说明：本文件来自“广州六中信息学基地”。
【execans.2 为深度优先搜索，广度优先搜索类,及技巧性题目题解】
【题目1】N皇后问题(八皇后问题的扩展)
【题目2】排球队员站位问题
【题目3】把自然数Ｎ分解为若干个自然数之和。
【题目4】把自然数Ｎ分解为若干个自然数之积。
【题目5】马的遍历问题。
【题目6】加法分式分解
【题目7】地图着色问题
【题目8】在n*n的正方形中放置长为2,宽为1的长条块,
【题目9】找迷宫的最短路径。（广度优先搜索算法）
【题目10】火车调度问题
【题目11】农夫过河
【题目12】七段数码管问题。
【题目13】把1-8这8个数放入下图8个格中,要求相邻的格(横,竖,对角线)上填的数不连续.
【题目14】在4×4的棋盘上放置8个棋,要求每一行,每一列上只能放置2个.
【题目15】迷宫问题.求迷宫的路径.(深度优先搜索法)
【题目16】一笔画问题
【题目17】城市遍历问题.
【题目18】棋子移动问题
【题目19】求集合元素问题（1,2x+1,3X+1类)
【题目】N皇后问题(含八皇后问题的扩展，规则同八皇后)：在N*N的棋盘上，放置N个皇后，要求每一横行
每一列，每一对角线上均只能放置一个皇后，问可能的方案及方案数。
const max=8;
var i,j:integer;
a:array[1..max] of 0..max; {放皇后数组}
b:array[2..2*max] of boolean; {/对角线标志数组}
c:array[-(max-1)..max-1] of boolean; {\对角线标志数组}
col:array[1..max] of boolean; {列标志数组}
total:integer; {统计总数}
procedure output; {输出}
var i:integer;
begin
write('No.':4,'[',total+1:2,']');
for i:=1 to max do write(a[i]:3);write(' ');
if (total+1) mod 2 =0 then writeln; inc(total);
end;
function ok(i,dep:integer):boolean; {判断第dep行第i列可放否}
begin
ok:=false;
if ( b[i+dep]=true) and ( c[dep-i]=true) {and (a[dep]=0)} and
(col[i]=true) then ok:=true
end;
procedure try(dep:integer);
var i,j:integer;
begin
for i:=1 to max do {每一行均有max种放法}
if ok(i,dep) then begin
a[dep]:=i;
b[i+dep]:=false; {/对角线已放标志}
c[dep-i]:=false; {\对角线已放标志}
col[i]:=false; {列已放标志}
if dep=max then output
else try(dep+1); {递归下一层}
a[dep]:=0; {取走皇后,回溯}
b[i+dep]:=true; {恢复标志数组}
c[dep-i]:=true;
col[i]:=true;
end;
end;
begin
for i:=1 to max do begin a[i]:=0;col[i]:=true;end;
for i:=2 to 2*max do b[i]:=true;
for i:=-(max-1) to max-1 do c[i]:=true;
total:=0;
try(1);
writeln('total:',total);
end.
