Bazen gerçekten sıkıldığımda, negatif olmayan bir tamsayı dizisinin toplamını almak isterim. Sadece ikisinin gücü olan uzunluk dizilerinin toplamını alıyorum. Ne yazık ki sık sık hata yapıyorum. Neyse ki, şu şekilde ilerledikçe çalışmalarımın kaydını tutuyorum:

Sadece bir tane kalana kadar bitişik sayıları çiftleri ekliyorum. Örneğin:

 6 + 18 + 9 + 6 + 6 + 3 + 8 + 10
=  24   +   15  +   9   +   18
=       39      +       27
=               66

İşin bir yerde bir hata mı yaptım belirlemek. Girişin işleve iletilmesini sağlayabilir veya standart içeriden okuyabilirsiniz. Çıktı yazdırılabilir veya iade edilebilir.

Giriş: Bir dizi / liste / etc. Negatif olmayan tamsayılar ve diliniz gerektiriyorsa muhtemelen bu dizinin uzunluğu. Bu dizi soldan sağa ve sonra yukarıdan aşağıya okunan tüm sayılar olacaktır. Örneğin, yukarıdaki dizi olur:
[[6, 18, 9, 6, 6, 3, 8, 10], [24, 15, 9, 18], [39, 27], [66]]
ya
[6, 18, 9, 6, 6, 3, 8, 10, 24, 15, 9, 18, 39, 27, 66]da istersen.

Çıktı: Bir hatanın yapılıp yapılmayacağını temsil eden tek bir boolean. Boolean, bir hata yapılan tüm girdilerin aynı sonucu döndürmesi / yazdırması ve hata içermeyen tüm girdinin aynı sonucu döndürmesi / yazdırması koşuluyla herhangi bir haritalama kullanılarak temsil edilebilir . Bu söylemeden geçmeli, ancak bu iki çıktı aynı olamaz.

Bazı Doğru Toplama Örnekleri:

6

5+6
=11

  3 + 2 + 4 + 5
=   5   +   9
=       14

[0, 1, 2, 3, 1, 5, 6]

[[1, 2, 4, 8], [3, 12], [15]]

Bazı Yanlış Toplama Örnekleri:

5+4
=8

4 + 4 + 4 + 4
= 9   +   7
=     16

[[1, 2, 3, 4], [7, 3], [10]]

[3, 4, 5, 6, 7, 8, 9]

Hata yapabileceğimi ve hala doğru cevabı alabileceğimi unutmayın. Eğer bir hata yaparsam, son dizide asla fazladan bir sayı veya eksik bir sayı olmaz, sadece yanlış bir sayı verilir.

Standart boşluklar yasaktır. Her dilde en kısa cevap kazanır. Eski cevap, beraberlik durumunda kazanacaktır. "Aynı dilin" ne olduğuna karar verme hakkını saklı tutarım, ancak açıkça söyleyeyim, hem Python 2 hem de Python 3'te bir puan kazanılamaz.

Jelly, 6 bytes

Ṗ+2/€ẇ

Try it online!

How it works

Ṗ+2/€ẇ  Main link. Argument: A (2D array)

Ṗ       Pop; yield A without its last element. Let's call the result B.
  2/    Pairwise reduce...
    €     each array in B...
 +          by addition.
     ẇ  Window exists; test if the result appears in A as a contiguous subarray.

Python 2, 51 bytes

lambda l:map(sum,zip(*[iter(l)]*2))==l[len(l)/2+1:]

Try it online! Thanks to Rod for the test cases.

Takes the whole list flat as input. Groups elements into adjacent pairs using the zip/iter trick, takes the sum of the pairs, and checks if the result equals the second half of the list.

A recursive method came close at 55 bytes:

f=lambda l:len(l)<2or l[0]+l[1]==l[len(l)/2+1]*f(l[2:])

This used that input integers are positive, which has since changed in the spec.

