Trabb Pardo–Knuth algorithm: Difference between revisions

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2.0 : 41.41421356237309
2.0 : 41.41421356237309
1.0 : 6.0</pre>
1.0 : 6.0</pre>

=={{header|ASIC}}==
{{works with|ASIC|5.0}} Compile with the option ''Decimal Math'' (<code>DEC</code> in command line).
Notes:
* ASIC does not have the fuction that calculates a square root. Thus, the program uses the added subprogram <code>CALCSQRT:</code>.
* ASIC does not scroll a text screen, but it wraps the text. Thus, the program clears the screen after the user enters numbers.
<lang>
REM Trabb Pardo-Knuth algorithm
REM Used "magic numbers" because of strict specification of the algorithm.
DIM S@(10)
PRINT "Enter 11 numbers."
FOR I = 0 TO 10
I1= I + 1
PRINT I1;
PRINT " => ";
INPUT TMP@
S@(I) = TMP@
NEXT I
PRINT
REM Reverse
HALFIMAX = 10 / 2
FOR I = 0 TO HALFIMAX
IREV = 10 - I
TMP@ = S@(I)
S@(I) = S@(IREV)
S@(IREV) = TMP@
NEXT I
REM Results
REM Leading spaces in printed numbers are removed
CLS
FOR I = 0 TO 10
PRINT "f(";
STMP$ = STR$(S@(I))
STMP$ = LTRIM$(STMP$)
PRINT STMP$;
PRINT ") = ";
N@ = S@(I)
GOSUB CALCF:
R@ = F@
IF R@ > 400 THEN
PRINT "overflow"
ELSE
STMP$ = STR$(R@)
STMP$ = LTRIM$(STMP$)
PRINT STMP$
ENDIF
NEXT I
END

CALCF:
REM Calculates f(N@)
REM Result in F@
X@ = ABS(N@)
GOSUB CALCSQRT:
F@ = 5.0 * N@
F@ = F@ * N@
F@ = F@ * N@
F@ = F@ + SQRT@
RETURN

CALCSQRT:
REM Calculates approximate (+- 0.00001) square root of X@ for X@ >= 0 (bisection method)
REM Result in SQRT@
A@ = 0.0
IF X@ >= 1.0 THEN
B@ = X@
ELSE
B@ = 1.0
ENDIF
L@ = B@ - A@
L@ = ABS(L@)
WHILE L@ > 0.00001
MIDDLE@ = A@ + B@
MIDDLE@ = MIDDLE@ / 2
MIDDLETO2@ = MIDDLE@ * MIDDLE@
IF MIDDLETO2@ < X@ THEN
A@ = MIDDLE@
ELSE
B@ = MIDDLE@
ENDIF
L@ = B@ - A@
L@ = ABS(L@)
WEND
SQRT@ = MIDDLE@
RETURN
</lang>
{{out}}
Enter the data.
<pre>
Enter 11 numbers.
1 => ?-5
2 => ?-3
3 => ?-2
4 => ?-1
5 => ?0
6 => ?1
7 => ?2
8 => ?3
9 => ?4
10 => ?5
11 => ?6
</pre>
Results.
<pre>
f(6.00000) = overflow
f(5.00000) = overflow
f(4.00000) = 321.99999
f(3.00000) = 136.73206
f(2.00000) = 41.41422
f(1.00000) = 5.99999
f(0.00000) = 0.00001
f(-1.00000) = -4.00001
f(-2.00000) = -38.58578
f(-3.00000) = -133.26794
f(-5.00000) = -622.76393
</pre>


=={{header|AutoIt}}==
=={{header|AutoIt}}==

Revision as of 00:35, 18 January 2022

Task
Trabb Pardo–Knuth algorithm
You are encouraged to solve this task according to the task description, using any language you may know.

The TPK algorithm is an early example of a programming chrestomathy. It was used in Donald Knuth and Luis Trabb Pardo's Stanford tech report The Early Development of Programming Languages. The report traces the early history of work in developing computer languages in the 1940s and 1950s, giving several translations of the algorithm.

From the wikipedia entry: 

ask for 11 numbers to be read into a sequence S
reverse sequence S
for each item in sequence S
    result := call a function to do an operation
    if result overflows
        alert user
    else
        print result

The task is to implement the algorithm:

  1. Use the function:    
  2. The overflow condition is an answer of greater than 400.
  3. The 'user alert' should not stop processing of other items of the sequence.
  4. Print a prompt before accepting eleven, textual, numeric inputs.
  5. You may optionally print the item as well as its associated result, but the results must be in reverse order of input.
  6. The sequence   S   may be 'implied' and so not shown explicitly.
  7. Print and show the program in action from a typical run here. (If the output is graphical rather than text then either add a screendump or describe textually what is displayed).



11l

Translation of: Python

<lang 11l>F f(x) 

  R sqrt(abs(x)) + 5 * x ^ 3

V s = Array(1..11) s.reverse() L(x) s 

  V result = f(x)
  I result > 400
     print(‘#.: #.’.format(x, ‘TOO LARGE!’))
  E
     print(‘#.: #.’.format(x, result))

print()</lang>

Output:
11: TOO LARGE!
10: TOO LARGE!
9: TOO LARGE!
8: TOO LARGE!
7: TOO LARGE!
6: TOO LARGE!
5: TOO LARGE!
4: 322
3: 136.732050808
2: 41.414213562
1: 6

Ada

<lang Ada>with Ada.Text_IO, Ada.Numerics.Generic_Elementary_Functions;

procedure Trabb_Pardo_Knuth is 

  type Real is digits 6 range -400.0 .. 400.0;

  package TIO renames Ada.Text_IO;
  package FIO is new TIO.Float_IO(Real);
  package Math is new  Ada.Numerics.Generic_Elementary_Functions(Real);

  function F(X: Real) return Real is
  begin
     return (Math.Sqrt(abs(X)) + 5.0 * X**3);
  end F;

  Values: array(1 .. 11) of Real;

begin 

  TIO.Put("Please enter 11 Numbers:");
  for I in Values'Range loop
     FIO.Get(Values(I));
  end loop;

  for I in reverse Values'Range loop
     TIO.Put("f(");
     FIO.Put(Values(I), Fore => 2, Aft => 3, Exp => 0);
     TIO.Put(")=");
     begin
        FIO.Put(F(Values(I)), Fore=> 4, Aft => 3, Exp => 0);
     exception
        when Constraint_Error => TIO.Put("-->too large<--");
     end;
     TIO.New_Line;
  end loop;

end Trabb_Pardo_Knuth;</lang>

Output:
> ./trabb_pardo_knuth 
Please enter 11 Numbers:10 -1 1 2 3 4 4.3 4.305 4.303 4.302 4.301
f( 4.301)= 399.886
f( 4.302)=-->too large<--
f( 4.303)=-->too large<--
f( 4.305)=-->too large<--
f( 4.300)= 399.609
f( 4.000)= 322.000
f( 3.000)= 136.732
f( 2.000)=  41.414
f( 1.000)=   6.000
f(-1.000)=  -4.000
f(10.000)=-->too large<--

Agena

Tested with Agena 2.9.5 Win32

Translation of: ALGOL W

<lang agena>scope # TPK algorithm in Agena 

 local y;
 local a := [];
 local f := proc( t :: number ) is return sqrt(abs(t))+5*t*t*t end;
 for i from 0 to 10 do a[i] := tonumber( io.read() ) od;
 for i from 10 to 0 by - 1 do
     y:=f(a[i]);
     if y > 400
     then print( "TOO LARGE" )
     else printf( "%10.4f\n", y )
     fi
 od

epocs</lang>

Output:
1
2
3
4
5
6
7
8
9
10
11
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
  322.0000
  136.7321
   41.4142
    6.0000

ALGOL 60

This is as close as possible to Pardo and Knuth's original but works with the GNU MARST ALGOL-to-C compiler. Note Pardo and Knuth did not insist on prompts or textual I/O as their report mostly concerned systems that predated even the idea of keyboard interaction.

<lang>begin 

 integer i; real y; real array a[0:10];
 real procedure f(t); value t; real t;
   f:=sqrt(abs(t))+5*t^3;
 for i:=0 step 1 until 10 do inreal(0, a[i]);
 for i:=10 step -1 until 0 do
   begin
      y:=f(a[i]);
      if y > 400 then outstring(1, "TOO LARGE")
                 else outreal(1,y);
      outchar(1, "\n", 1)
   end

end</lang>

Compilation and sample run: 

bash-3.2$ marst tpk.a60 -o tpk.c
bash-3.2$ gcc tpk.c -lalgol -lm -o tpk
bash-3.2$ ./tpk
1 2 3 4 5 6 7 8 9 10 11
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
322 
136.732050808 
41.4142135624 
6 
bash-3.2$

ALGOL 68

Translation of: ALGOL W

which was itself a Translation of ALGOL 60.

<lang algol68>[ 0 : 10 ]REAL a; PROC f = ( REAL t )REAL: 

   sqrt(ABS t)+5*t*t*t;

FOR i FROM LWB a TO UPB a DO read( ( a[ i ] ) ) OD; FOR i FROM UPB a BY -1 TO LWB a DO 

      REAL y=f(a[i]);
      IF y > 400 THEN print( ( "TOO LARGE", newline ) )
                 ELSE print( ( fixed( y, -9, 4 ), newline ) )
      FI

OD</lang>

Output:
1 2 3 4 5 6 7 8 9 10 11
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
 322.0000
 136.7321
  41.4142
   6.0000

ALGOL W

Translation of: ALGOL 60

<lang algolw>begin 

 real y; real array a( 0 :: 10 );
 real procedure f( real value t ); 
   sqrt(abs(t))+5*t*t*t;
 for i:=0 until 10 do read( a(i) );
 r_format := "A"; r_w := 9; r_d := 4;
 for i:=10 step -1 until 0 do
   begin
      y:=f(a(i));
      if y > 400 then write( "TOO LARGE" )
                 else write( y );
   end

end.</lang>

Output:
1 2 3 4 5 6 7 8 9 10 11
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
 322.0000
 136.7320
  41.4142
   6.0000

APL

Works with: Dyalog APL

<lang APL>∇ {res}←Trabb;f;S;i;a;y ⍝ define a function Trabb 

 f←{(0.5*⍨|⍵)+5×⍵*3}                ⍝ define a function f
 S←,⍎{⍞←⍵ ⋄ (≢⍵)↓⍞}'Please, enter 11 numbers: '
 :For i a :InEach (⌽⍳≢S)(⌽S)        ⍝ loop through N..1 and reversed S
     :If 400<y←f(a)
         ⎕←'Too large: ',⍕i
     :Else
         ⎕←i,y
     :EndIf
 :EndFor

∇</lang>

Arturo

<lang rebol>proc: function [x]-> 

   ((abs x) ^ 0.5) + 5 * x ^ 3

ask: function [msg][ 

   to [:floating] first.n: 11 split.words strip input msg

]

loop reverse ask "11 numbers: " 'n [ 

   result: proc n
   print [n ":" (result > 400)? -> "TOO LARGE!" -> result]

]</lang>

Output:
11 numbers: 1 2 3 4 5 6 7 8 9 10 11
11.0 : TOO LARGE! 
10.0 : TOO LARGE! 
9.0 : TOO LARGE! 
8.0 : TOO LARGE! 
7.0 : TOO LARGE! 
6.0 : TOO LARGE! 
5.0 : TOO LARGE! 
4.0 : 322.0 
3.0 : 136.7320508075689 
2.0 : 41.41421356237309 
1.0 : 6.0

ASIC

Works with: ASIC version 5.0

Compile with the option Decimal Math (DEC in command line).

Notes:

  • ASIC does not have the fuction that calculates a square root. Thus, the program uses the added subprogram CALCSQRT:.
  • ASIC does not scroll a text screen, but it wraps the text. Thus, the program clears the screen after the user enters numbers.