【测试数据】
n=8 八皇后问题
No.[ 1] 1 5 8 6 3 7 2 4 No.[ 2] 1 6 8 3 7 4 2 5
No.[ 3] 1 7 4 6 8 2 5 3 No.[ 4] 1 7 5 8 2 4 6 3
No.[ 5] 2 4 6 8 3 1 7 5 No.[ 6] 2 5 7 1 3 8 6 4
No.[ 7] 2 5 7 4 1 8 6 3 No.[ 8] 2 6 1 7 4 8 3 5
No.[ 9] 2 6 8 3 1 4 7 5 No.[10] 2 7 3 6 8 5 1 4
No.[11] 2 7 5 8 1 4 6 3 No.[12] 2 8 6 1 3 5 7 4
No.[13] 3 1 7 5 8 2 4 6 No.[14] 3 5 2 8 1 7 4 6
No.[15] 3 5 2 8 6 4 7 1 No.[16] 3 5 7 1 4 2 8 6
No.[17] 3 5 8 4 1 7 2 6 No.[18] 3 6 2 5 8 1 7 4
No.[19] 3 6 2 7 1 4 8 5 No.[20] 3 6 2 7 5 1 8 4
No.[21] 3 6 4 1 8 5 7 2 No.[22] 3 6 4 2 8 5 7 1
No.[23] 3 6 8 1 4 7 5 2 No.[24] 3 6 8 1 5 7 2 4
No.[25] 3 6 8 2 4 1 7 5 No.[26] 3 7 2 8 5 1 4 6
No.[27] 3 7 2 8 6 4 1 5 No.[28] 3 8 4 7 1 6 2 5
No.[29] 4 1 5 8 2 7 3 6 No.[30] 4 1 5 8 6 3 7 2
No.[31] 4 2 5 8 6 1 3 7 No.[32] 4 2 7 3 6 8 1 5
No.[33] 4 2 7 3 6 8 5 1 No.[34] 4 2 7 5 1 8 6 3
No.[35] 4 2 8 5 7 1 3 6 No.[36] 4 2 8 6 1 3 5 7
No.[37] 4 6 1 5 2 8 3 7 No.[38] 4 6 8 2 7 1 3 5
No.[39] 4 6 8 3 1 7 5 2 No.[40] 4 7 1 8 5 2 6 3
No.[41] 4 7 3 8 2 5 1 6 No.[42] 4 7 5 2 6 1 3 8
No.[43] 4 7 5 3 1 6 8 2 No.[44] 4 8 1 3 6 2 7 5
No.[45] 4 8 1 5 7 2 6 3 No.[46] 4 8 5 3 1 7 2 6
No.[47] 5 1 4 6 8 2 7 3 No.[48] 5 1 8 4 2 7 3 6
No.[49] 5 1 8 6 3 7 2 4 No.[50] 5 2 4 6 8 3 1 7
No.[51] 5 2 4 7 3 8 6 1 No.[52] 5 2 6 1 7 4 8 3
No.[53] 5 2 8 1 4 7 3 6 No.[54] 5 3 1 6 8 2 4 7
No.[55] 5 3 1 7 2 8 6 4 No.[56] 5 3 8 4 7 1 6 2
No.[57] 5 7 1 3 8 6 4 2 No.[58] 5 7 1 4 2 8 6 3
No.[59] 5 7 2 4 8 1 3 6 No.[60] 5 7 2 6 3 1 4 8
No.[61] 5 7 2 6 3 1 8 4 No.[62] 5 7 4 1 3 8 6 2
No.[63] 5 8 4 1 3 6 2 7 No.[64] 5 8 4 1 7 2 6 3
No.[65] 6 1 5 2 8 3 7 4 No.[66] 6 2 7 1 3 5 8 4
No.[67] 6 2 7 1 4 8 5 3 No.[68] 6 3 1 7 5 8 2 4
No.[69] 6 3 1 8 4 2 7 5 No.[70] 6 3 1 8 5 2 4 7
No.[71] 6 3 5 7 1 4 2 8 No.[72] 6 3 5 8 1 4 2 7
No.[73] 6 3 7 2 4 8 1 5 No.[74] 6 3 7 2 8 5 1 4
