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我想解决这个练习：http ://code.google.com/codejam/contest/351101/dashboard#s=p0使用 F#。

我是函数式编程和 F# 的新手，但我非常喜欢这个概念和语言。我也喜欢 codejam 练习，它看起来很简单，但却很真实。有人可以指出我的解决方案吗？

目前我已经编写了这段代码，它只是简单的命令式，从功能的角度来看很难看：

(*
    C - Credit
    L - Items
    I - List of Integer, wher P is single integer

    How does the data look like inside file
    N
    [...
    * Money
    * Items in store
    ...]
*)

    let lines = System.IO.File.ReadAllLines("../../../../data/A-small-practice.in")
    let CBounds c = c >= 5 && c <= 1000
    let PBounds p = p >= 1 && p <= 1000

    let entries = int(lines.[0]) - 1
    let mutable index = 1   (* First index is how many entries*)
    let mutable case = 1

    for i = 0 to entries do
        let index = (i*3) + 1
        let C = int(lines.[index])
        let L = int(lines.[index+1])
        let I = lines.[index+2]    
        let items = I.Split([|' '|]) |> Array.map int    
        // C must be the sum of some items

        // Ugly imperative way which contains duplicates
        let mutable nIndex = 0
        for n in items do
            nIndex <- nIndex + 1
            let mutable mIndex = nIndex
            for m in items.[nIndex..] do            
                mIndex <- mIndex + 1
                if n + m = C then do
                    printfn "Case #%A: %A %A" case nIndex mIndex
                    case <- case + 1

我想找出加起来等于 C 值但不是以通常的命令方式的项目 - 我想要功能方法。

4

1 回答 1

4

你没有具体说明你将如何解决问题，所以很难给出建议。

关于读取输入，您可以将其表示为Seq. Seq 模块的高阶函数非常方便：

let data = 
   "../../../../data/A-small-practice.in"
   |> System.IO.File.ReadLines
   |> Seq.skip 1
   |> Seq.windowed 3
   |> Seq.map (fun lines -> let C = int(lines.[0])
                            let L = int(lines.[1])
                            let items = lines.[2].Split([|' '|]) |> Array.map int
                            (C, L, items))

更新：

对于示例的其余部分，您可以使用sequence expression。它功能强大且易于表达嵌套计算：

 let results = 
     seq { 
          for (C, _, items) in data do
            for j in 1..items.Length-1 do
              for i in 0..j-1 do
                if items.[j] + items.[i] = C then yield (i, j)
     }

 Seq.iteri (fun case (i, j) -> printfn "Case #%A: %A %A" case i j) results
于 2013-01-20T16:30:14.070 回答