Subtract the next numbers

Question

Description

Subtract the next P numbers from a N number. The next number of N is N + 1.

Look at the examples to get what I mean.

Examples:

Input: N=2,P=3
Calculate: n - (n+1) - (n+2) - (n+3)     //Ending with 3, because P=3
Calculate: 2 -  2+1  -  2+2  - 2+3       //Replacing N with 2 from Input
Calculate: 2 -  3    -  4    - 5
Output: -10


Input: N=100,P=5
Calculate: n - (n+1) - (n+2) - (n+3) - (n+4) - (n+5)
Calculate: 100-  101 -  102  -  103  -  104  - 105
Output: -415


Input: N=42,P=0
Calculate: n
Calculate: 42
Output: 42


Input: N=0,P=3
Calculate: n - (n+1) - (n+2) - (n+3)
Calculate: 0 -  1    -  2    -  3
Output: -6


Input: N=0,P=0
Calulate: n
Calculate: 0
Output: 0

Input:

N: Integer, positive, negative or 0

P: Integer, positive or 0, not negative

Output:

Integer or String, leading 0 allowed, trailing newline allowed

Rules:

• No loopholes
• This is code-golf, so shortest code in bytes wins
• Input and Output must be as described

Answer 1

05AB1E, 5 3 bytes

Saved 2 bytes thanks to Adnan

Ý+Æ

Explanation

Takes P then N as input.

       # implicit input, ex 5, 100
Ý      # range(0,X): [0,1,2,3,4,5]
 +     # add: [100,101,102,103,104,105]
  Æ    # reduced subtraction: 100-101-102-103-104-105

Answer 2

Python 2, 26 24 23 bytes

-2 bytes thanks to @Adnan (replace p*(p+1)/2 with p*-~p/2)
-1 byte thanks to @MartinEnder (replace -p*-~p/2 with +p*~p/2

lambda n,p:n-p*n+p*~p/2

Tests are on ideone

Answer 3

CJam, 8 bytes

{),f+:-}

Test suite.

Too bad that the closed form solution is longer. :|

Explanation

),  e# Get range [0 1 ... P].
f+  e# Add N to each value to get [N N+1 ... N+P].
:-  e# Fold subtraction over the list, computing N - (N+1) - (N+2) - ... - (N+P).

Answer 4

Haskell, 21 bytes

a#b=foldl1(-)[a..a+b]

Answer 5

Javascript (ES6), 20 19 18 bytes

n=>p=>n+p*(~p/2-n)

Saved 1 byte by currying, as suggested by Zwei
Saved 1 byte thanks to user81655

Test

let f =
n=>p=>n+p*(~p/2-n)

console.log(f(2)(3))
console.log(f(100)(5))
console.log(f(42)(0))
console.log(f(0)(3))
console.log(f(0)(0))

Answer 6

Haskell, 19 18 bytes

n#p=n+sum[-n-p..n]

Previous 19 bytes solutions

n#p=n-n*p-(p*p+p)/2
n#p=n-sum[n+1..n+p]

Answer 7

Jelly, 4 bytes

r+_/

Try it online!

How it works

r+_/  Main link. Arguments: n, p

 +    Yield n+p.
r     Range; yield [n, ..., n+p].
  _/  Reduce by subtraction.

Answer 8

C#, 21 20 bytes

Edit: Saved one byte thanks to TheLethalCoder

N=>P=>N-P++*(N+P/2);

Try it online!

Full source, including test cases:

using System;

namespace substract
{
    class Program
    {
        static void Main(string[] args)
        {
            Func<int,Func<int,int>>s=N=>P=>N-P++*(N+P/2);
            Console.WriteLine(s(2)(3));     //-10
            Console.WriteLine(s(100)(5));   //-415
            Console.WriteLine(s(42)(0));    //42
            Console.WriteLine(s(0)(3));     //-6
            Console.WriteLine(s(0)(0));     //0

        }
    }
}

Answer 9

Perl, 23 22 bytes

Includes +1 for -p

Give n and p (in that order) on separate lines of STDIN:

subtract.pl
2
3
^D

subtract.pl:

#!/usr/bin/perl -p
$_-=eval"+2+\$_++"x<>

(using '' quotes to save the \ invokes a 2 byte penalty because it can't be combined with -e)

Same idea and length:

#!/usr/bin/perl -p
$_+=eval"-1-++\$_"x<>

Surprisingly doing the actual calculation is shorter than using the direct formula (these $'s really hurt for arithmetic)

Answer 10

C++, 54 51 bytes

  [](int N,int P){int F=N;while(P--)F-=++N;return F;}

[](int N,int P){int F;for(F=N;P;F-=++N,P--);return F;}

Test:

#include <iostream>
int main(void)
{
    int N, P;
    std::cin >> N >> P;
    auto f = [](int N,int P){int F=N;while(P--)F-=++N;return F;};
    std::cout << f(N,P) << std::endl;
    return 0;
}

Answer 11

Mathematica, 15 bytes

#2-##-#(#+1)/2&

An unnamed function that receives P and n as its parameters in that order.