No.[75] 6 3 7 4 1 8 2 5 No.[76] 6 4 1 5 8 2 7 3
No.[77] 6 4 2 8 5 7 1 3 No.[78] 6 4 7 1 3 5 2 8
No.[79] 6 4 7 1 8 2 5 3 No.[80] 6 8 2 4 1 7 5 3
No.[81] 7 1 3 8 6 4 2 5 No.[82] 7 2 4 1 8 5 3 6
No.[83] 7 2 6 3 1 4 8 5 No.[84] 7 3 1 6 8 5 2 4
No.[85] 7 3 8 2 5 1 6 4 No.[86] 7 4 2 5 8 1 3 6
No.[87] 7 4 2 8 6 1 3 5 No.[88] 7 5 3 1 6 8 2 4
No.[89] 8 2 4 1 7 5 3 6 No.[90] 8 2 5 3 1 7 4 6
No.[91] 8 3 1 6 2 5 7 4 No.[92] 8 4 1 3 6 2 7 5
total:92
对于N皇后:
┏━━━┯━━┯━━┯━━┯━━┯━━┯━━┯━━┓
┃皇后 N│ 4 │ 5 │ 6 │ 7 │ 8 │ 9 │ 10 ┃
┠───┼──┼──┼──┼──┼──┼──┼──┨
┃方案数│ 2 │ 10 │ 4 │ 40 │ 92 │352 │724 ┃
┗━━━┷━━┷━━┷━━┷━━┷━━┷━━┷━━┛
【题目】排球队员站位问题
┏━━━━━━━━┓图为排球场的平面图，其中一、二、三、四、五、六为位置编号，
┃ ┃二、三、四号位置为前排，一、六、五号位为后排。某队比赛时，
┃ ┃一、四号位放主攻手，二、五号位放二传手，三、六号位放副攻
┠──┬──┬──┨手。队员所穿球衣分别为１，２，３，４，５，６号，但每个队
┃ 四 │ 三 │ 二 ┃员的球衣都与他们的站位号不同。已知１号、６号队员不在后排，
┠──┼──┼──┨２号、３号队员不是二传手，３号、４号队员不在同一排，５号、
┃ 五 │ 六 │ 一 ┃６号队员不是副攻手。
┗━━┷━━┷━━┛ 编程求每个队员的站位情况。
【算法分析】本题可用一般的穷举法得出答案。也可用回溯法。以下为回溯解法。
【参考程序】
type sset=set of 1..6;
var a:array[1..6]of 1..6;
d:array[1..6]of sset;
i:integer;
procedure output; {输出}
begin
if not( (a[3]in [2,3,4])= (a[4] in[2,3,4])) then
begin { 3,4号队员不在同一排 }
write('number:');for i:=1 to 6 do write(i:8);writeln;
write('weizhi:');for i:=1 to 6 do write(a[i]:8);writeln;
end;
end;
procedure try(i:integer;s:sset); {递归过程 i:第i个人,s:哪些位置已安排人了}
var
j,k:integer;
begin
for j:=1 to 6 do begin {每个人都有可能站1-6这6个位置}
if (j in d[i]) and not(j in s) then begin
{j不在d[i]中,则表明第i号人不能站j位. j如在s集合中,表明j位已排人了}
a[i]:=j; {第 i 人可以站 j 位}
if i<6 then try(i+1,s+[j]) {未安排妥,则继续排下去}
else output; {6个人都安排完,则输出}
end;
end;
end;
begin
for i:=1 to 6 do d[i]:=[1..6]-[i]; {每个人的站位都与球衣的号码不同}
d[1]:=d[1]-[1,5,6];
d[6]:=d[6]-[1,5,6]; {1,6号队员不在后排}
d[2]:=d[2]-[2,5];
d[3]:=d[3]-[2,5]; {2,3号队员不是二传手}
d[5]:=d[5]-[3,6];
d[6]:=d[6]-[3,6]; {5,6号队员不是副攻手}
try(1,[]);
end.