<lang> REM Trabb Pardo-Knuth algorithm REM Used "magic numbers" because of strict specification of the algorithm. DIM S@(10) PRINT "Enter 11 numbers." FOR I = 0 TO 10 

 I1= I + 1
 PRINT I1;
 PRINT " => ";
 INPUT TMP@
 S@(I) = TMP@

NEXT I PRINT REM Reverse HALFIMAX = 10 / 2 FOR I = 0 TO HALFIMAX 

 IREV = 10 - I
 TMP@ = S@(I)
 S@(I) = S@(IREV)
 S@(IREV) = TMP@

NEXT I REM Results REM Leading spaces in printed numbers are removed CLS FOR I = 0 TO 10 

 PRINT "f(";
 STMP$ = STR$(S@(I))
 STMP$ = LTRIM$(STMP$)
 PRINT STMP$;
 PRINT ") = ";
 N@ = S@(I)
 GOSUB CALCF:
 R@ = F@
 IF R@ > 400 THEN
   PRINT "overflow"
 ELSE
   STMP$ = STR$(R@)
   STMP$ = LTRIM$(STMP$)
   PRINT STMP$
 ENDIF

NEXT I END

CALCF: REM Calculates f(N@) REM Result in F@ X@ = ABS(N@) GOSUB CALCSQRT: F@ = 5.0 * N@ F@ = F@ * N@ F@ = F@ * N@ F@ = F@ + SQRT@ RETURN

CALCSQRT: REM Calculates approximate (+- 0.00001) square root of X@ for X@ >= 0 (bisection method) REM Result in SQRT@ A@ = 0.0 IF X@ >= 1.0 THEN 

 B@ = X@

ELSE 

 B@ = 1.0

ENDIF L@ = B@ - A@ L@ = ABS(L@) WHILE L@ > 0.00001 

 MIDDLE@ = A@ + B@
 MIDDLE@ = MIDDLE@ / 2
 MIDDLETO2@ = MIDDLE@ * MIDDLE@
 IF MIDDLETO2@ < X@ THEN
   A@ = MIDDLE@
 ELSE
   B@ = MIDDLE@
 ENDIF
 L@ = B@ - A@
 L@ = ABS(L@)

WEND SQRT@ = MIDDLE@ RETURN </lang>

Output:

Enter the data. 

Enter 11 numbers.
     1 => ?-5
     2 => ?-3
     3 => ?-2
     4 => ?-1
     5 => ?0
     6 => ?1
     7 => ?2
     8 => ?3
     9 => ?4
    10 => ?5
    11 => ?6

Results. 

f(6.00000) = overflow
f(5.00000) = overflow
f(4.00000) = 321.99999
f(3.00000) = 136.73206
f(2.00000) = 41.41422
f(1.00000) = 5.99999
f(0.00000) = 0.00001
f(-1.00000) = -4.00001
f(-2.00000) = -38.58578
f(-3.00000) = -133.26794
f(-5.00000) = -622.76393

AutoIt

<lang AutoIt>; Trabb Pardo–Knuth algorithm

by James1337 (autoit.de)
AutoIt Version
3.3.8.1

Local $S, $i, $y

Do $S = InputBox("Trabb Pardo–Knuth algorithm", "Please enter 11 numbers:", "1 2 3 4 5 6 7 8 9 10 11") If @error Then Exit $S = StringSplit($S, " ") Until ($S[0] = 11)

For $i = 11 To 1 Step -1 $y = f($S[$i]) If ($y > 400) Then ConsoleWrite("f(" & $S[$i] & ") = Overflow!" & @CRLF) Else ConsoleWrite("f(" & $S[$i] & ") = " & $y & @CRLF) EndIf Next

Func f($x) Return Sqrt(Abs($x)) + 5*$x^3 EndFunc</lang>

Output:
Input: "1 2 3 4 5 6 7 8 9 10 11"

f(11) = Overflow!
f(10) = Overflow!
f(9) = Overflow!
f(8) = Overflow!
f(7) = Overflow!
f(6) = Overflow!
f(5) = Overflow!
f(4) = 322
f(3) = 136.732050807569
f(2) = 41.4142135623731
f(1) = 6

AWK

<lang AWK>

  1. syntax: GAWK -f TRABB_PARDO-KNUTH_ALGORITHM.AWK

BEGIN { 

   printf("enter 11 numbers: ")
   getline S
   n = split(S,arr," ")
   if (n != 11) {
     printf("%d numbers entered; S/B 11\n",n)
     exit(1)
   }
   for (i=n; i>0; i--) {
     x = f(arr[i])
     printf("f(%s) = %s\n",arr[i],(x>400) ? "too large" : x)
   }
   exit(0)

} function abs(x) { if (x >= 0) { return x } else { return -x } } function f(x) { return sqrt(abs(x)) + 5 * x ^ 3 } </lang>

Output:
enter 11 numbers: 1 2 3 -4 5 6 -7 8 9 10 11
f(11) = too large
f(10) = too large
f(9) = too large
f(8) = too large
f(-7) = -1712.35
f(6) = too large
f(5) = too large
f(-4) = -318
f(3) = 136.732
f(2) = 41.4142
f(1) = 6

BASIC256

<lang BASIC256>dim s(11) print 'enter 11 numbers' for i = 0 to 10 

  input i + ">" , s[i]

next i

for i = 10 to 0 step -1 

  print "f(" + s[i] + ")=";
  x = f(s[i])
  if x > 400 then
     print "--- too large ---"
  else
     print x
  endif

next i end

function f(n) 

  return sqrt(abs(n))+5*n^3

end function</lang>

Output:
enter 11 numbers
0>-4
1>-3
2>-4
3>-2
4>-1
5>-
6>1
7>2
8>3
9>4
10>5
f(5)=--- too large ---
f(4)=322
f(3)=136.7320508
f(2)=41.4142136
f(1)=6
f(0)=0
f(-1)=-4
f(-2)=-38.5857864
f(-4)=-318
f(-3)=-133.2679492
f(-4)=-318

C

<lang c>

  1. include<math.h>
  2. include<stdio.h>

int main () { 

 double inputs[11], check = 400, result;
 int i;

 printf ("\nPlease enter 11 numbers :");

 for (i = 0; i < 11; i++)
   {
     scanf ("%lf", &inputs[i]);
   }

 printf ("\n\n\nEvaluating f(x) = |x|^0.5 + 5x^3 for the given inputs :");

 for (i = 10; i >= 0; i--)
   {
     result = sqrt (fabs (inputs[i])) + 5 * pow (inputs[i], 3);

     printf ("\nf(%lf) = ");

     if (result > check)
       {
         printf ("Overflow!");
       }

     else
       {
         printf ("%lf", result);
       }
   }

 return 0;

} </lang>

Output:
Please enter 11 numbers :10 -1 1 2 3 4 4.3 4.305 4.303 4.302 4.301



Evaluating f(x) = |x|^0.5 + 5x^3 for the given inputs :
f(3.000000) = 399.886300
f(3.000000) = Overflow!
f(3.000000) = Overflow!
f(3.000000) = Overflow!
f(3.000000) = 399.608644
f(3.000000) = 322.000000
f(3.000000) = 136.732051
f(3.000000) = 41.414214
f(3.000000) = 6.000000
f(6.000000) = -4.000000
f(3.000000) = Overflow!

C++

<lang cpp>

  1. include <iostream>
  2. include <cmath>
  3. include <vector>
  4. include <algorithm>
  5. include <iomanip>

int main( ) { 

  std::vector<double> input( 11 ) , results( 11 ) ;
  std::cout << "Please enter 11 numbers!\n" ;
  for ( int i = 0 ; i < input.size( ) ; i++ ) 
     std::cin >> input[i];
     
  std::transform( input.begin( ) , input.end( ) , results.begin( ) ,

[ ]( double n )-> double { return sqrt( abs( n ) ) + 5 * pow( n , 3 ) ; } ) ; 

  for ( int i = 10 ; i > -1 ; i-- ) {
     std::cout << "f( " << std::setw( 3 ) << input[ i ] << " ) : " ; 
     if ( results[ i ] > 400 )

std::cout << "too large!" ; 

     else

std::cout << results[ i ] << " !" ; 

     std::cout << std::endl ;
  }
  return 0 ;

}</lang>

Output:
Please enter 11 numbers!
1
2
3
4
5
6
7
8
9
10
11
f(  11 ) : too large!
f(  10 ) : too large!
f(   9 ) : too large!
f(   8 ) : too large!
f(   7 ) : too large!
f(   6 ) : too large!
f(   5 ) : too large!
f(   4 ) : 322 !
f(   3 ) : 136.732 !
f(   2 ) : 41.4142 !
f(   1 ) : 6 !

Commodore BASIC

Translation of: XBasic
Works with: Commodore BASIC version 4.5

<lang basic> 10 REM TRABB PARDO-KNUTH ALGORITHM 20 REM USED "MAGIC NUMBERS" BECAUSE OF STRICT SPECIFICATION OF THE ALGORITHM. 30 DEF FNF(N)=SQR(ABS(N))+5*N*N*N 40 DIM S(10) 50 PRINT "ENTER 11 NUMBERS." 60 FOR I=0 TO 10 70 PRINT STR$(I+1); 80 INPUT S(I) 90 NEXT I 100 PRINT 110 REM REVERSE 120 FOR I=0 TO 10/2 130 TMP=S(I) 140 S(I)=S(10-I) 150 S(10-I)=TMP 160 NEXT I 170 REM RESULTS 180 FOR I=0 TO 10 190 PRINT "F(";STR$(S(I));") ="; 200 R=FNF(S(I)) 210 IF R>400 THEN PRINT " OVERFLOW":ELSE PRINT R 220 NEXT I 230 END </lang>

Output:
ENTER 11 NUMBERS. 
 1? -5
 2? -3
 3? -2
 4? -1
 5? 0
 6? 1
 7? 2
 8? 3
 9? 4
 10? 5
 11? 6

F( 6) = OVERFLOW
F( 5) = OVERFLOW
F( 4) = 322
F( 3) = 136.732051
F( 2) = 41.4142136
F( 1) = 6
F( 0) = 0
F(-1) =-4
F(-2) =-38.5857864
F(-3) =-133.267949
F(-5) =-622.763932

Common Lisp

<lang lisp>(defun read-numbers () 

 (princ "Enter 11 numbers (space-separated): ")
 (let ((numbers '()))
   (dotimes (i 11 numbers)
     (push (read) numbers))))

(defun trabb-pardo-knuth (func overflowp) 

 (let ((S (read-numbers)))
   (format T "~{~a~%~}"
           (substitute-if "Overflow!" overflowp (mapcar func S)))))

(trabb-pardo-knuth (lambda (x) (+ (expt (abs x) 0.5) (* 5 (expt x 3)))) 

                  (lambda (x) (> x 400)))</lang>
Output:
Enter 11 numbers (space-separated): 10 -1 1 2 3 4 4.3 4.305 4.303 4.302 4.301
399.88635
Overflow!
Overflow!
Overflow!
399.6087
322.0
136.73206
41.414215
6.0
-4.0
Overflow!

D

<lang d>import std.stdio, std.math, std.conv, std.algorithm, std.array;

double f(in double x) pure nothrow { 

   return x.abs.sqrt + 5 * x ^^ 3;

}

void main() { 

   double[] data;

   while (true) {
       "Please enter eleven numbers on a line: ".write;
       data = readln.split.map!(to!double).array;
       if (data.length == 11)
           break;
       writeln("Those aren't eleven numbers.");
   }
   foreach_reverse (immutable x; data) {
       immutable y = x.f;
       writefln("f(%0.3f) = %s", x, y > 400 ? "Too large" : y.text);
   }

}</lang>

Output:
Please enter eleven numbers on a line: 1 2 3 -4.55 5.1111 6 -7 8 9 10
Those aren't eleven numbers.
Please enter eleven numbers on a line: 1 2 3 -4.55 5.1111 6 -7 8 9 10 11
f(11.000) = Too large
f(10.000) = Too large
f(9.000) = Too large
f(8.000) = Too large
f(-7.000) = -1712.35
f(6.000) = Too large
f(5.111) = Too large
f(-4.550) = -468.849
f(3.000) = 136.732
f(2.000) = 41.4142
f(1.000) = 6

EchoLisp

<lang scheme> (define (trabb-fun n) (+ (* 5 n n n) (sqrt(abs n))))

(define (check-trabb n) (if (number? n) (if (<= (trabb-fun n) 400) (printf "🌱 f(%d) = %d" n (trabb-fun n)) (printf "❌ f(%d) = %d" n (trabb-fun n))) (error "not a number" n)))

(define (trabb (numlist null)) (while (< (length numlist) 11) (set! numlist (append numlist (or (read default: (shuffle (iota 11)) prompt: (format "Please enter %d more numbers" (- 11 (length numlist)))) (error 'incomplete-list numlist))))) ;; users cancel (for-each check-trabb (reverse (take numlist 11)))) </lang>

Output:

<lang scheme> (trabb)

input
(0 4 1 8 5 9 10 3 6 7 2)

🌱 f(2) = 41.41421356237309 ❌ f(7) = 1717.6457513110645 ❌ f(6) = 1082.4494897427833 🌱 f(3) = 136.73205080756887 ❌ f(10) = 5003.162277660168 ❌ f(9) = 3648 ❌ f(5) = 627.2360679774998 ❌ f(8) = 2562.828427124746 🌱 f(1) = 6 🌱 f(4) = 322 🌱 f(0) = 0

extra credit
let's find the threshold

(lib 'math) (define (g x) (- (trabb-fun x) 400)) (root g 0 10) 

