How to use the most concise method, the output from a to ZZZZ-CodePudding

In the most concise way output from a b c... Z aa ab, ac... Zz... ZZZZ (length of four as a demo only), please give the code,

CodePudding user response:

The cycle is ok, need others to give the code?

CodePudding user response:

Nested loop

CodePudding user response:

1. According to the length of letters ordering

 var 
I, j, k: Byte; 
The begin 
For I:=97 to 122 the do 
Mmo1. Lines. The Add (Char (I)); 
For I:=97 to 122 the do 
For j:=97 to 122 the do 
The begin 
Mmo1. Lines. The Add (Char (I) + Char (j)); 
For k:=97 to 122 the do 
end; 
For I:=97 to 122 the do 
For j:=97 to 122 the do 
For k:=97 to 122 the do 
The begin 
Mmo1. Lines. The Add (Char (I) + Char (j) + Char (k)); 
end; 
end;

2. According to the alphabetical order

 var 
I, j, k: Byte; 
The begin 
For I:=97 to 122 the do 
The begin 
Mmo1. Lines. The Add (Char (I)); 
For j:=97 to 122 the do 
The begin 
Mmo1. Lines. The Add (Char (I) + Char (j)); 
For k:=97 to 122 the do 
Mmo1. Lines. The Add (Char (I) + Char (j) + Char (k)); 
end; 
end; 
end;

CodePudding user response:

 var 
I, j: Integer; 
The begin 
For I=1 to 4 do//here is you want to output the length of the string 
The begin 
For j:=word to word (' a ') (' z ') do 
The begin 
Memo1. Lines. The Add (StringOfChar (CRH (j), I)); 
end; 
end; 
end;

CodePudding user response:

reference 4 floor hongss response:

Delphi/Pascal code? 1234567891011 var I, j: Integer; The begin for I=1 to 4 do//here is you want to output the length of the string the begin for j:=word to word (' a ') (' z ') do the begin Memo1. Lines. The Add (StringOfChar (CRH (j), I)); .

Wrong, and that is a single character cycle,,

CodePudding user response:

 Type 
TOutLineProc=Procedure (Const Line: String);//output interface, and need to be initialized before invoking 

//recursive each output 
Procedure _OutPutAZ (Byte: Integer; CurrentByte: integer; Const Line: String; The CallBack: TOutLineProc); 
Var 
I: integer; 
The begin 
If CurrentByte=Byte then 
For I:=word to word (' a ') (' z ') do the begin 
{$IFDEF CPU64} 
PChar (Int64 (Line) + CurrentByte) ^ :=Char (I); 
{$ELSE} 
PChar (Integer (Line) + CurrentByte) ^ :=Char (I); 
{$ENDIF} 
The CallBack (Line); 
End 
The else 
For I:=word to word (' a ') (' z ') do the begin 
{$IFDEF CPU64} 
PChar (Int64 (Line) + CurrentByte) ^ :=Char (I); 
{$ELSE} 
PChar (Integer (Line) + CurrentByte) ^ :=Char (I); 
{$ENDIF} 
_OutPutAZ (Byte, CurrentByte + 1, Line, the CallBack); 
end; 
end; 

//output, MaxByte for the largest number, the CallBack as the CallBack function, when there is a new will CallBack the process 
Procedure OutPutA2Z (MaxByte: Integer; The CallBack: TOutLineProc); 
Var 
I: integer; 
The Line: String; 
The begin 
For I:=1 to MaxByte do begin 
SetLength (Line, I); 
_OutPutAZ (I - 1, 0, the Line, the CallBack); 
end; 
end; 


//the following method to invoke 
//callback process execution is very frequent, so should consider memory Buffer to handle, whether write files or show 
Procedure OutLineProc (Const Line: String); 
The begin 
//comment out this sentence is very fast and slow slow right here 
Form1. Memo1. Lines. The Add (Line); 
end; 

Procedure TForm1. FormCreate (Sender: TObject); 
The begin 
OutPutA2Z (3, OutLineProc);//the time used to double-digit increase exponentially, please consider the biggest digits 
end;

CodePudding user response:

 
The program az. 
Type alpha26=string;//26 letters of the 26 hexadecimal number 

The function pow_l (x: qword; N: longint) : qword; 
Var I: longint; 
Re: qword; 
The begin 
Re:=1; 
I:=0; 
While i<> N do the begin 
Re:=re * x; 
Inc (I); 
end; 
Pow_l:=re; 

end; 

The function num2A26 (x: qword; Len: longint) : alpha26; 
Var I: longint; 
Re: alpha26; 
The begin 
Setlength (re, len); 
For I:=len downto do begin 1 
Re: [I]=CRH (26 + x mod word (' a ')); 
X: x=div 26; 
end; 
Num2a26:=re; 
end; 

Procedure printA26 (len: longint); 
Var I: qword; 
The begin 
I:=0; 
While i<> Pow_l (26, len) do the begin 
Write (num2A26 (I, len) : 8); 
Inc (I); 
end; 
end; 
Procedure printAllA26 (len: longint); 
Var I: longint; 
The begin 
For I:=1 to len do 
PrintA26 (I); 
end; 

Var n: qword; 
The begin 
Readln (n); 
While (n> 0) do the begin 
PrintAllA26 (n); 
Readln (n); 
end; 
End.

The violence, no code optimization, pow_l can be optimized,

CodePudding user response:

 procedure TForm1. Button1Click (Sender: TObject); 
Var 
I, j, k, o, m: Integer; 
The begin 
The for m:=1 to 4 do 
The begin 
For I:=65 to 90 the do 
The begin 
If m> 1 then 
The begin 
For j:=65 to 90 the do 
The begin 
If m & gt; 2 then 
The begin 
For k:=65 to 90 the do 
The begin 
If m & gt; 3 then 
The begin 
For: o=65 to 90 the do 
The begin 
Memo1. Lines. The Add (Char (I) + Char (j) + Char (k) + Char (o)); 
end; 
end; 
Memo1. Lines. The Add (Char (I) + Char (j) + Char (k)); 
end; 
end; 
Memo1. Lines. The Add (Char (I) + Char (j)); 
end; 
end; 
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