【题目】把自然数Ｎ分解为若干个自然数之和。
【参考答案】
n │ total
5 │ 7
6 │ 11
7 │ 15
10 │ 42
100 │ 190569291
【参考程序】
var n:byte; num:array[0..255] of byte; total:word;
procedure output(dep:byte);
var j:byte;
begin
for j:=1 to dep do write(num[j]:3);writeln; inc(total);
end;
procedure find(n,dep:byte); {N:待分解的数,DEP:深度}
var i,j,rest:byte;
begin
for i:=1 to n do {每一位从N到1去试}
if num[dep-1]<=i then {保证选用的数大于前一位}
begin
num[dep]:=i;
rest:=n - i; {剩余的数进行下一次递归调用}
if (rest>0) then begin find(rest,dep+1);end
else if rest=0 then output(dep);{刚好相等则输出}
num[dep]:=0;
end;
end;
begin {主程序}
writeln('input n:');readln(n);
fillchar(num,sizeof(num),0);
total:=0; num[0]:=0;
find(n,1);
writeln('sum=',total);
end.
【题目】把自然数Ｎ分解为若干个自然数之积。
【参考程序】
var path :array[1..1000] of integer;
total,n:integer;
procedure find(k,sum,dep:integer); {K:}
var b,d:Integer;
begin
if sum=n then {积等于N}
begin
write(n,'=',path[1]);
for d:=2 to dep-1 do write('*',path[d]);
writeln;inc(total);
exit;
end;
if sum>n then exit; {累积大于N}
for b:= trunc(n/sum)+1 downto k do {每一种可能都去试}
begin
path[dep]:=b;
find(b,sum*b,dep+1);
end;
end;
begin
readln(n); total:=0;
find(2,1,1);writeln('total:',total);
readln;
end.
【题目】马的遍历问题。在Ｎ＊Ｍ的棋盘中，马只能走日字。马从位置(x,y)处出发，把
棋盘的每一格都走一次，且只走一次。找出所有路径。
【参考程序】 {深度优先搜索法}
const n=5;m=4;
fx:array[1..8,1..2]of -2..2=((1,2),(2,1),(2,-1),(1,-2),(-1,-2),(-2,-1),
(-2,1),(-1,2)); {八个方向增量}
var
dep,i:byte; x,y:byte;
cont:integer; {统计总数}
a:array[1..n,1..m]of byte; {记录走法数组}
procedure output; {输出,并统计总数}
var x,y:byte;
begin
cont:=cont+1; writeln;
writeln('count=',cont);
for y:=1 to n do begin
for x:=1 to m do write(a[y,x]:3); writeln;
end; { readln; halt;}
end;
procedure find(y,x,dep:byte);
var i,xx,yy:integer;
begin
for i:=1 to 8 do
begin
xx:=x+fx[i,1];yy:=y+fx[i,2]; {加上方向增量,形成新的坐标}
if ((xx in [1..m])and(yy in [1..n]))and(a[yy,xx]=0) then
{判断新坐标是否出界,是否已走过?}
begin
a[yy,xx]:=dep; {走向新的坐标}
if (dep=n*m) then output
else find(yy,xx,dep+1); {从新坐标出发,递归下一层}
a[yy,xx]:=0 {回溯,恢复未走标志}
end;
end;
end;
begin
cont:=0;
fillchar(a,sizeof(a),0);
dep:=1;
writeln('input y,x');readln(y,x);
{ x:=1;y:=1;}
if (y>n) or(x>m) then begin writeln('x,y error!');halt;end;
a[y,x]:=1;
find(y,x,2);
if cont=0 then writeln('No answer!') else write('The End!');
readln;
end.