   → 4.301409367213084

</lang>

Ela

Translation of OCaml version:

<lang ela>open monad io number string

IO

take_numbers 0 xs = do 

 return $ iter xs
 where f x = sqrt (toSingle x) + 5.0 * (x ** 3.0)
       p x = x < 400.0
       iter [] = return ()
       iter (x::xs) 
         | p res = do
             putStrLn (format "f({0}) = {1}" x res)
             iter xs
         | else = do
             putStrLn (format "f({0}) :: Overflow" x)
             iter xs
         where res = f x

take_numbers n xs = do 

 x <- readAny
 take_numbers (n - 1) (x::xs)

do 

 putStrLn "Please enter 11 numbers:"
 take_numbers 11 []</lang>
Output:
Please enter 11 numbers:
1
2
3
4
5
6
7
8
9
10
11
f(11) :: Overflow
f(10) :: Overflow
f(9) :: Overflow
f(8) :: Overflow
f(7) :: Overflow
f(6) :: Overflow
f(5) :: Overflow
f(4) = 322
f(3) = 136.732050807569
f(2) = 41.4142135623731
f(1) = 6

Elena

Translation of: C

ELENA 5.0 : <lang elena>import extensions; import extensions'math;

public program() { 

   real[] inputs := new real[](11);
   console.printLine("Please enter 11 numbers :");
   for(int i := 0, i < 11, i += 1)
   {
       inputs[i] := console.readLine().toReal()
   };

   console.printLine("Evaluating f(x) = |x|^0.5 + 5x^3 for the given inputs :");
   for(int i := 10, i >= 0, i -= 1)
   {
       real result := sqrt(abs(inputs[i])) + 5 * power(inputs[i], 3);

       console.print("f(", inputs[i], ")=");

       if (result > 400)
       {
           console.printLine("Overflow!")
       }
       else
       {
           console.printLine(result)
       }
   }

}</lang>

Output:
Please enter 11 numbers :
1
2
3
4
5
6
7
8
9
10
11	
Evaluating f(x) = |x|^0.5 + 5x^3 for the given inputs :
f(11.0)=Overflow!
f(10.0)=Overflow!
f(9.0)=Overflow!
f(8.0)=Overflow!
f(7.0)=Overflow!
f(6.0)=Overflow!
f(5.0)=Overflow!
f(4.0)=322.0
f(3.0)=136.7320508076
f(2.0)=41.41421356237
f(1.0)=6.0

Elixir

Translation of: Erlang

<lang elixir>defmodule Trabb_Pardo_Knuth do 

 def task do
   Enum.reverse( get_11_numbers )
   |> Enum.each( fn x -> perform_operation( &function(&1), 400, x ) end )
 end
 
 defp alert( n ), do: IO.puts "Operation on #{n} overflowed"
 
 defp get_11_numbers do
   ns = IO.gets( "Input 11 integers.  Space delimited, please: " )
        |> String.split
        |> Enum.map( &String.to_integer &1 )
   if 11 == length( ns ), do: ns, else: get_11_numbers
 end
 
 defp function( x ), do: :math.sqrt( abs(x) ) + 5 * :math.pow( x, 3 )
 
 defp perform_operation( fun, overflow, n ), do: perform_operation_check_overflow( n, fun.(n), overflow )
 
 defp perform_operation_check_overflow( n, result, overflow ) when result > overflow, do: alert( n )
 defp perform_operation_check_overflow( n, result, _overflow ), do: IO.puts "f(#{n}) => #{result}"

end

Trabb_Pardo_Knuth.task</lang>

Output:
Input 11 integers.  Space delimited, please: 0 1 2 3 4 5 6 7 8 9 10
Operation on 10 overflowed
Operation on 9 overflowed
Operation on 8 overflowed
Operation on 7 overflowed
Operation on 6 overflowed
Operation on 5 overflowed
f(4) => 322.0
f(3) => 136.73205080756887
f(2) => 41.41421356237309
f(1) => 6.0
f(0) => 0.0

Erlang

<lang Erlang> -module( trabb_pardo_knuth ).

-export( [task/0] ).

task() -> Sequence = get_11_numbers(), S = lists:reverse( Sequence ), [perform_operation( fun function/1, 400, X) || X <- S].


alert( N ) -> io:fwrite( "Operation on ~p overflowed~n", [N] ).

get_11_numbers() -> {ok, Ns} = io:fread( "Input 11 integers. Space delimited, please: ", "~d ~d ~d ~d ~d ~d ~d ~d ~d ~d ~d" ), 11 = erlang:length( Ns ), Ns.

function( X ) -> math:sqrt( erlang:abs(X) ) + 5 * math:pow( X, 3 ).

perform_operation( Fun, Overflow, N ) -> perform_operation_check_overflow( N, Fun(N), Overflow ).

perform_operation_check_overflow( N, Result, Overflow ) when Result > Overflow -> alert( N ); perform_operation_check_overflow( N, Result, _Overflow ) -> io:fwrite( "f(~p) => ~p~n", [N, Result] ). </lang>

Output:
5> trabb_pardo_knuth:task().
Input 11 integers.  Space delimited, please:  1 2 3 4 5 6 7 8 9 10 11
Operation on 11 overflowed
Operation on 10 overflowed
Operation on 9 overflowed
Operation on 8 overflowed
Operation on 7 overflowed
Operation on 6 overflowed
Operation on 5 overflowed
f(4) => 322.0
f(3) => 136.73205080756887
f(2) => 41.41421356237309
f(1) => 6.0

ERRE

<lang ERRE> !Trabb Pardo-Knuth algorithm PROGRAM TPK !VAR I%,Y DIM A[10]

FUNCTION F(T) 

   F=SQR(ABS(T))+5*T^3

END FUNCTION

BEGIN 

 DATA(10,-1,1,2,3,4,4.3,4.305,4.303,4.302,4.301)
 FOR I%=0 TO 10 DO
     READ(A[I%])
 END FOR
 FOR I%=10 TO 0 STEP -1 DO
      Y=F(A[I%])
      PRINT("F(";A[I%];")=";)
      IF Y>400 THEN PRINT("--->too large<---")
               ELSE PRINT(Y)
      END IF
 END FOR

END PROGRAM </lang> Numbers to be elaborated is included in the program with a DATA statement. You can substitute this with an input keyboard like this 

   FOR I%=0 TO 10 DO
    PRINT("Number #";I%;)
    INPUT(A[I%])
   END FOR

F#

<lang fsharp> module ``Trabb Pardo - Knuth`` open System let f (x: float) = sqrt(abs x) + (5.0 * (x ** 3.0))

Console.WriteLine "Enter 11 numbers:" [for _ in 1..11 -> Convert.ToDouble(Console.ReadLine())] |> List.rev |> List.map f |> List.iter (function | n when n <= 400.0 -> Console.WriteLine(n) | _ -> Console.WriteLine("Overflow")) </lang>

Output:
fsharpi Program.fsx
[Loading Program.fsx]
Enter 11 numbers:
1
2
3
4
5
6
7
8
9
10
11
Overflow
Overflow
Overflow
Overflow
Overflow
Overflow
Overflow
322
136.732050807569
41.4142135623731
6

Factor

<lang factor>USING: formatting io kernel math math.functions math.parser prettyprint sequences splitting ; IN: rosetta-code.trabb-pardo-knuth

CONSTANT: threshold 400 CONSTANT: prompt "Please enter 11 numbers: "

fn ( x -- y ) [ abs 0.5 ^ ] [ 3 ^ 5 * ] bi + ;
overflow? ( x -- ? ) threshold > ;
get-input ( -- seq )
   prompt write flush readln " " split dup length 11 =
   [ drop get-input ] unless ;
?result ( ..a quot: ( ..a -- ..b ) -- ..b )
   [ "f(%u) = " sprintf ] swap bi dup overflow?
   [ drop "overflow" ] [ "%.3f" sprintf ] if append ; inline
main ( -- )
   get-input reverse
   [ string>number [ fn ] ?result print ] each ;

MAIN: main</lang>

Output:
Please enter 11 numbers: 1 2 3
Please enter 11 numbers: 10 -1 1 2 3 4 4.3 4.305 4.303 4.302 4.301
f(4.301) = 399.886
f(4.302) = overflow
f(4.303) = overflow
f(4.305) = overflow
f(4.3) = 399.609
f(4) = 322.000
f(3) = 136.732
f(2) = 41.414
f(1) = 6.000
f(-1) = -4.000
f(10) = overflow

Forth

<lang forth>: f(x) fdup fsqrt fswap 3e f** 5e f* f+ ;

4e2 fconstant f-too-big

11 Constant #Elements

float-array ( compile: n -- / run: n -- addr )
   create 
       floats allot
   does>
       swap floats + ;
  1. Elements float-array vec
get-it ( -- )
   ." Enter " #Elements . ." numbers:" cr
   #Elements 0 DO
       ." > " pad 25 accept cr
       pad swap >float 0= abort" Invalid Number"
       i vec F!
   LOOP ;
reverse-it ( -- )
   #Elements 2/  0 DO
       i vec F@  #Elements i - 1- vec F@
       i vec F!  #Elements i - 1- vec F!
   LOOP ;
do-it ( -- )
   #Elements 0 DO
       i vec F@ fdup f. [char] : emit space

f(x) fdup f-too-big f> IF 

           fdrop ." too large"
       ELSE
           f.
       THEN cr
   LOOP ;
tpk ( -- )
   get-it reverse-it do-it ;</lang>
Output:
Gforth 0.7.2, Copyright (C) 1995-2008 Free Software Foundation, Inc.
Gforth comes with ABSOLUTELY NO WARRANTY; for details type `license'
Type `bye' to exit
tpk Enter 11 numbers:
> 1 
> 2 
> 3 
> 4 
> 5 
> 6 
> 2.71828 
> 3.14159 
> 76 
> 7 
> 8 
8. : too large
7. : too large
76. : too large
3.14159 : 156.80344365595 
2.71828 : 102.07620267347 
6. : too large
5. : too large
4. : 322. 
3. : 136.732050807569 
2. : 41.4142135623731 
1. : 6. 
 ok

Fortran

Fortran 95

Works with: Fortran version 95 and later

<lang fortran>program tpk 

 implicit none
 
 real, parameter :: overflow = 400.0
 real :: a(11), res
 integer :: i

 write(*,*) "Input eleven numbers:"
 read(*,*) a

 a = a(11:1:-1)
 do i = 1, 11
   res = f(a(i))
   write(*, "(a, f0.3, a)", advance = "no") "f(", a(i), ") = "
   if(res > overflow) then
     write(*, "(a)") "overflow!"
   else
      write(*, "(f0.3)") res
   end if
 end do

contains

real function f(x) 

 real, intent(in) :: x
 
 f = sqrt(abs(x)) + 5.0*x**3

end function end program</lang>

Output:
 Input eleven numbers:
10 -1 1 2 3 4 4.3 4.305 4.303 4.302 4.301
f(4.301) = 399.886
f(4.302) = overflow!
f(4.303) = overflow!
f(4.305) = overflow!
f(4.300) = 399.609
f(4.000) = 322.000
f(3.000) = 136.732
f(2.000) = 41.414
f(1.000) = 6.000
f(-1.000) = -4.000
f(10.000) = overflow!

