a b a a c a b
1   2 3   4    occurrence of a
  1         2  occurrence of b
        1      occurrence of c
1+1+2+3+1+4+2 = 14

input -> output
"abaacab" -> 14
"Programming Puzzles & Code Golf" -> 47
"" -> 0
"       " -> 28
"abcdefg" -> 7
"aA" -> 2

&=Rz

&=   % Implicit input. Matrix of all equality comparisons
     % STACK: [1 0 0 1;
               0 1 1 0;
               0 1 1 0;
               1 0 0 1]
R    % Upper triangular part
     % STACK: [1 0 0 1;
               0 1 1 0;
               0 0 1 0;
               0 0 0 1]
z    % Number of nonzeros. Implicitly display
     % STACK: 6

declare @ char(876),@x int,@v int=0Select @=q+CHAR(9)from z X:seleCT @x=len(@),@=REPLACE(@,LEFT(@,1),''),@v+=(@x-LEN(@))*(@x-LEN(@)+1)/2IF LEN(@)>0GOTO X prINT @v-1

declare @ char(876), @x int, @v int=0
Select @=q+CHAR(9)from z
X:
    seleCT @x=len(@)
          ,@=REPLACE(@,LEFT(@,1),'')
          ,@v+=(@x-LEN(@))*(@x-LEN(@)+1)/2
IF LEN(@)>0 GOTO X
prINT @v-1

s=>s.replace(d=/./g,z=>q+=d[z]=-~d[z],q=0)&&q

f=([c,...s],d={})=>c?(d[c]=-~d[c])+f(s,d):0

s(O`.
M&!`^|(?<=(.))1*
.

s(O`.

M&!s`^|(?<=(.))1*

.

z;g(char*s){int y[128]={z=0};while(*s)z+=--y[*s++];z=-z;}

f(a:b)=1+sum[1|c<-b,c==a]+f b;f _=0

;ċ"⁸S

+F/V._

+F/V._
+F/V._QQ  implicit input
  /V      vectorize count: for each element in the first argument,
                           count the number of occurrences of the
                           second argument:
    ._Q       all prefixes of input
       Q      input
+F        fold (reduce) on +, base case 0.

1#.,@(*+/"1)&=

1#.,@(*+/"1)&=
              =  Self-classify: bit matrix of equality between input
                 and its unique elements.
     (      )&   Apply verb in parentheses to it:
       +/         running sums
          "1       of each row
      *            multiplied with original matrix.
                 This causes the i'th 1 on each row to be replaced by i.
   ,@            Flatten the resulting matrix
1#.              and interpret as a base-1 number, computing its sum.

ĠJ€ẎS

{γ€gLO

Tr[#(#+1)/2&/@Last/@Tally@Characters[#]]&

s`(.)(?<=((1)|.)+)
$#3$*
1

(loop with w =(fill(make-list 128)0)as z across(read)sum(incf(elt w(char-code z))))

(loop with w = (fill (make-list 128) 0)  ; create a list to count characters
   as z across (read)                   ; for each character of input
   sum (incf (elt w (char-code z))))     ; increase count in list and sum

#(apply +(for[[k j](frequencies %)](*(inc j)j 0.5)))

;╗╔⌠╜cRΣ⌡MΣ

;╗╔⌠╜cRΣ⌡MΣ  (implicit input: S)
;╗           save a copy of S in register 0
  ╔          uniquify S (call it A)
   ⌠╜cRΣ⌡M   for each unique character in A:
    ╜c         count the number of occurrences in S
      R        range(1, count+1)
       Σ       sum
          Σ  sum

$-:@+[:+/,/@(=/~)

let x(y:bool)=System.Convert.ToInt32(y)
let rec p(k:string)q=
 let j=k.Length
 if(j=1)then(x(k.[0]=q))else p k.[0..(j-2)] q+x(k.[j-1]=q)
let rec d(a:string)=
 let z=a.Length
 if(z<2)then z else d a.[0..(z-2)]+p a (a.[z-1])

let x (y:bool)=
    System.Convert.ToInt32(y)
let rec e (c:string) b=
    let j=c.Length
    if(j=1)then
        (x(c.[0]=b))
    else
        e c.[0..(j-2)] b+x (c.[j-1]=b)
let rec d (a:string)=
    let h=a.Length
    if(h<2)then
        h
    else
        d a.[0..(h-2)]+e a (a.[h-1])

$x=$_;$b+=++$a{$_}for(split//,$x);$_=$b

$z=@{}
$ARGS|% getE*|%{$u+=($Z.$_+=1)};$U

PS C:> D:unique-is-cheap.ps1 (gc D:unique-is-cheap.ps1 -raw)
64