【题目】加法分式分解。如：1/2=1/4+1/4.找出所有方案。
输入：Ｎ Ｍ Ｎ为要分解的分数的分母
Ｍ为分解成多少项
【参考程序】
program fenshifenjie;
const nums=5;
var
t,m,dep:integer;
n,maxold,max,j:longint;
path:array[0..nums] of longint;
maxok,p:boolean;
sum,sum2:real;
procedure print;
var i:integer;
begin
t:=t+1;
if maxok=true then begin maxold:=path[m];maxok:=false;end;
write ('NO.',t);
for i:=1 to m do write(' ',path[i]:4); writeln;
if path[1]=path[m] then begin writeln('Ok! total:',t:4);readln;halt;end;
end;
procedure input;
begin
writeln ('input N:'); readln(n);
writeln ('input M(M<=',nums:1,'):'); readln(m);
if (n<=0) or (m<=0) or (m>4) or (n>maxlongint)
then begin writeln('Invalid Input!');readln;halt;end;
end;
function sum1(ab:integer):real;
var a,b,c,d,s1,s2:real;
i:integer;
begin
if ab=1 then
sum1:=1/path[1]
else
begin
a:=path[1];
b:=1 ;
c:=path[2];
d:=1;
for i:=1 to ab-1 do
begin
s2:=(c*b+a*d);
s1:=(a*c);
a:=s1;
b:=s2;
c:=path[i+2];
end;
sum1:=s2/s1;
end;
end;
procedure back;
begin
dep:=dep-1;
if dep<=m-2 then max:=maxold;
sum:=sum-1/path[dep];
j:=path[dep];
end;
procedure find;
begin
repeat
dep:=dep+1;
j:=path[dep-1]-1;
p:=false;
repeat
j:=j+1;
if (dep<>m) and (j<=max) then
if (sum+1/j) >=1/n then p:=false
else begin
p:=true;
path[dep]:=j;
sum:=sum+1/path[dep];
end
else if j>max then back;
if dep=m then begin
path[dep]:=j;
sum2:=sum1(m);
if (sum2)>1/n then p:=false;
if (sum2)=1/n then begin print;
max:=j;
back;
end;
if (sum2<1/n) then back;
if (j>=max) then back;
end;
until p
until dep=0;
end;
begin
INPUT;
maxok:=true;
for t:=0 to m do path[t]:=n;
dep:=0; t:=0; sum:=0;
max:=maxlongint;
find;
readln;
end.
【题目】地图着色问题
【参考程序１】
const lin:array[1..12,1..12] of 0..1 {区域相邻数组,1表示相邻}
=((0,1,1,1,1,1,0,0,0,0,0,0),
(1,0,1,0,0,1,1,1,0,0,0,0),
(1,1,0,1,0,0,0,1,1,0,0,0),
(1,0,1,0,1,0,1,0,1,1,0,0),
(1,0,0,1,0,1,0,0,0,1,1,0),
(1,1,0,0,1,0,1,0,0,0,1,0),
(0,1,0,0,0,1,0,1,0,0,1,1),
(0,1,1,0,0,0,1,0,1,0,0,1),
(0,0,1,1,0,0,0,1,0,1,0,1),
(0,0,0,1,1,0,0,0,1,0,1,1),
(0,0,0,0,1,1,1,0,0,1,0,1),
(0,0,0,0,0,0,1,1,1,1,1,1));
var color:array[1..12] of byte; {color数组放已填的颜色}
total:integer;
function ok(dep,i:byte):boolean; {判断选用色i是否可用}
var k:byte; {条件:相邻的区域颜色不能相同}
begin
for k:=1 to dep do
if (lin[dep,k]=1) and (i=color[k]) then begin ok:=false;exit;end;
ok:=true;
end;
procedure output; {输出}
var k:byte;
begin
for k:=1 to 12 do write(color[k],' ');writeln;
total:=total+1;
end;
procedure find(dep:byte); {参数dep:当前正在填的层数}
var i:byte;
begin
for i:=1 to 4 do begin {每个区域都可能是1-4种颜色}
if ok(dep,i) then begin
color[dep]:=i;
if dep=12 then output else find(dep+1);
color[dep]:=0; {恢复初始状态,以便下一次搜索}
end;
end;
end;
begin
total:=0; {总数初始化}
fillchar(color,sizeof(color),0);
find(1);
writeln('total:=',total);
end.