Uses the closed form solution n - n*p - p(p+1)/2.

Answer 12

Pyke, 6 bytes

m+mhs-

Try it here!

m+     -    map(range(input_2), +input_1)
  mh   -   map(^, +1)
    s  -  sum(^)
     - - input_1 - ^

Answer 13

Brachylog, 19 17 bytes

hHyL,?+y:Lx+$_:H+

Explanation

hH                  Input = [H, whatever]
 HyL,               L = [0, …, H]
     ?+             Sum the two elements in the Input
       y            Yield the range from 0 to the result of the sum
        :Lx         Remove all elements of L from that range
           +        Sum the remaining elements
            $_      Negate the result
              :H+   Add H

Answer 14

MATL, 5 bytes

:y+s-

Inputs are P and then N.

Try it at MATL Online!

Explanation

:     % Take P implicitly. Range [1 2 ... P]
      %     Stack: [1 2 ... P]
y     % Take N implicitly at the bottom of the stack, and push another copy
      %     Stack: N, [1 2 ... P], N
+     % Add the top two arrays in the stack , element-wise
      %     Stack: N, [N+1 N+2 ... N+P]
s     % Sum of array
      %     Stack: N, N+1+N+2+...+N+P
-     % Subtract the top two numbers
      %     Stack: N-(N+1+N+2+...+N+P)
      % Implicitly display

Answer 15

Forth, 36 bytes

Simple computation of n - (p*n + (p^2+p) / 2)

: f 2dup dup dup * + 2/ -rot * + - ;

Try it online

Answer 16

Fourier, 34 bytes

I~No~OI~P>0{1}{@P+N(N^~NO-N~ON)Oo}

Try it online!

Answer 17

Batch, 30 bytes

@cmd/cset/a%1-(%1*2+%2+1)*%2/2

Takes n and p as command-line parameters and prints the result without a trailing newline.

Answer 18

S.I.L.O.S, 80 bytes

GOTO b
lbla
n+1
m-n
i-1
GOTO d
lblb
readIO
n=i
m=n
readIO
lbld
if i a
printInt m

Try it online with test cases:
2,3
100,5
42,0
0,3
0,0

Answer 19

R, 17 14 bytes

N-N*P-sum(0:P)

Thanks to billywob for golfing away 3 bytes. Previous answer:

N-sum(N+if(P)1:P)

Note that 1:0 expands to the vector (1,0) so we need the if(P) condition (or to use seq_len, but that is more bytes). Without the condition, we would get the wrong output if P=0.

If P is zero, then the sum expands to sum(N+NULL), then to sum(numeric(0)), which is zero.

Answer 20

Java, 67, 63 bytes

Golfed:

int x(int n,int p){return-((p%2<1)?p*p/2+p:p/2*(p+2)+1)+n-p*n;}

Ungolfed:

int x(int n, int p)
{
    return -((p%2<1) ? p*p/2+p : p/2 * (p+2) + 1) + n - p*n;
}

Basically I did some math on the formula. The n - p*n part takes care of the all n's in the formula. Then I used a super fun property of summing together linearly increasing set of integers (arithmetic series): I used the sum of first and last integer and then multiply it by set.length / 2 (I also check for the parity and handle it appropriately).

Try it: https://ideone.com/DEd85A

Answer 21

Java 7, 43 40 bytes

int c(int n,int p){return n-p*n+p*~p/2;}

Java 8, 19 bytes

(n,p)->n-p*n+p*~p/2

Shamelessly stolen from @JonathanAllan's amazing Python 2 formula.

Original answer (61 60 bytes):

int c(int n,int p){int r=n,i=1;for(;i<p;r-=n+++i);return r;}

Ungolfed & test cases:

Try it here.

class M{
  static int c(int n, int p){
    return n - p*n + p*~p / 2;
  }

  public static void main(String[] a){
    System.out.println(c(2, 3));
    System.out.println(c(100, 5));
    System.out.println(c(42, 0));
    System.out.println(c(0, 3));
    System.out.println(c(0, 0));
  }
}

Output:

-10
-415
42
-6
0

Answer 22

PHP, 33 Bytes

$n-=$n*$p+array_sum(range(0,$p));

Answer 23

Jelly, 7 bytes

RS+×_×-

Arguments are P, N
Test it on TryItOnline

How?