Fortran I

Written in FORTRAN I (1957), the original language quoted in the 1976 Donald Knuth & Luis Trabb Pardo’s study. Let’ note: no type declarations (INTEGER, REAL), no subprogram FUNCTION (only statement function), no logical IF, no END statement, and only Hollerith strings. The input data are on 2 80-column punched cards, only 1 to 72 columns are used so 6 values are read on the first card and 5 on the second card, so even input data could be numbered in the 73-80 area. <lang fortran>C THE TPK ALGORITH - FORTRAN I - 1957 TPK00010 

     FTPKF(X)=SQRTF(ABSF(X))+5.0*X**3                                  TPK00020
     DIMENSION A(11)                                                   TPK00030
     READ 100,A                                                        TPK00040
100  FORMAT(6F12.4/)                                                   TPK00050
     DO 3 I=1,11                                                       TPK00060
     J=12-I                                                            TPK00070           
     Y=FTPKF(A(J))                                                     TPK00080
     IF (Y-400.0)2,2,1                                                 TPK00090
  1  PRINT 301,I,A(J)                                                  TPK00100
301  FORMAT(I10,F12.7,18H *** TOO LARGE ***)                           TPK00110
     GO TO 10                                                          TPK00120
  2  PRINT 302,I,A(J),Y                                                TPK00130
302  FORMAT(I10,2F12.7)                                                TPK00140
  3  CONTINUE                                                          TPK00150
     STOP 0                                                            TPK00160

</lang>

FreeBASIC

<lang freebasic>' version 22-07-2017 ' compile with: fbc -s console

Function f(n As Double) As Double 

   return Sqr(Abs(n)) + 5 * n ^ 3

End Function

' ------=< MAIN >=------

Dim As Double x, s(1 To 11) Dim As Long i

For i = 1 To 11 

   Print Str(i);
   Input " => ", s(i)

Next

Print Print String(20,"-")

i -= 1 Do 

   Print "f(" + Str(s(i)) + ") = ";
  x = f(s(i))
  If x > 400 Then 
      Print "-=< overflow >=-"
  Else 
      Print x
  End If
  i -= 1

Loop Until i < 1

' empty keyboard buffer While InKey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>

Output:
1 => -5
2 => -3
3 => -2
4 => -1
5 => 0
6 => 1
7 => 2
8 => 3
9 => 4
10 => 5
11 => 6

--------------------
f(6) = -=< overflow >=-
f(5) = -=< overflow >=-
f(4) =  322
f(3) =  136.7320508075689
f(2) =  41.41421356237309
f(1) =  6
f(0) =  0
f(-1) = -4
f(-2) = -38.58578643762691
f(-3) = -133.2679491924311
f(-5) = -622.7639320225002

Go

Task/Wikipedia

This solution follows the task description by reversing the sequence. It also rejects non-numeric input until 11 numbers are entered. <lang go>package main

import ( 

   "fmt"
   "log"
   "math"

)

func main() { 

   // prompt
   fmt.Print("Enter 11 numbers: ")
   // accept sequence
   var s [11]float64
   for i := 0; i < 11; {
       if n, _ := fmt.Scan(&s[i]); n > 0 {
           i++
       }
   }
   // reverse sequence
   for i, item := range s[:5] {
       s[i], s[10-i] = s[10-i], item
   }
   // iterate
   for _, item := range s {
       if result, overflow := f(item); overflow {
           // send alerts to stderr
           log.Printf("f(%g) overflow", item)
       } else {
           // send normal results to stdout
           fmt.Printf("f(%g) = %g\n", item, result)
       }
   }

}

func f(x float64) (float64, bool) { 

   result := math.Sqrt(math.Abs(x)) + 5*x*x*x
   return result, result > 400

}</lang>

Output:

The input is chosen to show some interesting boundary cases. 

Enter 11 numbers: 0 1 4.3 4.4 -1 -5 non-number -1e102 -1e103 -Inf Inf NaN 
f(NaN) = NaN
2016/04/15 18:38:29 f(+Inf) overflow
f(-Inf) = NaN
f(-1e+103) = -Inf
f(-1e+102) = -5e+306
f(-5) = -622.7639320225002
f(-1) = -4
2016/04/15 18:38:29 f(4.4) overflow
f(4.3) = 399.6086441353327
f(1) = 6
f(0) = 0

TPK paper

The original paper had no requirement to reverse the sequence in place, but instead processed the sequence in reverse order. <lang go>package main

import ( 

   "fmt"
   "math"

)

func f(t float64) float64 { 

   return math.Sqrt(math.Abs(t)) + 5*math.Pow(t, 3)

}

func main() { 

   var a [11]float64
   for i := range a {
       fmt.Scan(&a[i])
   }
   for i := len(a) - 1; i >= 0; i-- {
       if y := f(a[i]); y > 400 {
           fmt.Println(i, "TOO LARGE")
       } else {
           fmt.Println(i, y)
       }
   }

}</lang>

Haskell

<lang Haskell>import Control.Monad (replicateM, mapM_)

f :: Floating a => a -> a f x = sqrt (abs x) + 5 * x ** 3

main :: IO () main = do 

 putStrLn "Enter 11 numbers for evaluation"
 x <- replicateM 11 readLn
 mapM_
   ((\x ->
        if x > 400
          then putStrLn "OVERFLOW"
          else print x) .
    f) $
   reverse x</lang>
Output:
Enter 11 numbers for evaluation
1
2
3
4
5
6
7
8
9
10
11
OVERFLOW
OVERFLOW
OVERFLOW
OVERFLOW
OVERFLOW
OVERFLOW
OVERFLOW
322.0
136.73205080756887
41.41421356237309
6.0

Icon and Unicon

The following Unicon-specific solution can be implemented in Icon by replaces reverse(S) with S[*S to 1 by -1].

<lang unicon>procedure main() 

   S := []
   writes("Enter 11 numbers: ")
   read() ? every !11 do (tab(many(' \t'))|0,put(S, tab(upto(' \t')|0)))
   every item := !reverse(S) do
       write(item, " -> ", (400 >= f(item)) | "overflows")

end

procedure f(x) 

  return abs(x)^0.5 + 5*x^3

end</lang>

Sample run: 

->tpk
Enter 11 numbers: 1 2 3 4 5 6 7 8 9 10 11
11 -> overflows
10 -> overflows
9 -> overflows
8 -> overflows
7 -> overflows
6 -> overflows
5 -> overflows
4 -> 322.0
3 -> 136.7320508075689
2 -> 41.41421356237309
1 -> 6.0
->

Io

<lang Io> // Initialize objects to be used in_num := File standardInput() nums := List clone result := Number

// Prompt the user and get numbers from standard input "Please enter 11 numbers:" println 11 repeat(nums append(in_num readLine() asNumber()))

// Reverse the numbers received nums reverseInPlace

// Apply the function and tell the user if the result is above // our limit. Otherwise, tell them the result. nums foreach(v, 

 // v needs parentheses around it for abs to properly convert v to its absolute value
 result = (v) abs ** 0.5 + 5 * v ** 3
 if (result > 400,
   "Overflow!" println
 ,
    result println
 )

) </lang>

Output:
io tpk.io
Please enter 11 numbers:
1
2
3
4
5
6
7
8
9
10
11
Overflow!
Overflow!
Overflow!
Overflow!
Overflow!
Overflow!
Overflow!
322
136.7320508075688679
41.4142135623730923
6

J

Input and output in J is done using "foreigns", in this case it is reading from the keyboard. The calculations are straightforward and applied to the whole set simultaneously. Similarly, overflow detection and changing the value to 'user alert' is also done once for all values.

No checks are done if the input is actually numbers and if there are actually eleven of them. This doesn't affect the algorithm. Additional checks can be done separately. <lang J>tpk=: 3 :0 

 smoutput 'Enter 11 numbers: '
 t1=: ((5 * ^&3) + (^&0.5@* *))"0 |. _999&".;._1 ' ' , 1!:1 [ 1
 smoutput 'Values of functions of reversed input: ' , ": t1
 ; <@(,&' ')@": ` ((<'user alert ')&[) @. (>&400)"0 t1

)</lang> A possible use scenario: <lang J> tpk Enter 11 numbers: 1 2 3 4 5 6 7 8.8 _9 10.123 0 Values of functions of reversed input: 0 5189.96 _3642 3410.33 1717.65 1082.45 627.236 322 136.732 41.4142 6 0 user alert _3642 user alert user alert user alert user alert 322 136.732 41.4142 6 </lang>

Note that the result of tpk is persisted in t1 and is also its explicit result rather than being an explicit output.

Here's an alternative approach:

<lang J>get11numbers=: 3 :0 

 smoutput 'Enter 11 numbers: '
 _&". 1!:1]1

)

f_x=: %:@| + 5 * ^&3

overflow400=: 'user alert'"_`":@.(<:&400)"0

tpk=: overflow400@f_x@|.@get11numbers</lang>

And, here's this alternative in action:

<lang J> tpk Enter 11 numbers: 1 2 3 4 5 6 7 8.8 _9 10.123 0 0 user alert _3642 user alert user alert user alert user alert 322 136.732 41.4142 6</lang>

(clearly, other alternatives are also possible).

Note that no error is reported if something other than 11 numbers are provided, since it's not clear what should be done for that case -- we just process all of them.

Java

<lang java>/** 

* Alexander Alvonellos 
*/

import java.util.*; import java.io.*;

public class TPKA { public static void main(String... args) { double[] input = new double[11]; double userInput = 0.0; Scanner in = new Scanner(System.in); for(int i = 0; i < 11; i++) { System.out.print("Please enter a number: "); String s = in.nextLine(); try { userInput = Double.parseDouble(s); } catch (NumberFormatException e) { System.out.println("You entered invalid input, exiting"); System.exit(1); } input[i] = userInput; } for(int j = 10; j >= 0; j--) { double x = input[j]; double y = f(x); if( y < 400.0) { System.out.printf("f( %.2f ) = %.2f\n", x, y); } else { System.out.printf("f( %.2f ) = %s\n", x, "TOO LARGE"); } } }

private static double f(double x) { return Math.pow(Math.abs(x), 0.5) + (5*(Math.pow(x, 3))); } } </lang>

Output:
Please enter a number: 1
Please enter a number: 2
Please enter a number: 3
Please enter a number: 4
Please enter a number: 5
Please enter a number: 6
Please enter a number: 7
Please enter a number: 8
Please enter a number: 9
Please enter a number: 10
Please enter a number: 11
f( 11.00 ) = TOO LARGE
f( 10.00 ) = TOO LARGE
f( 9.00 ) = TOO LARGE
f( 8.00 ) = TOO LARGE
f( 7.00 ) = TOO LARGE
f( 6.00 ) = TOO LARGE
f( 5.00 ) = TOO LARGE
f( 4.00 ) = 322.00
f( 3.00 ) = 136.73
f( 2.00 ) = 41.41
f( 1.00 ) = 6.00

JavaScript

Spidermonkey

<lang javascript>#!/usr/bin/env js

function main() { 

   var nums = getNumbers(11);
   nums.reverse();
   for (var i in  nums) {
       pardoKnuth(nums[i], fn, 400);
   }

}

function pardoKnuth(n, f, max) { 

   var res = f(n);
   putstr('f(' + String(n) + ')');
   if (res > max) {
       print(' is too large');
   } else {
       print(' = ' + String(res));
   }

}

function fn(x) { 

   return Math.pow(Math.abs(x), 0.5) + 5 * Math.pow(x, 3);

}

function getNumbers(n) { 

   var nums = [];
   print('Enter', n, 'numbers.');
   for (var i = 1; i <= n; i++) {
       putstr('   ' + i + ': ');
       var num = readline();
       nums.push(Number(num));        
   }
   return nums;

}

main(); </lang>

Results: 

Enter 11 numbers.
   1: 1
   2: 2
   3: 3
   4: 4
   5: 5
   6: 6
   7: 7
   8: 8
   9: 9
   10: 10
   11: 11
f(11)  is too large
f(10)  is too large
f(9)  is too large
f(8)  is too large
f(7)  is too large
f(6)  is too large
f(5)  is too large
f(4) = 322
f(3) = 136.73205080756887
f(2) = 41.41421356237309
f(1) = 6

jq

jq does not currently have an interactive mode allowing a prompt to be issued first, and so the initial prompt is implemented here using "echo", in keeping with the jq approach of dovetailing with other command-line tools. <lang jq>def f: 

 def abs: if . < 0 then -. else . end; 
 def power(x): (x * log) | exp;
 . as $x | abs | power(0.5) + (5 * (.*.*. ));

. as $in | split(" ") | map(tonumber) | if length == 11 then 

   reverse | map(f | if . > 400 then "TOO LARGE" else . end)
 else error("The number of numbers was not 11.")
 end

| .[] # print one result per line</lang>

Output:

<lang sh>$ echo "Enter 11 numbers on one line; when done, enter the end-of-file character:" ;\ jq -M -s -R -f Trabb_Pardo-Knuth_algorithm.jq > Enter 11 numbers on one line; when done, enter the end-of-file character: 1 2 3 4 5 6 7 8 9 10 11 "TOO LARGE" "TOO LARGE" "TOO LARGE" "TOO LARGE" "TOO LARGE" "TOO LARGE" "TOO LARGE" 322 136.73205080756887 41.41421356237309 6</lang>

Julia

<lang julia>f(x) = √abs(x) + 5x^3 for i in reverse(split(readline())) 

   v = f(parse(Int, i))
   println("$(i): ", v > 400 ? "TOO LARGE" : v)

end</lang>

Output:
1 2 3 4 5 6 7 8 9 10 11
11: TOO LARGE
10: TOO LARGE
9: TOO LARGE
8: TOO LARGE
7: TOO LARGE
6: TOO LARGE
5: TOO LARGE
4: 322.0
3: 136.73205080756887
2: 41.41421356237309
1: 6.0

Kotlin

<lang scala>// version 1.1.2

fun f(x: Double) = Math.sqrt(Math.abs(x)) + 5.0 * x * x * x

fun main(args: Array<String>) { 

   val da = DoubleArray(11)
   println("Please enter 11 numbers:")
   var i = 0
   while (i < 11) {
       print("  ${"%2d".format(i + 1)}: ")
       val d = readLine()!!.toDoubleOrNull()
       if (d == null)
           println("Not a valid number, try again")
       else
           da[i++] = d
   }
   println("\nThe sequence you just entered in reverse is:")
   da.reverse()
   println(da.contentToString())
   println("\nProcessing this sequence...")
   for (j in 0..10) {
       val v = f(da[j])
       print("  ${"%2d".format(j + 1)}: ") 
       if (v > 400.0) 
           println("Overflow!")
       else
           println(v)
   }

}</lang>

Output:

Sample session: 

Please enter 11 numbers:
   1: 10
   2: -1
   3: 1
   4: 2
   5: 3
   6: 4
   7: 4.3
   8: 4.305
   9: 4.303
  10: 4.302
  11: 4.301

The sequence you just entered in reverse is:
[4.301, 4.302, 4.303, 4.305, 4.3, 4.0, 3.0, 2.0, 1.0, -1.0, 10.0]

Processing this sequence...
   1: 399.88629974772687
   2: Overflow!
   3: Overflow!
   4: Overflow!
   5: 399.6086441353327
   6: 322.0
   7: 136.73205080756887
   8: 41.41421356237309
   9: 6.0
  10: -4.0
  11: Overflow!