Röda, 40 bytes

{[0]if{|x|[[x()|[_+_]]=y]if tryPeek y}_}

Try it online!

It's an anonymous function that returns 0 if there are no mistakes and nothing if there are mistakes.

Explanation:

{[0]if{|x|[[x()|[_+_]]=y]if tryPeek y}_}
{                                      } /* Anonymous function */
      {|x|                           }_  /* Loop over lists in the stream */
                         if tryPeek y    /* If there are lists left after this */
            x()                          /* Push values in list x to the stream */
               |[_+_]                    /* Sum every pair of numbers in x */
           [         ]                   /* Create a list of sums */
                      =y                 /* If the list equals to the next list */
          [             ]                /* Push the result */
    if                                   /* If all results are TRUE */
 [0]                                     /* Return 0 */
                                         /* Otherwise return nothing */

Here's a version that is shorter (35 bytes) but against the rules (I think):

{{|x|[[x()|[_+_]]=y]if tryPeek y}_}

Try it online!

It's an anonymous function that reads values from the stream and pushes TRUE or FALSE for each correct line.

I'm not sure if this (multiple return values) is accepted in the rules. Here's my defence: in Röda, conditions of if and while blocks are not boolean values, but streams. A "truthy" stream is either empty or contains only TRUEs, and a "falsy" stream contains one or more FALSEs. In this way, this function returns a "boolean" value. And can be used as a condition of an if statement without any reduce-operations, etc.

Python 2, 69 65 bytes

lambda x:reduce(lambda a,b:(b==map(sum,zip(a[::2],a[1::2])))*b,x)

Try it online!
Returns :
Empty List as Falsy
Total sum as Truthy

Mathematica, 36 bytes

Most[Tr/@#~Partition~2&/@#]==Rest@#&

Pure function taking a nested list as input and returning True or False. The function Tr/@#~Partition~2& takes the pairwise sums of a list, which is then applied (/@#) to each sublist of the input list. The first, second, ... sublists in the resulting list are supposed to equal the second, third, ... sublists in the original input; Most[...]==Rest@# tests for this property.

Python 2, 80 bytes

lambda l:all(l[i+1]==map(sum,zip(l[i][::2],l[i][1::2]))for i in range(len(l)-1))

Try it online!

Not quite as good as the other python answer, but I felt like posting it anyway. This just goes to show why I'm not as good at golfing in regular languages.

JavaScript (ES6), 54 bytes

a=>!a.slice(-a.length/2).some((n,i)=>a[i+=i]+a[i+1]-n)

Takes a flattened array.

05AB1E, 15 12 bytes

¬svyQy2ôO}\P

Try it online!

Explanation

¬             # get the first element from input without popping
 sv      }    # for each element y in input
   yQ         # compare y to the top of the stack 
              # leaves 1 on the stack if equal and otherwise 0
     y2ô      # split y in pieces of 2
        O     # sum each pair in the list
          \   # discard the top of the stack (the left over final element of the list)
           P  # product of stack (all the 1's and/or 0's from the comparisons)

Haskell, 82 79 65 bytes

-14 bytes thanks to nimi!

p(x:y:z)=x+y:p z
f x=and.zipWith(==)(drop(length x`div`2+1)x)$p x

Works by comparing the sum each pair of elements to the corresponding element on the next line down. Some bytes can probably be golfed from f, but I can't figure out where.

Python 3, 69 68 bytes

lambda v:any(x+v[i-1]-v[(len(v)+i)//2]for i,x in enumerate(v)if i%2)

I know there are already two other python answers... but this one's in python 3, so it's exotic.

This works on a flattened input.

Output:

False if there is no mistake,

True if there is a mistake.

Ruby, 50 bytes

->x{a=b=0;b&&=x[a/2]==x[a]+x[a-1]while x[-a-=2];b}

Reversing the array, any element of the first half (position n) must be the sum of the elements in position n*2 and n*2+1.

