Google Treasure Hunt 2008 走格式题解

[ahjoe]

题目: 34x68的格子，从左上角到右下角，有多少种走法。规则：只能向右或向下行走。
解题方法: 从一个格子到右下角的走法数量，等于它右边格子与下边格子的走法数量之和，从最下边的格子或最右边的格子到右下角只有一种走法。
结果: 0294692427022540894366527900
计算用时: 瞬间
注: 用String做大数计算。

unit GoogleHunt2u;

interface

uses
  Windows, Messages, SysUtils, Classes, Graphics, Controls, Forms, Dialogs,
  StdCtrls, Spin;

type
  TForm1 = class(TForm)
    SpinEdit1: TSpinEdit;
    Label1: TLabel;
    Label2: TLabel;
    SpinEdit2: TSpinEdit;
    Button1: TButton;
    Edit1: TEdit;
    Label3: TLabel;
    procedure Button1Click(Sender: TObject);
    procedure FormCreate(Sender: TObject);
  private
    { Private declarations }
  public
    { Public declarations }
  end;

var
  Form1: TForm1;

implementation

{$R *.DFM}

var
  VArr: array [0..100, 0..100] of string;

function BigIntAdd(V1, V2: string): string;
var
  i: integer;
  max1, max2: integer;
  max: integer;
  v: integer;
begin
  if V1[1] = '0' then
    max1 := Length(V1) - 1
  else
    max1 := Length(V1);
  if V2[1] = '0' then
    max2 := Length(V2) - 1
  else
    max2 := Length(V2);
  if max1 > max2 then
    max := max1 + 1
  else
    max := max2 + 1;

  while Length(V1) < max do
  begin
    V1 := '0' + V1;
  end;
  while Length(V2) < max do
  begin
    V2 := '0' + V2;
  end;

  SetLength(Result, max);
  v := 0;
  for i := max downto 1 do
  begin
    v := Byte(V1[i]) + Byte(V2[i])-Byte('0')*2 + v;
    Result[i] := Char((v mod 10) + Byte('0'));
    v := v div 10;
  end;
end;

function CountRoute(W, H: integer): string;
begin
  if (W = 1) or (H = 1) then
    Result := '1'
  else
  begin
    if VArr[W-1][H] = '' then
    begin
      VArr[W-1][H] := CountRoute(W-1, H);
      VArr[H][W-1] := VArr[W-1][H];
    end;
    if VArr[W][H-1] = '' then
    begin
      VArr[W][H-1] := CountRoute(W, H-1);
      VArr[H-1][W] := VArr[W][H-1];
    end;
    Result := BigIntAdd(VArr[W-1][H], VArr[W][H-1]);
  end;
end;

procedure TForm1.Button1Click(Sender: TObject);
var
  W, H: integer;
begin
  W := SpinEdit1.Value;
  H := SpinEdit2.Value;
  Edit1.Text := CountRoute(W, H);
end;

procedure TForm1.FormCreate(Sender: TObject);
begin
  FillChar(VArr, sizeof(VArr), 0);
end;

end.

[ahjoe]

小技巧: 用一个2维数组保存特定宽高格子的走法数量，避免重复计算。

[ahjoe]

标题打错了，版主帮我。"走格子题解"

[rcaicc]

棒~~
收藏了.
这种问题给我直接就是递归.估计34*68能算死