Ksh

<lang ksh>

  1. !/bin/ksh
  1. Trabb Pardo–Knuth algorithm
  1. # Variables:

integer NUM_ELE=11 typeset -F FUNC_LIMIT=400

  1. # Functions:
  3. # Function _input(_arr) - Ask for user input, build array

function _input { typeset _arr ; nameref _arr="$1" typeset _i ; integer _i

clear ; print "Please input 11 numbers..." for ((_i=1 ; _i<=NUM_ELE ; _i++)); do read REPLY?"${_i}: " [[ $REPLY != {,1}(-)+(\d){,1}(.)*(\d) ]] && ((_i--)) && continue _arr+=( $REPLY ) done }

  1. # Function _function() - Apply |x|^0.5 + 5x^3
  2. # note: >400 creates an overflow situation

function _function { typeset _x ; _x=$1

(( _result = sqrt(abs(${_x})) + 5 * _x * _x * _x )) (( _result <= $FUNC_LIMIT )) && echo ${_result} && return 0 return 1 } 

######
  1. main #
######

typeset -a inputarr _input inputarr integer i printf "%s\n\n" "Evaluating f(x) = |x|^0.5 + 5x^3 for the given inputs :" for (( i=NUM_ELE-1; i>=0; i-- )); do result=$(_function ${inputarr[i]}) if (( $? )); then printf "%s\n" "Overflow" else printf "%s\n" "${result}" fi done</lang>

Output:


Please input 11 numbers...
1: 10
2: -1
3: 1
4: 2
5: 3
6: 4
7: 4.3
8: 4.305
9: 4.303
10: 4.302
11: 4.301
Evaluating f(x) = |x|^0.5 + 5x^3 for the given inputs :

399.8862997477268
Overflow
Overflow
Overflow
399.608644135332772
322
136.732050807568877
41.414213562373095
6
-4


Overflow

Liberty BASIC

Translation of: XBasic
Works with: Just BASIC version any

<lang lb> ' Trabb Pardo-Knuth algorithm ' Used "magic numbers" because of strict specification of the algorithm. dim s(10) print "Enter 11 numbers." for i = 0 to 10 

 print i + 1;
 input " => "; s(i)

next i print ' Reverse for i = 0 to 10 / 2 

 tmp = s(i)
 s(i) = s(10 - i)
 s(10 - i) = tmp

next i 'Results for i = 0 to 10 

 print "f("; s(i); ") = ";
 r = f(s(i))
 if r > 400 then
   print "overflow"
 else
   print r
 end if

next i end

function f(n) 

 f = sqr(abs(n)) + 5 * n * n * n

end function </lang>

Output:
Enter 11 numbers.
1 => -5
2 => -3
3 => -2
4 => -1
5 => 0
6 => 1
7 => 2
8 => 3
9 => 4
10 => 5
11 => 6

f(6) = overflow
f(5) = overflow
f(4) = 322
f(3) = 136.732051
f(2) = 41.4142136
f(1) = 6
f(0) = 0
f(-1) = -4
f(-2) = -38.5857864
f(-3) = -133.267949
f(-5) = -622.763932

Lua

Implementation of task description

<lang Lua>function f (x) return math.abs(x)^0.5 + 5*x^3 end

function reverse (t) 

   local rev = {}
   for i, v in ipairs(t) do rev[#t - (i-1)] = v end
   return rev

end

local sequence, result = {} print("Enter 11 numbers...") for n = 1, 11 do 

   io.write(n .. ": ")
   sequence[n] = io.read()

end for _, x in ipairs(reverse(sequence)) do 

   result = f(x)
   if result > 400 then print("Overflow!") else print(result) end

end</lang>

Output:
Enter 11 numbers...
1: 1
2: 2
3: 3
4: 4
5: 5
6: 6
7: 7
8: 8
9: 9
10: 10
11: 11
Overflow!
Overflow!
Overflow!
Overflow!
Overflow!
Overflow!
Overflow!
322
136.73205080757
41.414213562373
6

Line-for-line from TPK paper

<lang Lua>local a, y = {} function f (t) 

   return math.sqrt(math.abs(t)) + 5*t^3

end for i = 0, 10 do a[i] = io.read() end for i = 10, 0, -1 do 

   y = f(a[i])
   if y > 400 then print(i, "TOO LARGE")
              else print(i, y) end

end</lang>

Output:
1
2
3
4
5
6
7
8
9
10
11
10      TOO LARGE
9       TOO LARGE
8       TOO LARGE
7       TOO LARGE
6       TOO LARGE
5       TOO LARGE
4       TOO LARGE
3       322
2       136.73205080757
1       41.414213562373
0       6

M2000 Interpreter

<lang M2000 Interpreter> Module Input11 { 

     Flush ' empty stack
     For I=1 to 11 {
           Input "Give me a number ", a
           Data a   ' add to bottom of stack, use: Push a to add to top, to get reverse order here
     }

} Module Run { 

     Print "Trabb Pardo–Knuth algorithm"
     Print "f(x)=Sqrt(Abs(x))+5*x^3"
     if not match("NNNNNNNNN") then Error "Need 11 numbers" 
     Shiftback 1, -11 ' reverse  order 11 elements of stack of values
     Def f(x)=Sqrt(Abs(x))+5*x^3
     For i=1 to 11 {
           Read pop
           y=f(pop)
           if y>400 Then {
                 Print format$("f({0}) = Overflow!", pop)
           }  Else {
                 Print format$("f({0}) = {1}", pop, y)
           }
     }

} Run 10, -1, 1, 2, 3, 4, 4.3, 4.305, 4.303, 4.302, 4.301 Run 1, 2, 3, -4.55,5.1111, 6, -7, 8, 9, 10, 11 Input11 Run </lang>

To collect the output in clipboard. Global variables need <= to assign values, and document append values using = or <= (for globals)

Output with "," for decimals (Locale 1032). We can change this using statement Locale 1033

<lang M2000 Interpreter> Global a$ Document a$ ' make a$ as a document - string with paragraphs Module Run { 

     a$<={Trabb Pardo–Knuth algorithm
           f(x)=Sqrt(Abs(x))+5*x^3
           }
     if not match("NNNNNNNNN") then Error "Need 11 numbers" 
     Shiftback 1, -11 ' reverse  order 11 elements of stack of values
     Def f(x)=Sqrt(Abs(x))+5*x^3
     For i=1 to 11 {
           Read pop
           y=f(pop)
           if y>400 Then {
                a$<=format$("f({0}) = Overflow!", pop)+{
                 }
           }  Else {
                 a$<=format$("f({0}) = {1}", pop, y)+{
                 }
           }
     }

} Run 10, -1, 1, 2, 3, 4, 4.3, 4.305, 4.303, 4.302, 4.301 Run 1, 2, 3, -4.55,5.1111, 6, -7, 8, 9, 10, 11 Clipboard a$ </lang>

Output:
Trabb Pardo–Knuth algorithm
f(x)=Sqrt(Abs(x))+5*x^3
f(4,301) = 399,886299747727
f(4,302) = Overflow!
f(4,303) = Overflow!
f(4,305) = Overflow!
f(4,3) = 399,608644135333
f(4) = 322
f(3) = 136,732050807569
f(2) = 41,4142135623731
f(1) = 6
f(-1) = -4
f(10) = Overflow!
Trabb Pardo–Knuth algorithm
f(x)=Sqrt(Abs(x))+5*x^3
f(11) = Overflow!
f(10) = Overflow!
f(9) = Overflow!
f(8) = Overflow!
f(-7) = -1712,35424868894
f(6) = Overflow!
f(5,1111) = Overflow!
f(-4,55) = -468,84880209923
f(3) = 136,732050807569
f(2) = 41,4142135623731
f(1) = 6

Maple

<lang Maple>seqn := ListTools:-Reverse([parse(Maplets[Display](Maplets:-Elements:-Maplet(Maplets:-Elements:-InputDialog['ID1']("Enter a sequence of numbers separated by comma", 'onapprove' = Maplets:-Elements:-Shutdown(['ID1']), 'oncancel' = Maplets:-Elements:-Shutdown())))[1])]): f:= x -> abs(x)^0.5 + 5*x^3: for item in seqn do result := f(item): if (result > 400) then print("Alert: Overflow."): else print(result): end if: end do:</lang>

Usage:

Input:1,2,3,4,5,6,7,8,9,10,11 

                       "Alert: Overflow."
                       "Alert: Overflow."
                       "Alert: Overflow."
                       "Alert: Overflow."
                       "Alert: Overflow."
                       "Alert: Overflow."
                       "Alert: Overflow."
                          322.0000000
                          136.7320508
                          41.41421356
                               6.

Mathematica/Wolfram Language

<lang Mathematica>numbers=RandomReal[{-2,6},11] tpk[numbers_,overflowVal_]:=Module[{revNumbers}, 

revNumbers=Reverse[numbers];
f[x_]:=Abs[x]^0.5+5 x^3;
Do[
 If[f[i]>overflowVal,
  Print["f[",i,"]= Overflow"]
 ,
  Print["f[",i,"]= ",f[i]]
 ]
 ,
 {i,revNumbers}
]

] tpk[numbers,400]</lang>

Output:
{0.470145,1.18367,2.36984,4.86759,2.40274,5.48793,3.30256,5.34393,4.21944,2.23501,-0.0200707}
f[-0.0200707]= 0.141631
f[2.23501]= 57.3176
f[4.21944]= 377.663
f[5.34393]= Overflow
f[3.30256]= 181.921
f[5.48793]= Overflow
f[2.40274]= 70.9068
f[4.86759]= Overflow
f[2.36984]= 68.0859
f[1.18367]= 9.38004
f[0.470145]= 1.20527

min

Works with: min version 0.19.3

<lang min>((0 <) (-1 *) when) :abs (((abs 0.5 pow) (3 pow 5 * +)) cleave) :fn "Enter 11 numbers:" puts! (gets float) 11 times (fn (400 <=) (pop "Overflow") unless puts!) 11 times</lang>

Output:
Enter 11 numbers:
1
2
3
4
5
6
7
8
9
10
11
Overflow
Overflow
Overflow
Overflow
Overflow
Overflow
Overflow
322.0
136.7320508075689
41.41421356237309
6.0

Nim

Translation of: Python

<lang nim>import math, rdstdin, strutils, algorithm, sequtils

proc f(x: float): float = x.abs.pow(0.5) + 5 * x.pow(3)

proc ask: seq[float] = 

 readLineFromStdin("11 numbers: ").strip.split[0..10].map(parseFloat)

var s = ask() reverse s for x in s: 

 let result = f(x)
 echo x, ": ", if result > 400: "TOO LARGE!" else: $result</lang>
Output:
11 numbers: 1 2 3 4 5 6 7 8 9 10 11
11 numbers: 1 2 3 4 5 6 7 8 9 10 11
11.0: TOO LARGE!
10.0: TOO LARGE!
9.0: TOO LARGE!
8.0: TOO LARGE!
7.0: TOO LARGE!
6.0: TOO LARGE!
5.0: TOO LARGE!
4.0: 322.0
3.0: 136.7320508075689
2.0: 41.41421356237309
1.0: 6.0