Brachylog, 16 13 bytes

s₂ᵘ{{ġ₂+ᵐ}ᵈ}ᵐ

Try it online!

This is just dreadfully long! There has to be some way to not nest inline predicates here.

The predicate succeeds (printing true. as a program) if no mistakes were made and fails (printing false. as a program) otherwise.

s₂ᵘ              Every length 2 substring of the input
   {       }ᵐ    for every element satisfies the following:
    {ġ₂          the pairs of elements of the input
       +ᵐ        when each pair is summed is the output
         }ᵈ      where the input is the first item and the output is the second.

Python 2, 64 bytes

lambda a:[map(int.__add__,x[::2],x[1::2])for x in a[:-1]]==a[1:]

Try it online!

An unnamed function that takes a list of lists (one per line of working as it were), and returns True if no mistakes were made and False otherwise.

It works by using the input without the last entry, a[:-1], to form what the input without the first entry should be and checking that is what was input, ==a[1:].

This formation is achieved by mapping the integer type's addition function, int.__add__, over the pairs of numbers as yielded by two "slices", one slice being every other item starting at the 0th index, x[::2], the other slice being every other item starting at the 1st index, x[1::2].

Pip, 20 19 bytes

$*{{b=$+*Ya<>2}MPa}

This is an anonymous function that takes one argument, a list of lists (e.g. [[1 2 3 4] [3 7] [10]]). Verify all test cases: Try it online!

Explanation

In a Pip function, the first two arguments are assigned to a and b.

  {               }  Anonymous function:
   {          }MPa    To each pair of sublists from a, map this helper function:
          a<>2         Group the 1st member of the pair into 2-item sublists
         Y             Yank that value (no-op used to override precedence order)
      $+*              Map (*) the fold ($) on addition (+) operator
    b=                 If the 2nd member of the pair is = to the result, 1; else 0
$*                   Modify the outside function by folding its return value on *
                     (makes a list containing all 1's into 1 and any 0's into 0)

For example:

a
[[1 2 3 4] [7 3] [10]]

{...}MP
a:[1 2 3 4] b:[7 3]
a:[7 3]     b:[10]

a<>2
[[1 2] [3 4]]
[[7 3]]

$+*
[3 7]
[10]

b=
0
1

Final result of {...}MPa
[0 1]

$*
0

PHP, 96 95 bytes:

using builtins:

function f($a){return!$a[1]||array_pop($a)==array_map(array_sum,array_chunk(end($a),2))&f($a);}
// or
function f($a){return!$a[1]||array_map(array_sum,array_chunk(array_shift($a),2))==$a[0]&f($a);}

recursive functions return true or false.

breakdown for first function:

function f($a){
    return!$a[1]||      // true if array has no two rows ... or
    array_pop($a)==     // remove last row, true if equal to
    array_map(array_sum,    // 3. sum up every chunk
        array_chunk(        // 2. split to chunks of 2
            end($a)         // 1. new last row
        ,2))
    &f($a);             // and recursion returns true
}

older solutions (96 bytes each) using loops:

function f($a){foreach($a[0]as$k=>$v)$b[$k/2]+=$v;return$b==$a[1]&&!$a[2]|f(array_slice($a,1));}
//or
function f($a,$y=0){foreach($a[$y]as$k=>$v)$b[$k/2]+=$v;return$b==$a[++$y]&&!$a[$y+1]|f($a,$y);}

breakdown for last function:

function f($a,$y=0){
    foreach($a[$y]as$k=>$v)$b[$k/2]+=$v;    // build $b with correct sums from current row
    return$b==$a[++$y]                      // true if $b equals next row
    &&!$a[$y+1]                             // and (finished
        |f($a,$y);                          //      or recursion returns true)
}

iterative snippets, 81 bytes

for(;$a[1];)if(array_pop($a)!=array_map(array_sum,array_chunk(end($a),2)))die(1);
for(;$a[1];)if(array_map(array_sum,array_chunk(array_shift($a),2))!=$a[0])die(1);
for(;$a[++$y];$b=[]){foreach($a[$y-1]as$k=>$v)$b[$k/2]+=$v;if($a[$y]!=$b)die(1);}

assume array predefined in $a; exits with error if it is incorrect.

