全排列

 1 解法一：
 2 program ex5_2; 框架[一]
 3 VAR  s: set of 1..100;
 4 n,r,num:integer;
 5 a:array [1..100] of integer;
 6 PROCEDURE print;              //输出方案
 7 var i:integer;
 8 begin
 9    num:=num+1;
10    for i:=1 to r do
11      write(a[i]:3);
12    writeln;
13 end;
14 PROCEDURE Search(k:integer);                    //回溯过程
15 VAR  i:integer;
16 begin
17    for i:=1 to n do
18      if i in s then                                                 //判断i是否可用
19        begin
20          a[k]:=i;                                                     //保存结果
21          s:=s-[i];                                                    //从s集合中删除i
22          if k=r then print
23           else Search(k+1);
24          s:=s+[i];                                                   //把i放回s集合中
25        end;
26 end;
27 BEGIN
28    write(‘Input n,r:‘);readln(n,r);
29    s:=[1..n];num:=0;
30    Search(1);
31    writeln(‘number=‘,num);                                //输出方案总数
32 END.
33 
34 
35 解法二：
36 program ex5_2; 框架[二]
37 type se=set of 1..100;
38 var  s:se;
39        n,r,num,k:integer;
40        a:array [1..100] of integer;
41 PROCEDURE print;          
42                                   //输出方案
43 var i:integer;
44 begin
45    num:=num+1;
46    for i:=1 to r do
47      write(a[i]:3);
48    writeln;
49 end;
50 PROCEDURE Search(s:se;k:integer);  
51                                     //回溯过程
52 VAR i:integer;
53 begin
54    if k>r then print
55     else 
56 for i:=1 to n do
57           if i in s then             //判断i是否可用
58              begin
59                  a[k]:=i;             //保存结果
60                  Search(s-[i],k+1); //进一步递归，i已放过的信息通过参数进行传递
61              end;
62 end;
63 BEGIN
64    write(‘Input n,r:‘); readln(n,r);
65    s:=[1..n];num:=0;
66    Search(s,1);
67    writeln(‘number=‘,num); //输出方案总数
68    readln;
69 END.