###Explanation:

Explanation:

R, 62 bytes

function(a,l)diffinv(matrix(a,max(a)*l+1,l))[cbind(a+2,1:l)]-a

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Explanation to follow.

R, 62 58 bytes

function(a,l)diag(diffinv(matrix(a,max(a)*l+1,l))[a+2,])-a

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###Explanation:

Given input array a, let M=max(a) and l=length(a).

Observe that M+l is the maximum possible index we'd need to access, and that M+l<=M*l+1, since if M,l>1,M+l<=M*l (with equality only when M=l=2) and if l==1 or M==1, then M+l==M*l+1.

By way of example, let a=c(4,3,2,1). Then M=l=4.

We construct the M*l+1 x l matrix in R by matrix(a,max(a)*l+1,l). Because R recycles a in column-major order, we end up with a matrix repeating the elements of a as such:

      [,1] [,2] [,3] [,4]
 [1,]    4    3    2    1
 [2,]    3    2    1    4
 [3,]    2    1    4    3
 [4,]    1    4    3    2
 [5,]    4    3    2    1
 [6,]    3    2    1    4
 [7,]    2    1    4    3
 [8,]    1    4    3    2
 [9,]    4    3    2    1
[10,]    3    2    1    4
[11,]    2    1    4    3
[12,]    1    4    3    2
[13,]    4    3    2    1
[14,]    3    2    1    4
[15,]    2    1    4    3
[16,]    1    4    3    2
[17,]    4    3    2    1

Each column is the cyclic successors of each element of a, with a across the first row; this is due to the way that R recycles its arguments in a matrix.

Next, we take the inverse "derivative" with diffinv, essentially the cumulative sum of each column with an additional 0 as the first row, generating the matrix

      [,1] [,2] [,3] [,4]
 [1,]    0    0    0    0
 [2,]    4    3    2    1
 [3,]    7    5    3    5
 [4,]    9    6    7    8
 [5,]   10   10   10   10
 [6,]   14   13   12   11
 [7,]   17   15   13   15
 [8,]   19   16   17   18
 [9,]   20   20   20   20
[10,]   24   23   22   21
[11,]   27   25   23   25
[12,]   29   26   27   28
[13,]   30   30   30   30
[14,]   34   33   32   31
[15,]   37   35   33   35
[16,]   39   36   37   38
[17,]   40   40   40   40
[18,]   44   43   42   41

In the first column, entry 6=4+2 is equal to 14=4 + (3+2+1+4), which is the cyclic successor sum (CSS) plus a leading 4. Similarly, in the second column, entry 5=3+2 is equal to 10=3 + (4+1+2), and so forth.

So in column i, the a[i]+2nd entry is equal to CSS(i)+a[i]. Hence, we take rows indexed by a+2, yielding a square matrix:

     [,1] [,2] [,3] [,4]
[1,]   14   13   12   11
[2,]   10   10   10   10
[3,]    9    6    7    8
[4,]    7    5    3    5

The entries along the diagonal are equal to the cyclic successor sums plus a, so we extract the diagonal and subtract a, returning the result as the cyclic successor sums.

R, 62 bytes

function(a,l)diffinv(matrix(a,max(a)*l+1,l))[cbind(a+2,1:l)]-a

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An alternative to the other R solution. In the comments JayCe made a mention of cumsum which triggered something in my brain to use diffinv and matrix recycling instead.

R, 62 bytes

function(a,l)diffinv(matrix(a,max(a)*l+1,l))[cbind(a+2,1:l)]-a

Try it online!

An alternative to the other R solution. In the comments JayCe made a mention of cumsum which triggered something in my brain to use diffinv and matrix recycling instead of rep.

Explanation to follow.