C, 54 bytes:

f(int*s,int*e){return e-s>1?*s+s[1]-*e||f(s+2,e+1):0;}

Ungolfed:

int f(int*s,int*e) {
    if(e-s>1) {
        return *s+s[1] != *e || f(s+2,e+1);
    } else {
        return 0;
    }
}

Test with

#include <assert.h>
int main() {
    int input1[15] = {6, 18, 9, 6, 6, 3, 8, 10, 24, 15, 9, 18, 39, 27, 66};
    assert(!f(input1, input1+8));

    int input2[7] = {3, 4, 5, 6, 7, 8, 9};
    assert(f(input2, input2+4));
}

As you see, f() returns true for invalid inputs and false (= 0) for valid ones.

As always, recursion is less bytes than iteration, so f() is recursive, even though it takes two iterators as arguments. It works by repeatedly comparing the sum of two integers at s to one integer at e, ignoring the level boundaries, and carrying on until the two iterators meet. I have also used some boolean zen together with the fact that any non-zero integer value is considered true in C to further shorten the code.

R, 92 77 bytes

Anonymous function which takes a flat sequence of numbers as input. Returns TRUE or FALSE as appropriate. Uses the same approach conceptually as xnor's python answer.

function(x,l=sum(1|x)/2)all(rowSums(cbind(x[1:l*2-1],x[1:l*2]))==tail(x,l-1))

Previous solution, using the rollapply function from the zoo package, and taking input as a list, e.g. list(c(6, 18, 9, 6, 6, 3, 8, 10), c(24, 15, 9, 18), c(39, 27), c(66)):

function(l,T=1){for(i in 2:length(l))T=T&all(zoo::rollapply(l[[i-1]],2,sum,by=2)==l[[i]]);T}

JavaScript (ES6), 46 44 bytes

Takes input as a flattened array. Returns NaN for valid or 0 for invalid.

f=([a,b,...c])=>a+b==c[c.length>>1]?f(c):b-b

Test

f=([a,b,...c])=>a+b==c[c.length>>1]?f(c):b-b

console.log(f([6, 18, 9, 6, 6, 3, 8, 10, 24, 15, 9, 18, 39, 27, 66])) // NaN
console.log(f([0, 1, 2, 3, 1, 5, 6]))                                 // NaN
console.log(f([1, 2, 4, 8, 3, 12, 15]))                               // NaN
console.log(f([1, 2, 3, 4, 7, 3, 10]))                                // 0
console.log(f([3, 4, 5, 6, 7, 8, 9]))                                 // 0

PHP, 102 Bytes

input als url parameter in this format ?0=[1,2,3]&1=[3,3]&2=[6] use this input [[int,int],[int]]

<?$r=[$_GET[0]];for(;count(end($r))>1;)$r[]=array_map(array_sum,array_chunk(end($r),2));echo$r==$_GET;

Breakdown

$r=[$_GET[0]]; # Set the first item of the input in the result array 
for(;count(end($r))>1;) # till the last item in the result array has only one int
$r[]=array_map(array_sum,array_chunk(end($r),2));# add the next item to the result array
echo$r==$_GET; # compare the input array with the result array

Japt, 10 bytes

Takes input as a 2-D array.

äÏeXò mxÃe

Try it

äÏeXò mxÃe     :Implicit input of 2-D array
ä              :Reduce each consecutive pair of sub-arrays
 Ï             :By passing them through the following function as X & Y, respectively
  e            :  Test Y for equality with
   Xò          :    Split X on every 2nd element
      m        :    Map
       x       :      Reduce by addition
        Ã      :End function
         e     :All true?