【参考程序２】
const {lin数组:代表区域相邻情况}
lin:array[1..12] of set of 1..12 =
([2,3,4,5,6],[1,3,6,7,8],[1,2,4,8,9],[1,3,5,9,10],[1,4,6,10,11],
[1,2,5,7,11],[12,8,11,6,2],[12,9,7,2,3],[12,8,10,3,4],
[12,9,11,4,5],[12,7,10,5,6],[7,8,9,10,11]);
color:array[1..4] of char=('r','y','b','g');
var a:array[1..12] of byte; {因有12个区域,故a数组下标为1-12}
total:integer;
function ok(dep,i:integer):boolean; {判断第dep块区域是否可填第i种色}
var j:integer; { j 为什么设成局部变量?}
begin
ok:=true;
for j:=1 to 12 do
if (j in lin[dep]) and (a[j]=i) then ok:=false;
end;
procedure output; {输出过程}
var j:integer; { j 为什么设成局部变量?}
begin
inc(total); {方案总数加1}
write(total:4); {输出一种方案}
for j:=1 to 12 do write(color[a[j]]:2);writeln;
end;
procedure find(dep:byte);
var i:byte; { i 为什么设成局部变量?}
begin
for i:=1 to 4 do {每一区域均可从4种颜色中选一}
begin
if ok(dep,i) then begin {可填该色}
a[dep]:=i; {第dep块区域填第i种颜色}
if (dep=12) then output {填完12个区域}
else find(dep+1); {未填完}
a[dep]:=0; {取消第dep块区域已填的颜色}
end;
end;
end;
begin {主程序}
fillchar(a,sizeof(a),0); {记得要给变量赋初值!}
total:=0;
find(1);
writeln('End.');
end.
【题目】在n*n的正方形中放置长为2,宽为1的长条块,问放置方案如何
【参考程序１】
const n=4;
var k,u,v,result:integer;
a:array[1..n,1..n]of char;
procedure printf; {输出}
begin
result:=result+1; {方案总数加1}
writeln('--- ',result,' ---');
for v:=1 to n do begin
for u:=1 to n do write(a[u,v]); writeln end; writeln;
end;
procedure try; {填放长条块}
var i,j,x,y:integer; full:boolean;
begin
full:=true;
if k<>trunc(n*n/2) then full:=false;{测试是否已放满}
if full then printf; {放满则可输出}
if not full then begin {未满}
x:=0;y:=1; {以下先搜索未放置的第一个空位置}
repeat
x:=x+1;
if x>n then begin x:=1;y:=y+1 end
until a[x,y]=' ';
{找到后,分两种情况讨论}
if a[x+1,y]=' ' then begin {第一种情况:横向放置长条块}
k:=k+1; {记录已放的长条数}
a[x,y]:=chr(k+ord('@')); {放置}
a[x+1,y]:=chr(k+ord('@'));
try; {递归找下一个空位置放}
k:=k-1;
a[x,y]:=' '; {回溯,恢复原状}
a[x+1,y]:=' '
end;
if a[x,y+1]=' ' then begin {第二种情况:竖向放置长条块}
k:=k+1; {记录已放的长条数}
a[x,y]:=chr(k+ord('0')); {放置}
a[x,y+1]:=chr(k+ord('0'));
try; {递归找下一个空位置放}
k:=k-1;
a[x,y]:=' '; {回溯,恢复原状}
a[x,y+1]:=' '
end;
end;
end;
begin {主程序}
fillchar(a,sizeof(a),' '); {记录放置情况的字符数组,初始值为空格}
result:=0; k:=0; {k记录已放的块数,如果k=n*n/2,则说明已放满}
try; {每找到一个空位置,把长条块分别横放和竖放试验}
end.
