## Scala Hacking: Computing min and max of a List

There are several ways to accomplish this. Next code snippet shows how to compute min and max using `reduceLeft`.

```val ls = List(1,2,3,4,5)

ls.reduceLeft(_ min _) // is equivalent to ls.min

ls.reduceLeft(_ max _) // is equivalent to ls.max
```

Same can be accomplished via `foldLeft` or `foldRight`.

```ls.foldLeft(Int.MaxValue) (_ min _)

ls.foldLeft(Int.MinValue) (_ max _)
```

However, can we compute both min and max in one line? Check out the following snippet.

```ls.map(
x=>(x,x)).reduceLeft(
(x,y) => (x._1 min y._1, x._2 max y._2)
```

An alternative is to use `foldLeft`:

```ls.foldLeft
((Int.MaxValue, Int.MinValue))
((acc:(Int,Int),y:Int) => (acc._1 min y, acc._2 max y))
```

## Problem Definition

You are asked to calculate factorials of some small positive integers.

### Input

An integer t, 1<=t<=100, denoting the number of testcases, followed by t lines, each containing a single integer n, 1<=n<=100.

### Output

For each integer n given at input, display a line with the value of n!

More details about the problem is available here.

## Solution

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 open System open System.Numerics let init() = Array.init 158 (fun x -> if x = 0 then 1 else 0) let minLength (a:int array) : int = let rec m' (a:int array) i = match a.[i-1] with | 0 -> m' a (i-1) | _ -> i in m' a a.Length let toString' (a:int array) = let s = System.Text.StringBuilder() let mLength = minLength a for i=(mLength-1) downto 0 do s.Append(a.[i])|>ignore s.ToString() let (^*) multiplier (multiplicand:int array)= let rec multiplyOPT (a:int array) b i acc length= let l = Math.Min(length,a.Length) if i< l then let r = a.[i] * b + acc a.[i] <- r%10 multiplyOPT a b (i+1) (r/10) length else a in multiplyOPT multiplicand multiplier 0 0 (multiplicand |>minLength |> (+) 2) let factorial n = let rec fac n r = match n with | 0 -> toString' r | _ -> fac (n-1) (n^*r) fac n (init()) let intFactorial (i : uint64) = let rec f (i:uint64) (acc:uint64) = match i with | 0UL -> acc | i -> f (i-1UL) (acc*i) in f i 1UL let computeFactorial (n:int) = if n > 20 then factorial n else (intFactorial (n |> uint64)).ToString(); let solveSpoj24() = let rec solveLines currentLine maxLines = if currentLine < maxLines then System.Console.ReadLine() |> int |> computeFactorial |> printfn "%s" solveLines (currentLine+1) maxLines in match Console.ReadLine() |> Int32.TryParse with | (true, i) when i > 0 -> solveLines 0 i | _ -> () solveSpoj24()
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## Problem Definition

The Antique Comedians of Malidinesia prefer comedies to tragedies. Unfortunately, most of the ancient plays are tragedies. Therefore the dramatic advisor of ACM has decided to transfigure some tragedies into comedies. Obviously, this work is very hard because the basic sense of the play must be kept intact, although all the things change to their opposites. For example the numbers: if any number appears in the tragedy, it must be converted to its reversed form before being accepted into the comedy play.

Reversed number is a number written in arabic numerals but the order of digits is reversed. The first digit becomes last and vice versa. For example, if the main hero had 1245 strawberries in the tragedy, he has 5421 of them now. Note that all the leading zeros are omitted. That means if the number ends with a zero, the zero is lost by reversing (e.g. 1200 gives 21). Also note that the reversed number never has any trailing zeros.

ACM needs to calculate with reversed numbers. Your task is to add two reversed numbers and output their reversed sum. Of course, the result is not unique because any particular number is a reversed form of several numbers (e.g. 21 could be 12, 120 or 1200 before reversing). Thus we must assume that no zeros were lost by reversing (e.g. assume that the original number was 12).

### Input

The input consists of N cases (equal to about 10000). The first line of the input contains only positive integerN. Then follow the cases. Each case consists of exactly one line with two positive integers separated by space. These are the reversed numbers you are to add.

### Output

For each case, print exactly one line containing only one integer – the reversed sum of two reversed numbers. Omit any leading zeros in the output.

## Solution

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 // problem definition : https://www.spoj.com/problems/ADDREV/ open System let parseTuple() : int*int = let line = Console.ReadLine().Split() line. |> int , line. |> int let reverseInt n = let rec reverseInt' n acc = let acc' = acc + (n%10) match n with | n when n >= 10 -> reverseInt' ((int)(n/10)) (acc'*10) | n -> acc' reverseInt' n 0 let solveCase (n1,n2) = reverseInt (reverseInt n1 + reverseInt n2) let rec solveCases i maxCases = if i < maxCases then parseTuple() |> solveCase |> printfn "%d" solveCases (i + 1) maxCases // solving next case let spoj42() = match Console.ReadLine() |> Int32.TryParse with | (true, i) when i > 0 -> solveCases 0 i | _ -> () spoj42()
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More details of this problem is available here