RS+×_×-  - takes two arguments: P, N
R        - range(P): [1,2,3, ... ,P]
 S       - sum: 1+2+3+ ... +P
   ×     - multiply: P*N
  +      - add: 1+2+3+ ... +P + P*N
    _    - subtract: 1+2+3+ ... +P + P*N - N
      -  - -1
     ×   - multiply: (1+2+3+ ... +P + P*N - N)*-1
                   = -1-2-3- ... -P - P*N + N
                   = N - (N+1) - (N+2) - (N+3) - ... - (N+P)

Answer 24

Pyth - 6 bytes

-F}Q+E

Test Suite.

Answer 25

Burlesque, 12 bytes

perz?+{.-}r[

Try it online!

pe     #Read input as P, N
rz     #Range (0,N)
?+     #Add P to each
{.-}r[ #Reduce via subtraction

Answer 26

Labyrinth, 15 bytes

?:?:}*-{:)*#/-!

or

??:}`)*{:)*#/-!

Uses the closed form solution n - n*P - P*(P+1)/2.

Answer 27

php, 38 bytes

<?=$argv[1]*(1-$a=$argv[2])-$a++*$a/2;

Answer 28

Excel, 20 bytes

Subtract the next B1 integers from A1:

=A1-B1*(A1+(B1+1)/2)

Answer 29

GolfScript, 11 bytes

Formula stolen elsewhere...

~1$1$)2/+*-

Try it online!

Explanation

This just boils down to

A-B*(A+(B+1)/2)

Converted to postfix.

GolfScript, 15 bytes

~),{1$+}%{-}*\;

Try it online!

Explanation

~               # Evaluate the input
 ),             # Generate inclusive range
   {1$+}%       # Add each item by the other input
         {-}*   # Reduce by difference
             \; # Discard the extra other input

Answer 30

Keg, -hr, 8 6 bytes

+ɧ^∑$-

Try it online!

No formula, just a for loop!

Answer 31

W d, 6 4 bytes

Surprised to see that it's not short at all. (Although it at least ties with Jelly...)

7▄i←

Explanation

Uncompressed:

+.@-R
+     % Generate b+a
 .    % Generate range(a,b+a)
  @-R % Reduce via subtraction

Answer 32

Pyth, 11 bytes

Ms+Gm_+GdSH

A function g that takes input of n and p via argument and prints the result. It can be called in the form gn p.

Try it online

How it works

Ms+Gm_+GdSH  Function g. Inputs: G, H
M            g=lambda G,H:
         SH   1-indexed range, yielding [1, 2, 3, ..., H]
    m_+Gd     Map lambda d:-(G+d) over the above, yielding [-(G+1), -(G+2), -(G+3),
              ..., -(G+H)]
  +G          Add G to the above, yielding [G, -(G+1), -(G+2), -(G+3), ..., -(G+H)]
 s            Reduce on addition with base case 0, yielding G-(G+1)-(G+2)-(G+3)...
              -(G+H)
              Implicitly print

Answer 33

C89, 38, 35, 33 bytes

h(n,p,r)int*r;{*r=n-p++*(n+p/2);}

Test it on Coliru.

Answer 34

Maple, 19 bytes

n-sum(i,i=n+1..n+p)

Usage:

> f:=(n,p)->n-sum(i,i=n+1..n+p);
> f(2, 3);
  -10
> f(100,5);
  -415
> f(42,0);
  42

Answer 35

Perl 6, 21 bytes

{$^n-[+] $n^..$n+$^p}

Explanation:

# bare block lambda with two placeholder parameters ｢$n｣ and ｢$p｣
{
  $^n -
      # reduce using ｢&infix:<+>｣
      [+]
          # a Range that excludes ｢$n｣ and has ｢$p｣ values after it
          $n ^.. ($n + $^p)
}

Answer 36

Java 8, 25 bytes

(n,p)->n-(p*n+p*(p+1)/2);

Ungolfed test program

public static void main(String[] args) {

    BiFunction<Integer, Integer, Integer> f = (n, p) -> n - (p * n + p * (p + 1) / 2);
    System.out.println(f.apply(100, 5));
}

Answer 37

Scala, 41 bytes

def?(n:Int,p:Int)=n-(1 to p).map{n+_}.sum

Testing code:

println(?(2,3))
println(?(100,5))
println(?(42,0))
println(?(0,3))
println(?(0,0))

// Output
-10
-415
42
-6
0

Answer 38

Clojure/Clojurescript, 30 bytes

#(reduce -(range %(+ 1 % %2)))

The straightforward approach is shorter than the mathematically clever ones.

Answer 39

Julia: 17-18 bytes (13 as program with terminal inputs)

As per suggestion in comments that "function or program" form is needed:

• as function: 17 characters, 18 bytes if you count ∘ as multibyte

  n∘r=2n-sum(n:n+r)  
  

  usage: 5∘3 outputs -16

• as program, passed initial parameters from terminal: 13 bytes:

  2n-sum(n:n+r)  
  

  usage: julia -E 'n,r=5,3;include("codegolf.jl")'