【参考程序２】
const dai:array [1..2,1..2]of integer=((0,1),(1,0));
type node=record
w,f:integer;
end;
var a:array[1..20,1..20]of integer;
path:array[0..200]of node;
s,m,n,nn,i,j,x,y,dx,dy,dep:integer;
p,px:boolean;
procedure inputn;
begin
{ write('input n');readln(n);}
n:=4;
nn:=n*n;m:=nn div 2;
end;
procedure print;
var i,j:integer;
begin
inc(s);writeln('no',s);
for i:=1 to n do begin
for j:=1 to n do
write(a[i,j]:3);writeln;
end;
writeln;
end;
function fg(h,v:integer):boolean;
var p:boolean;
begin
p:=false;
if (h<=n) and (v<=n) then
if a[h,v]=0 then p:=true;
fg:=p;
end;
procedure back;
begin
dep:=dep-1;
if dep=0 then begin p:=true ;px:=true;end
else begin
i:=path[dep].w;j:=path[dep].f;
x:=((i-1)div n )+1;y:=i mod n;
if y=0 then y:=n;
dx:=x+dai[j,1];dy:=y+dai[j,2];
a[x,y]:=0;a[dx,dy]:=0;
end;
end;
begin
inputn;
s:=0;
fillchar(a,sizeof(a),0);
x:=0;y:=0;dep:=0;
path[0].w:=0;path[0].f:=0;
repeat
dep:=dep+1;
i:=path[dep-1].w;
repeat
i:=i+1;x:=((i-1)div n)+1;
y:=i mod n;if y=0 then y:=n;
px:=false;
if fg(x,y)
then begin
j:=0;p:=false;
repeat
inc(j);
dx:=x+dai[j,1];dy:=y+dai[j,2];
if fg(dx,dy) and (j<=2) then begin
a[x,y]:=dep;a[dx,dy]:=dep;
path[dep].w:=i;path[dep].f:=j;
if dep=m then begin print;dep:=m+1;back;end
else begin p:=true;px:=true;end;
end
else if j>=2 then back
else p:=false;
until p;
end
else if i>=nn then back
else px:=false;
until px;
until dep=0;
readln;
end.
【题目】找迷宫的最短路径。（广度优先搜索算法）
【参考程序】
uses crt;
const
migong:array [1..5,1..5] of integer=((0,0,-1,0,0), (0,-1,0,0,-1),
(0,0,0,0,0), (0,-1,0,0,0), (-1,0,0,-1,0));
{迷宫数组}
fangxiang:array [1..4,1..2] of -1..1=((1,0),(0,1),(-1,0),(0,-1));
{方向增量数组}
type node=record
lastx:integer; {上一位置坐标}
lasty:integer;
nowx:integer; {当前位置坐标}
nowy:integer;
pre:byte; {本结点由哪一步扩展而来}
dep:byte; {本结点是走到第几步产生的}
end;
var
lujing:array[1..25] of node; {记录走法数组}
closed,open,x,y,r:integer;
procedure output;
var i,j:integer;
begin
for i:=1 to 5 do begin
for j:=1 to 5 do
write(migong[i,j]:4); writeln;end;
i:=open;
repeat
with lujing[i] do
write(nowy:2,',',nowx:2,' <--');
i:=lujing[i].pre;
until lujing[i].pre=0;
with lujing[i] do
write(nowy:2,',',nowx:2);
end;
begin
clrscr;
with lujing[1] do begin {初始化第一步}
lastx:=0; lasty:=0; nowx:=1;nowy:=1;pre:=0;dep:=1;end;
closed:=0;open:=1;migong[1,1]:=1;
repeat
inc(closed); {队列首指针加1,取下一结点}
for r:=1 to 4 do begin {以4个方向扩展当前结点}
x:=lujing[closed].nowx+fangxiang[r,1]; {扩展形成新的坐标值}
y:=lujing[closed].nowy+fangxiang[r,2];
if not ((x>5)or(y>5) or (x<1) or (y<1) or (migong[y,x]<>0)) then begin
{未出界,未走过则可视为新的结点}
inc(open); {队列尾指针加1}
with lujing[open] do begin {记录新的结点数据}
nowx:=x; nowy:=y;
lastx:=lujing[closed].nowx;{新结点由哪个坐标扩展而来}
lasty:=lujing[closed].nowy;
dep:=lujing[closed].dep+1; {新结点走到第几步}
pre:=closed; {新结点由哪个结点扩展而来}
end;
migong[y,x]:=lujing[closed].dep+1; {当前结点的覆盖范围}
if (x=5) and (y=5) then begin {输出找到的第一种方案}
writeln('ok,thats all right');output;halt;end;
end;
end;
until closed>=open; {直到首指针大于等于尾指针,即所有结点已扩展完}
end.