Objective-C

Works with: Mac OS X version 10.6+

<lang objc>// // TPKA.m // RosettaCode // // Created by Alexander Alvonellos on 5/26/12. // Trabb Pardo-Knuth algorithm //

  1. import <Foundation/Foundation.h>

double f(double x);

double f(double x) { 

   return pow(abs(x), 0.5) + 5*(pow(x, 3));

}

int main (int argc, const char * argv[]) { 

   @autoreleasepool {
       NSMutableArray *input = [[NSMutableArray alloc] initWithCapacity:0];

       printf("%s", "Instructions: please enter 11 numbers.\n");
       for(int i = 0; i < 11; i++) {
           double userInput = 0.0;
           printf("%s", "Please enter a number: ");
           scanf("%lf", &userInput);
           [input addObject: @(userInput)];
       }
       
       for(int i = 10; i >= 0; i--) {
           double x = [input[i] doubleValue];
           double y = f(x);
           printf("f(%.2f) \t=\t", x);
           if(y < 400.0) {
               printf("%.2f\n", y);
           } else {
               printf("%s\n", "TOO LARGE");
           }
       }
   }
   return 0;

} </lang>

Output:
Instructions: please enter 11 numbers.
Please enter a number: 1
Please enter a number: 2
Please enter a number: 3
Please enter a number: 4
Please enter a number: 5
Please enter a number: 6
Please enter a number: 7
Please enter a number: 8
Please enter a number: 9
Please enter a number: 10
Please enter a number: 11
f(11.00) 	=	TOO LARGE
f(10.00) 	=	TOO LARGE
f(9.00) 	=	TOO LARGE
f(8.00) 	=	TOO LARGE
f(7.00) 	=	TOO LARGE
f(6.00) 	=	TOO LARGE
f(5.00) 	=	TOO LARGE
f(4.00) 	=	322.00
f(3.00) 	=	136.73
f(2.00) 	=	41.41
f(1.00) 	=	6.00

OCaml

<lang ocaml>let f x = sqrt x +. 5.0 *. (x ** 3.0) let p x = x < 400.0

let () = 

 print_endline "Please enter 11 Numbers:";
 let lst = Array.to_list (Array.init 11 (fun _ -> read_float ())) in
 List.iter (fun x ->
   let res = f x in
   if p res
   then Printf.printf "f(%g) = %g\n%!" x res
   else Printf.eprintf "f(%g) :: Overflow\n%!" x
 ) (List.rev lst)</lang>
Output:
$ ocaml trabb_pardo_knuth.ml
Please enter 11 Numbers:
1
2
3
4
5
6
7
8
9
10
11
f(11) :: Overflow
f(10) :: Overflow
f(9) :: Overflow
f(8) :: Overflow
f(7) :: Overflow
f(6) :: Overflow
f(5) :: Overflow
f(4) = 322
f(3) = 136.732
f(2) = 41.4142
f(1) = 6

We output error messages on stderr. We flush outputs with "%!" so that results and error messages do not appear separated.

PARI/GP

<lang parigp>{ 

 print("11 numbers: ");
 v=vector(11, n, eval(input()));
 v=apply(x->x=sqrt(abs(x))+5*x^3;if(x>400,"overflow",x), v);
 vector(11, i, v[12-i])

}</lang>

Output:
11 numbers:
1
2
3
4
5
6
7
8
9
10
11
%1 = ["overflow", "overflow", "overflow", "overflow", "overflow", "overflow",
"overflow", 322.0000000000000000000000000, 136.7320508075688772935274463, 41.414
21356237309504880168872, 6.000000000000000000000000000]

Perl

<lang Perl>print "Enter 11 numbers:\n"; for ( 1..11 ) { 

  $number = <STDIN>;
  chomp $number;
  push @sequence, $number;

}

for $n (reverse @sequence) { 

  my $result = sqrt( abs($n) ) + 5 * $n**3;
  printf "f( %6.2f ) %s\n", $n, $result > 400 ? "  too large!" : sprintf "= %6.2f", $result

}</lang>

Output:
Enter 11 numbers:
2
1.2
3
3.4
4
4.5
5 
7.8
2.7
13
11.2
f(  11.20 )   too large!
f(  13.00 )   too large!
f(   2.70 ) = 100.06
f(   7.80 )   too large!
f(   5.00 )   too large!
f(   4.50 )   too large!
f(   4.00 ) = 322.00
f(   3.40 ) = 198.36
f(   3.00 ) = 136.73
f(   1.20 ) =   9.74
f(   2.00 ) =  41.41

Phix

function f(atom x)
    return sqrt(abs(x))+5*power(x,3)
end function

procedure test(string s, bool fake_prompt=true)
    if fake_prompt then printf(1,"Enter 11 numbers:%s\n",{s}) end if
    s = substitute(s,","," ")
    sequence S = scanf(s,"%f %f %f %f %f %f %f %f %f %f %f")
    if length(S)!=1 then puts(1,"not 11 numbers") abort(0) end if
    S = reverse(S[1])
    for i=1 to length(S) do
        atom result = f(S[i])
        if result>400 then
            printf(1,"f(%g):overflow\n",{S[i]})
        else
            printf(1,"f(%g):%g\n",{S[i],result})
        end if
    end for
    puts(1,"\n")
end procedure

--test(prompt_string("Enter 11 numbers:"),false)
constant tests = {"10 -1 1 2 3 4 4.3 4.305 4.303 4.302 4.301","1,2,3,4,5,6,7,8,9,10,11",
"0.470145,1.18367,2.36984,4.86759,2.40274,5.48793,3.30256,5.34393,4.21944,2.23501,-0.0200707"}
papply(tests,test)
Output:
Enter 11 numbers:10 -1 1 2 3 4 4.3 4.305 4.303 4.302 4.301
f(4.301):399.886
f(4.302):overflow
f(4.303):overflow
f(4.305):overflow
f(4.3):399.609
f(4):322
f(3):136.732
f(2):41.4142
f(1):6
f(-1):-4
f(10):overflow

Enter 11 numbers:1,2,3,4,5,6,7,8,9,10,11
f(11):overflow
f(10):overflow
f(9):overflow
f(8):overflow
f(7):overflow
f(6):overflow
f(5):overflow
f(4):322
f(3):136.732
f(2):41.4142
f(1):6

Enter 11 numbers:0.470145,1.18367,2.36984,4.86759,2.40274,5.48793,3.30256,5.34393,4.21944,2.23501,-0.0200707
f(-0.0200707):0.141631
f(2.23501):57.3174
f(4.21944):377.662
f(5.34393):overflow
f(3.30256):181.921
f(5.48793):overflow
f(2.40274):70.9071
f(4.86759):overflow
f(2.36984):68.0862
f(1.18367):9.38002
f(0.470145):1.20527

PicoLisp

<lang PicoLisp>(de f (X) 

  (+ (sqrt (abs X)) (* 5 X X X)) )

(trace 'f)

(in NIL 

  (prin "Input 11 numbers: ")
  (for X (reverse (make (do 11 (link (read)))))
     (when (> (f X) 400)
        (prinl "TOO LARGE") ) ) )</lang>

Test: <lang PicoLisp>Input 11 numbers: 1 2 3 4 5 6 7 8 9 10 11 

f : 11
f = 6658

TOO LARGE 

f : 10
f = 5003

TOO LARGE 

f : 9
f = 3648

TOO LARGE 

f : 8
f = 2562

TOO LARGE 

f : 7
f = 1717

TOO LARGE 

f : 6
f = 1082

TOO LARGE 

f : 5
f = 627

TOO LARGE 

f : 4
f = 322
f : 3
f = 136
f : 2
f = 41
f : 1
f = 6</lang>

PL/I

<lang PL/I> Trabb: Procedure options (main); /* 11 November 2013 */ 

  declare (i, n) fixed binary;
  declare s fixed (5,1) controlled;
  declare g fixed (15,5);

  put ('Please type 11 values:');
  do i = 1 to 11;
     allocate s;
     get (s);
     put (s);
  end;
  put skip(2) ('Results:');
  do i = 1 to 11;
     g = f(s); put skip list (s);
     if g > 400 then put ('Too large'); else put (g);
     free s;
  end;

f: procedure (x) returns (fixed(15,5)); 

  declare x fixed (5,1);
  return (sqrt(abs(x)) + 5*x**3);

end f;

end Trabb; </lang>

Output:
Please type 11 values: 
     1.0 
     3.0 
     2.0 
    -4.0 
    -5.0 
     6.0 
     7.0 
     9.0 
    11.0 
     1.5 
     2.4 

Results: 
     2.4                          70.66920 
     1.5                          18.09974 
    11.0                Too large 
     9.0                Too large 
     7.0                Too large 
     6.0                Too large 
    -5.0                        -622.76391 
    -4.0                        -318.00000 
     2.0                          41.41421 
     3.0                         136.73205 
     1.0                           6.00000

PL/M

Assuming the existence of suitable external library routines. <lang plm>TPK: DO; 

   /* external I/O and real mathematical routines */
   WRITE$STRING: PROCEDURE( S )      EXTERNAL; DECLARE S POINTER; END;
   WRITE$REAL:   PROCEDURE( R )      EXTERNAL; DECLARE R REAL;    END;
   WRITE$NL:     PROCEDURE           EXTERNAL;                    END;
   READ$REAL:    PROCEDURE( R ) REAL EXTERNAL; DECLARE R POINTER; END;
   REAL$ABS:     PROCEDURE( R ) REAL EXTERNAL; DECLARE R REAL;    END;
   REAL$SQRT:    PROCEDURE( R ) REAL EXTERNAL; DECLARE R REAL;    END;
   /* end external routines */

   F: PROCEDURE( T ) REAL;
       DECLARE T REAL;
       RETURN REAL$SQRT(REAL$ABS(T))+5*T*T*T;
   END F;
   MAIN: PROCEDURE;
       DECLARE Y REAL, A( 11 ) REAL, I INTEGER;
       DO I = 0 TO 10;
          CALL READ$REAL( @A( I ) );
       END;
       DO I = 10 TO 0 BY -1;
          Y = F( A( I ) );
          IF Y > 400.0 THEN CALL WRITE$STRING( @( 'TOO LARGE', 0 ) );
                       ELSE CALL WRITE$REAL( Y );
          CALL WRITE$NL();
       END;
   END MAIN;

END TPK;</lang>

Output:
1 2 3 4 5 6 7 8 9 10 11
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
TOO LARGE
322.0000
136.7321
 41.4142
  6.0000

PowerShell

<lang PowerShell> function Get-Tpk { 

   [CmdletBinding()]
   [OutputType([PSCustomObject])]
   Param
   (
       [Parameter(Mandatory=$true,
                  ValueFromPipeline=$true,
                  ValueFromPipelineByPropertyName=$true,
                  Position=0)]
       [double]
       $Number
   )

   Begin
   {
       function Get-TpkFunction ([double]$Number)
       {
           [Math]::Pow([Math]::Abs($Number),(0.5)) + 5 * [Math]::Pow($Number,3)
       }

       [object[]]$output = @()
   }
   Process
   {
       $Number | ForEach-Object {
           $n = Get-TpkFunction $_

           if ($n -le 400)
           {
               $result = $n
           }
           else
           {
               $result = "Overflow"
           }
       }

       $output += [PSCustomObject]@{
           Number = $Number
           Result = $result
       }
   }
   End
   {
       [Array]::Reverse($output)
       $output
   }

} </lang> <lang PowerShell> $tpk = 1..11 | Get-Tpk $tpk </lang>

Output:
Number Result          
------ ------          
    11 Overflow        
    10 Overflow        
     9 Overflow        
     8 Overflow        
     7 Overflow        
     6 Overflow        
     5 Overflow        
     4 322             
     3 136.732050807569
     2 41.4142135623731
     1 6

Sort back to ascending order ignoring Overflow results: <lang PowerShell> $tpk | where result -ne overflow | sort number </lang>

Output:
Number           Result
------           ------
     1                6
     2 41.4142135623731
     3 136.732050807569
     4              322

PureBasic

<lang purebasic>Procedure.d f(x.d) 

 ProcedureReturn Pow(Abs(x), 0.5) + 5 * x * x * x

EndProcedure

Procedure split(i.s, delimeter.s, List o.d()) 

 Protected index = CountString(i, delimeter) + 1 ;add 1 because last entry will not have a delimeter
 
