 71

(reduce + (filter pos? (map #(Integer/parseInt %) (next (line-seq *in*)))))

Answer 11

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.

Answer 12

PowerShell, 44

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

Answer 13

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))

Answer 14

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

Answer 15

Q,12

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

sample usage

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

Answer 16

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.

Answer 17

PYTHON 2.x, 50 chars

r=input
print sum(i for i in (r() for j in range(r())) if i>0)

Answer 18

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).

Answer 19

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 ?

Answer 20

PHP, 71

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

Answer 21

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]))

Answer 22

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace ConsoleApplication1
{
 static class Program
 {
   static void Main(string[] args)
   {
     Console.WriteLine(Console.In.ReadToEnd().Split('\n').Sum(l => { Int32 v; return Int32.TryParse(l, out v) ? v.To(1000).Where(x => x > 0).Sum() : 0; }));

    }
    static IEnumerable<int> To(this int a, int b) { for (var i = a; a <= b; a++) { yield return i; } }
  }
}

Answer 23

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);}

Answer 24

scala 55 54:

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

Answer 25

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);
}

Answer 26

excel, 27

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

count t in A1, rest of data a2 and down

Answer 27

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

Answer 28

Python, 21

y=0
for z in x:
    if z>0:
        y+=z

Where x is your given set and z is the sum
(Wasn't sure of the conditions of input and output)

Alternate Solution (courtesy of some guy I know):

sum([max(0,i) for i in k])