 While index > 0
   AddElement(o())
   o() = ValD(Trim(StringField(i, index, delimeter)))
   index - 1
 Wend

 ProcedureReturn ListSize(o())

EndProcedure

Define i$, entriesAreValid = 0, result.d, output$ NewList numbers.d()

If OpenConsole() 

 Repeat 
   PrintN(#crlf$ + "Enter eleven numbers that are each separated by spaces or commas:")
   
   i$ = Input(
   i$ = Trim(i$)
   If split(i$, ",", numbers.d()) < 11
     ClearList(numbers())
     If split(i$, " ", numbers.d()) < 11
       PrintN("Not enough numbers were supplied.")
       ClearList(numbers())
     Else 
       entriesAreValid = 1
     EndIf 
   Else
     entriesAreValid = 1
   EndIf
 Until entriesAreValid = 1
 
 ForEach numbers()
   output$  = "f(" + RTrim(RTrim(StrD(numbers(), 3), "0"), ".") + ") = "
   result.d = f(numbers())
   If result > 400
     output$ + "Too Large"
   Else
     output$ + RTrim(RTrim(StrD(result, 3), "0"), ".")
   EndIf
   PrintN(output$)
 Next
 
 Print(#crlf$ + #crlf$ + "Press ENTER to exit"): Input()
 CloseConsole()

EndIf</lang>

Output:
Enter eleven numbers that are each separated by spaces or commas:
10, -1, 1, 2, 3, 4, 4.3, 4.305, 4.303, 4.302, 4.301
f(4.301) = 399.886
f(4.302) = Too Large
f(4.303) = Too Large
f(4.305) = Too Large
f(4.3) = 399.609
f(4) = 322
f(3) = 136.732
f(2) = 41.414
f(1) = 6
f(-1) = -4
f(10) = Too Large

Python

Functional

<lang python>Python 3.2.2 (default, Sep 4 2011, 09:51:08) [MSC v.1500 32 bit (Intel)] on win32 Type "copyright", "credits" or "license()" for more information. >>> def f(x): return abs(x) ** 0.5 + 5 * x**3

>>> print(', '.join('%s:%s' % (x, v if v<=400 else "TOO LARGE!") for x,v in ((y, f(float(y))) for y in input('\nnumbers: ').strip().split()[:11][::-1])))

11 numbers: 1 2 3 4 5 6 7 8 9 10 11 11:TOO LARGE!, 10:TOO LARGE!, 9:TOO LARGE!, 8:TOO LARGE!, 7:TOO LARGE!, 6:TOO LARGE!, 5:TOO LARGE!, 4:322.0, 3:136.73205080756887, 2:41.41421356237309, 1:6.0 >>> </lang>

Procedural

<lang python>def f(x): 

   return abs(x) ** 0.5 + 5 * x**3

def ask(): 

   return [float(y)
           for y in input('\n11 numbers: ').strip().split()[:11]]

if __name__ == '__main__': 

   s = ask()
   s.reverse()
   for x in s:
       result = f(x)
       if result > 400:
           print(' %s:%s' % (x, "TOO LARGE!"), end=)
       else:
           print(' %s:%s' % (x, result), end=)
   print()</lang>
Sample output:
11 numbers: 1 2 3 4 5 6 7 8 9 10 11
 11.0:TOO LARGE! 10.0:TOO LARGE! 9.0:TOO LARGE! 8.0:TOO LARGE! 7.0:TOO LARGE! 6.0:TOO LARGE! 5.0:TOO LARGE! 4.0:322.0 3.0:136.73205080756887 2.0:41.41421356237309 1.0:6.0

R

<lang R>S <- scan(n=11)

f <- function(x) sqrt(abs(x)) + 5*x^3

for (i in rev(S)) { 

 res <- f(i)
 if (res > 400)
   print("Too large!")
 else
   print(res)

}</lang>

Sample output:
> source("~/tpk.R")
1: 1 2 3 4 5
6: 6 7 8 9 10
11: 11
Read 11 items
[1] "Too large!"
[1] "Too large!"
[1] "Too large!"
[1] "Too large!"
[1] "Too large!"
[1] "Too large!"
[1] "Too large!"
[1] 322
[1] 136.7321
[1] 41.41421
[1] 6

Racket

<lang racket>

  1. lang racket

(define input 

 (for/list ([i 11])
   (printf "Enter a number (~a of 11): " (+ 1 i))
   (read)))

(for ([x (reverse input)]) 

 (define res (+ (sqrt (abs x)) (* 5 (expt x 3))))
 (if (> res 400)
     (displayln "Overflow!")
     (printf "f(~a) = ~a\n" x res)))

</lang>

Output:
Enter a number (1 of 11): 1
Enter a number (2 of 11): 2
Enter a number (3 of 11): 3
Enter a number (4 of 11): 4
Enter a number (5 of 11): 5
Enter a number (6 of 11): 6
Enter a number (7 of 11): 7
Enter a number (8 of 11): 8
Enter a number (9 of 11): 9
Enter a number (10 of 11): 10
Enter a number (11 of 11): 11
Overflow!
Overflow!
Overflow!
Overflow!
Overflow!
Overflow!
Overflow!
f(4) = 322
f(3) = 136.73205080756887
f(2) = 41.41421356237309
f(1) = 6

Raku

(formerly Perl 6) <lang perl6>my @nums = prompt("Please type 11 space-separated numbers: ").words 

   until @nums == 11;

for @nums.reverse -> $n { 

   my $r = $n.abs.sqrt + 5 * $n ** 3;
   say "$n\t{ $r > 400 ?? 'Urk!' !! $r }";

}</lang>

Output:
Please type 11 space-separated numbers: 10 -1 1 2 3 4 4.3 4.305 4.303 4.302 4.301
4.301	399.88629974772681
4.302	Urk!
4.303	Urk!
4.305	Urk!
4.3	399.60864413533278
4	322
3	136.73205080756887
2	41.414213562373092
1	6
-1	-4
10	Urk!

REXX

The REXX language doesn't have a   sqrt   function, so a RYO version is included here.     [RYO   =   Roll Your Own.]

It could be noted that almost half of this program is devoted to prompting, parsing and validating of the (input) numbers,
not to mention some hefty code to support right-justified numbers such that they are aligned when displayed. <lang rexx>/*REXX program implements the Trabb─Pardo-Knuth algorithm for N numbers (default is 11).*/ numeric digits 200 /*the number of digits precision to use*/ parse arg N .; if N== | N=="," then N=11 /*Not specified? Then use the default.*/ maxValue= 400 /*the maximum value f(x) can have. */ 

    wid=  20                                    /*  ··· but only show this many digits.*/
   frac=   5                                    /*  ··· show this # of fractional digs.*/

say ' _____' /* ◄─── this SAY displays a vinculum.*/ say 'function: ƒ(x) ≡ √ │x│ + (5 * x^3)' prompt= 'enter ' N " numbers for the Trabb─Pardo─Knuth algorithm: (or Quit)" 

 do ask=0;    say;  /*░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░*/
 say prompt;  say;      pull $;     say                                             /*░*/
 if abbrev('QUIT',$,1)  then do;    say 'quitting.';    exit 1;   end               /*░*/
 ok=0                                                                               /*░*/
                        select                  /*validate there're N numbers.*/    /*░*/
                        when $=        then say  "no numbers entered"             /*░*/
                        when words($)<N  then say  "not enough numbers entered"     /*░*/
                        when words($)>N  then say  "too many numbers entered"       /*░*/
                        otherwise        ok=1                                       /*░*/
                        end   /*select*/                                            /*░*/
 if \ok  then iterate                                    /* [↓]  W=max width. */    /*░*/
 w=0;                   do v=1  for N;       _=word($, v);   w=max(w, length(_) )   /*░*/
                        if datatype(_, 'N')  then iterate          /*numeric ?*/    /*░*/
                        say _  "isn't numeric";   iterate ask                       /*░*/
                        end   /*v*/                                                 /*░*/
 leave                                                                              /*░*/
 end   /*ask*/      /*░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░*/

say 'numbers entered: ' $ say 

       do i=N  by -1  for N;  #=word($, i) / 1  /*process the numbers in reverse.      */
       g =  fmt(   f( # )        )              /*invoke  function  ƒ  with arg number.*/
       gw=right(  'ƒ('#") ",  w+7)              /*nicely formatted  ƒ(number).         */
       if g>maxValue  then say gw    "is >  "     maxValue      ' ['space(g)"]"
                      else say gw     "   = "     g
       end   /*i*/                              /* [↑]  display the result to terminal.*/

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ f: procedure; parse arg x; return sqrt( abs(x) ) + 5 * x**3 /*──────────────────────────────────────────────────────────────────────────────────────*/ fmt: z=right(translate(format(arg(1), wid, frac), 'e', "E"), wid) /*right adjust; use e*/ 

     if pos(.,z)\==0 then z=left(strip(strip(z,'T',0),"T",.),wid) /*strip trailing 0 &.*/
     return right(z, wid - 4*(pos('e', z)==0) )                   /*adjust: no exponent*/

/*──────────────────────────────────────────────────────────────────────────────────────*/ sqrt: procedure; parse arg x; if x=0 then return 0; d=digits(); m.=9; numeric form; h=d+6 

     numeric digits; parse value format(x,2,1,,0) 'E0'  with  g 'E' _ .;  g=g *.5'e'_ % 2
       do j=0  while h>9;      m.j=h;               h=h % 2  + 1;  end  /*j*/
       do k=j+5  to 0  by -1;  numeric digits m.k;  g=(g+x/g)*.5;  end  /*k*/;   return g</lang>
output   when prompted, using the input of:     5   3.3   3   2e-1   1   0   -1   -222   -33   4.0004   +5 
                           _____
function:        ƒ(x)  ≡  √ │x│   +   (5 * x^3)

enter  11  numbers for the Trabb─Pardo─Knuth algorithm:     (or Quit)

5   3.3   3   2e-1   1   0   -1   -222   -33   4.0004   +5   ◄■■■■■■■■■■■ this is what the user entered.

numbers entered:  5   3.3   3   2E-1   1   0   -1   -222   -33   4.0004   +5

        ƒ(5)  is >   400  [627.23607]
   ƒ(4.0004)     =         322.09611
      ƒ(-33)     =     -179679.25544
     ƒ(-222)     =   -54705225.10034
       ƒ(-1)     =          -4
        ƒ(0)     =           0
        ƒ(1)     =           6
      ƒ(0.2)     =           0.48721
        ƒ(3)     =         136.73205
      ƒ(3.3)     =         181.50159
        ƒ(5)  is >   400  [627.23607]

Ring

<lang ring>

  1. Project : Trabb Pardo–Knuth algorithm

decimals(3) x = list(11) for n=1 to 11 

   x[n] = n

next

s = [-5, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6] for i = 1 to 11 

   see string(i) + " => " + s[i] + nl

next see copy("-", 20) + nl i = i - 1

while i > 0 

     see "f(" + string(s[i]) + ") = " 
     x = f(s[i])
     if x > 400 
        see "-=< overflow >=-" + nl
     else 
        see x + nl
     ok
     i = i - 1

end

func f(n) 

    return sqrt(fabs(n)) + 5 * pow(n, 3)

</lang> Output: 

1 => -5
2 => -3
3 => -2
4 => -1
5 => 0
6 => 1
7 => 2
8 => 3
9 => 4
10 => 5
11 => 6
--------------------
f(6) = -=< overflow >=-
f(5) = -=< overflow >=-
f(4) = 322
f(3) = 136.732
f(2) = 41.414
f(1) = 6
f(0) = 0
f(-1) = -4
f(-2) = -38.586
f(-3) = -133.268
f(-5) = -622.764

Ruby

<lang ruby>def f(x) x.abs ** 0.5 + 5 * x ** 3 end

puts "Please enter 11 numbers:" nums = 11.times.map{ gets.to_f }

nums.reverse_each do |n| 

 print "f(#{n}) = "
 res = f(n)
 puts res > 400 ? "Overflow!" : res

end</lang>

Output:
ruby tpk.rb
Please enter 11 numbers:
1
2
3
4
5
6
7
8
9
-1
-4
f(-4.0) = -318.0
f(-1.0) = -4.0
f(9.0) = Overflow!
f(8.0) = Overflow!
f(7.0) = Overflow!
f(6.0) = Overflow!
f(5.0) = Overflow!
f(4.0) = 322.0
f(3.0) = 136.73205080756887
f(2.0) = 41.41421356237309
f(1.0) = 6.0

Rust

<lang rust> use std::io::{self, BufRead};

fn op(x: f32) -> Option<f32> { 

   let y = x.abs().sqrt() + 5.0 * x * x * x;
   if y < 400.0 {
       Some(y)
   } else {
       None
   }

}

fn main() { 

   println!("Please enter 11 numbers (one number per line)");
   let stdin = io::stdin();

   let xs = stdin
       .lock()
       .lines()
       .map(|ox| ox.unwrap().trim().to_string())
       .flat_map(|s| str::parse::<f32>(&s))
       .take(11)
       .collect::<Vec<_>>();

   for x in xs.into_iter().rev() {
       match op(x) {
           Some(y) => println!("{}", y),
           None => println!("overflow"),
       };
   }

} </lang>

Output:
Enter 11 numbers (one number per line)
1
2
3
4
5
6
7
8
9
10
11
overflow
overflow
overflow
overflow
overflow
overflow
overflow
322
136.73206
41.414215
6

Scala

<lang scala>object TPKa extends App { 

   final val numbers = scala.collection.mutable.MutableList[Double]()
   final val in = new java.util.Scanner(System.in)
   while (numbers.length < CAPACITY) {
       print("enter a number: ")
       try {
           numbers += in.nextDouble()
       }
       catch {
           case _: Exception =>
               in.next()
               println("invalid input, try again")
       }
   }

   numbers reverseMap { x =>
       val fx = Math.pow(Math.abs(x), .5D) + 5D * (Math.pow(x, 3))
       if (fx < THRESHOLD)
           print("%8.3f -> %8.3f\n".format(x, fx))
       else
           print("%8.3f -> %s\n".format(x, Double.PositiveInfinity.toString))
   }

   private final val THRESHOLD = 400D
   private final val CAPACITY = 11

}</lang>

Sidef

Translation of: Raku

<lang ruby>var nums; do { 

   nums = Sys.readln("Please type 11 space-separated numbers: ").nums

} while(nums.len != 11)

nums.reverse.each { |n| 

   var r = (n.abs.sqrt + (5 * n**3));
   say "#{n}\t#{ r > 400 ? 'Urk!' : r }";

}</lang>

Output:
Please type 11 space-separated numbers: 10 -1 1 2 3 4 4.3 4.305 4.303 4.302 4.301
4.301	399.886299747726800445468371077898575778355
4.302	Urk!
4.303	Urk!
4.305	Urk!
4.3	399.608644135332772087455898679984992632401
4	322
3	136.732050807568877293527446341505872366943
2	41.41421356237309504880168872420969807857
1	6
-1	-4
10	Urk!

Sinclair ZX81 BASIC

Works with the unexpanded (1k RAM) ZX81 <lang basic> 10 DIM A(11) 

20 PRINT "ENTER ELEVEN NUMBERS:"
30 FOR I=1 TO 11
40 INPUT A(I)
50 NEXT I
60 FOR I=11 TO 1 STEP -1
70 LET Y=SQR ABS A(I)+5*A(I)**3
80 IF Y<=400 THEN GOTO 110
90 PRINT A(I),"TOO LARGE"

100 GOTO 120 110 PRINT A(I),Y 120 NEXT I</lang>

Output:
ENTER ELEVEN NUMBERS:
2.8             111.43332
3.333           186.95529
1.01            6.1564926
2.55            84.503747
11              TOO LARGE
6               TOO LARGE
5               TOO LARGE
4               322
3               136.73205
2               41.414214
1               6

Swift

Works with: Swift 2.0

<lang swift>import Foundation

print("Enter 11 numbers for the Trabb─Pardo─Knuth algorithm:")

let f: (Double) -> Double = { sqrt(fabs($0)) + 5 * pow($0, 3) }

(1...11) 

   .generate()
   .map { i -> Double in
       print("\(i): ", terminator: "")
       guard let s = readLine(), let n = Double(s) else { return 0 }
       return n
   }
   .reverse()
   .forEach {
       let result = f($0)
       print("f(\($0))", result > 400.0 ? "OVERFLOW" : result, separator: "\t")
   }

</lang>

Output:
Enter 11 numbers for the Trabb─Pardo─Knuth algorithm:
1: 1
2: 2
3: 3
4: 4
5: 5
6: 6
7: 7
8: 8
9: 9
10: 10
11: 11
f(11.0)	OVERFLOW
f(10.0)	OVERFLOW
f(9.0)	OVERFLOW
f(8.0)	OVERFLOW
f(7.0)	OVERFLOW
f(6.0)	OVERFLOW
f(5.0)	OVERFLOW
f(4.0)	322.0
f(3.0)	136.732050807569
f(2.0)	41.4142135623731
f(1.0)	6.0

Symsyn

<lang symsyn> |Trabb Pardo–Knuth algorithm

a : 11 0 

i
if i LE 10
   [] $s
   ~ $s w
   w a.i
  + i
   goif
endif
10 i
if i GE 0
   call f
   if x GT 400
      'too large' $s
   else
      ~ x $s
   endif
   ~ i $r
   + ' ' $r
   + $r $s.1
   $s []
   - i
   goif
endif
stop

f a.i t 

 * t t x
 * x t x
 * 5 x
 abs t
 sqrt t y
 + y x
 return

</lang>

Tcl

<lang tcl># Helper procedures proc f {x} {expr {abs($x)**0.5 + 5*$x**3}} proc overflow {y} {expr {$y > 400}}

  1. Read in 11 numbers, with nice prompting

fconfigure stdout -buffering none for {set n 1} {$n <= 11} {incr n} { 

   puts -nonewline "number ${n}: "
   lappend S [scan [gets stdin] "%f"]

}

  1. Process and print results in reverse order

foreach x [lreverse $S] { 

   set result [f $x]
   if {[overflow $result]} {

puts "${x}: TOO LARGE!" 

   } else {

puts "${x}: $result" 

   }

}</lang>

Sample run:
number 1: 0
number 2: 1
number 3: 2
number 4: 3
number 5: 4
number 6: 5
number 7: 6
number 8: 7
number 9: 8
number 10: 9
number 11: 10
10.0: TOO LARGE!
9.0: TOO LARGE!
8.0: TOO LARGE!
7.0: TOO LARGE!
6.0: TOO LARGE!
5.0: TOO LARGE!
4.0: 322.0
3.0: 136.73205080756887
2.0: 41.41421356237309
1.0: 6.0
0.0: 0.0

VBScript

<lang vb> Function tpk(s) arr = Split(s," ") For i = UBound(arr) To 0 Step -1 n = fx(CDbl(arr(i))) If n > 400 Then WScript.StdOut.WriteLine arr(i) & " = OVERFLOW" Else WScript.StdOut.WriteLine arr(i) & " = " & n End If Next End Function

Function fx(x) fx = Sqr(Abs(x))+5*x^3 End Function

'testing the function WScript.StdOut.Write "Please enter a series of numbers:" list = WScript.StdIn.ReadLine tpk(list) </lang>

Output:

The number series was derived from the example of C. 

C:\>cscript /nologo tpk.vbs
Please enter 10 numbers:10 -1 1 2 3 4 4.3 4.305 4.303 4.302 4.301
4.301 = 399.886299747727
4.302 = OVERFLOW
4.303 = OVERFLOW
4.305 = OVERFLOW
4.3 = 399.608644135333
4 = 322
3 = 136.732050807569
2 = 41.4142135623731
1 = 6
-1 = -4
10 = OVERFLOW

Wren

Library: Wren-fmt

<lang ecmascript>import "io" for Stdin, Stdout import "/fmt" for Fmt

var f = Fn.new { |x| x.abs.sqrt + 5*x*x*x }

var s = List.filled(11, 0) System.print("Please enter 11 numbers:") var count = 0 while (count < 11) { 

   Fmt.write("  Number $-2d : ", count + 1)
   Stdout.flush()
   var number = Num.fromString(Stdin.readLine())
   if (!number) {
       System.print("Not a valid number, try again.")
   } else {
       s[count] = number
       count = count + 1
   }

} s = s[-1..0] System.print("\nResults:") for (item in s) { 

   var fi = f.call(item)
   if (fi <= 400) {
       Fmt.print("  f($6.3f) = $7.3f", item, fi)
   } else {
       Fmt.print("  f($6.3f) = overflow", item)
   }

}</lang>

Output:

Entering the same numbers as the Ada example: 

Please enter 11 numbers:
  Number 1  : 10
  Number 2  : -1
  Number 3  : 1
  Number 4  : 2
  Number 5  : 3
  Number 6  : 4
  Number 7  : 4.3
  Number 8  : 4.305
  Number 9  : 4.303
  Number 10 : 4.302
  Number 11 : 4.301

Results:
  f( 4.301) = 399.886
  f( 4.302) = overflow
  f( 4.303) = overflow
  f( 4.305) = overflow
  f( 4.300) = 399.609
  f( 4.000) = 322.000
  f( 3.000) = 136.732
  f( 2.000) =  41.414
  f( 1.000) =   6.000
  f(-1.000) =  -4.000
  f(10.000) = overflow

XBasic

Works with: Windows XBasic

<lang xbasic> ' Trabb Pardo-Knuth algorithm PROGRAM "tpkalgorithm" VERSION "0.0001"

IMPORT "xma"

DECLARE FUNCTION Entry () INTERNAL FUNCTION SINGLE F(SINGLE n)

FUNCTION Entry () 

 ' Used "magic numbers" because of strict specification of the algorithm.
 SINGLE s[10]
 SINGLE tmp, r
 UBYTE i
 PRINT "Enter 11 numbers."
 FOR i = 0 TO 10
   PRINT i + 1;
   s[i] = SINGLE(INLINE$(" => "))
 NEXT i
 PRINT
 ' Reverse
 FOR i = 0 TO 10 / 2
   tmp = s[i]
   s[i] = s[10 - i]
   s[10 - i] = tmp
 NEXT i
 'Results
 FOR i = 0 TO 10
   PRINT "f("; LTRIM$(STR$(s[i])); ") =";
   r = F(s[i])
   IF r > 400 THEN
     PRINT " overflow"
   ELSE
     PRINT r
   END IF
 NEXT i

END FUNCTION

FUNCTION SINGLE F(SINGLE n) 

 RETURN SQRT(ABS(n)) + 5 * n * n *n

END FUNCTION END PROGRAM </lang>

Output:
Enter 11 numbers.
 1 => -5
 2 => -3
 3 => -2
 4 => -1
 5 => 0
 6 => 1
 7 => 2
 8 => 3
 9 => 4
 10 => 5
 11 => 6

f(6) = overflow
f(5) = overflow
f(4) = 322
f(3) = 136.732
f(2) = 41.4142
f(1) = 6
f(0) = 0
f(-1) =-4
f(-2) =-38.5858
f(-3) =-133.268
f(-5) =-622.764

XPL0

<lang XPL0>include c:\cxpl\codes;

func real F(X); real X; return sqrt(abs(X)) + 5.0*X*X*X;

real Result, S(11); int I; [Text(0, "Please enter 11 numbers: "); for I:= 0 to 11-1 do S(I):= RlIn(0);

for I:= 11-1 downto 0 do 

       [RlOut(0, S(I));
       Result:= F(S(I));
       if Result > 400.0 then
               Text(0, "  overflows")
       else    RlOut(0, Result);
       CrLf(0)];

]</lang>

Output:
Please enter 11 numbers: 10 -1 1 2 3 4 4.3 4.305 4.303 4.302 4.301
    4.30100  399.88630
    4.30200  overflows
    4.30300  overflows
    4.30500  overflows
    4.30000  399.60864
    4.00000  322.00000
    3.00000  136.73205
    2.00000   41.41421
    1.00000    6.00000
   -1.00000   -4.00000
   10.00000  overflows

zkl

<lang zkl>fcn f(x) { x.abs().pow(0.5) + x.pow(3)*5 } reg ns; do{ 

  ns=ask("11 numbers seperated by spaces: ");
  try{ ns=ns.split(" ").filter().apply("toFloat") } catch{}

}while(not ns.isType(List) or ns.len()!=11); ns.reverse().apply(fcn(x){ 

  fx:=f(x); "f(%7.3f)-->%s".fmt(x, if(fx>400)"Overflow" else fx) })

.pump(Console.println);</lang>

Output:
11 numbers seperated by spaces: 10 -1 1     2 3 4 4.3 4.305 4.303 4.302 4.301
f(  4.301)-->399.886
f(  4.302)-->Overflow
f(  4.303)-->Overflow
f(  4.305)-->Overflow
f(  4.300)-->399.609
f(  4.000)-->322
f(  3.000)-->136.732
f(  2.000)-->41.4142
f(  1.000)-->6
f( -1.000)-->-4
f( 10.000)-->Overflow
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