【题目】火车调度问题
【参考程序】
const max=10;
type shuzu=array[1..max] of 0..max;
var stack,exitout:shuzu;
n,total:integer;
procedure output(exitout:shuzu);
var i:integer;
begin
for i:=1 to n do write(exitout[i]:2);writeln;
inc(total);
end;
procedure find(dep,have,rest,exit_weizhi:integer;stack,exitout:shuzu);
{dep:步数,have:入口处有多少辆车;rest:车站中有多少车;}
{exit_weizhi:从车站开出后,排在出口处的位置;}
{stack:车站中车辆情况数组;exitout:出口处车辆情况数组}
var i:integer;
begin {分入站,出站两种情况讨论}
if have>0 then begin {还有车未入站}
stack[rest+1]:=n+1-have; {入站}
if dep=2*n then output(exitout)
else find(dep+1,have-1,rest+1,exit_weizhi,stack,exitout);
end;
if rest>0 then begin {还有车可出站}
exitout[exit_weizhi+1]:=stack[rest]; {出站}
if dep=2*n then output(exitout) {经过2n步后,输出一种方案}
else find(dep+1,have,rest-1,exit_weizhi+1,stack,exitout);
end;
end;
begin
writeln('input n:');
readln(n);
fillchar(stack,sizeof(stack),0);
fillchar(exitout,sizeof(exitout),0);
total:=0;
find(1,n,0,0,stack,exitout);
writeln('total:',total);
readln;
end.
【解法2】用穷举二进制数串的方法完成.
uses crt;
var i,n,m,t:integer;
a,s,c:array[1..1000] of integer;
procedure test;
var t1,t2,k:integer;
notok:boolean;
begin
t1:=0;k:=0;t2:=0;
i:=0;
notok:=false;
repeat {二进制数串中,0表示出栈,1表示入栈}
i:=i+1; {数串中第I位}
if a[i]=1 then begin {第I位为1,则表示车要入栈}
inc(k); {栈中车数}
inc(t1); {入栈记录,T1为栈指针,S为栈数组}
s[t1]:=k;
end
else {第I位为0,车要出栈}
if t1<1 then notok:=true {已经无车可出,当然NOT_OK了}
else begin inc(t2);c[t2]:=s[t1];dec(t1);end;
{栈中有车,出栈,放到C数组中去,T2为C的指针,栈指针T1下调1}
until (i=2*n) or notok; {整个数串均已判完,或中途出现不OK的情况}
if (t1=0) and not notok then begin {该数串符合出入栈的规律则输出}
inc(m);write('[',m,']');
for i:=1 to t2 do write(c[i]:2);
writeln;
end;
end;
begin
clrscr; write('N=');readln(n);
m:=0;
for i:=1 to 2*n do a[i]:=0; {
repeat {循环产生N位二进制数串}
test; {判断该数串是否符合车出入栈的规律}
t:=2*n;
a[t]:=a[t]+1; {产生下一个二进制数串}
while (t>1) and (a[t]>1) do begin
a[t]:=0;dec(t);a[t]:=a[t]+1;
end;
until a[1]=2;
readln;
end.
N: 4 6 7 8
TOTAL: 14 132 429 1430
【题目】农夫过河。一个农夫带着一只狼，一只羊和一些菜过河。河边只有一条一船，由
于船太小，只能装下农夫和他的一样东西。在无人看管的情况下，狼要吃羊，羊
要吃菜，请问农夫如何才能使三样东西平安过河。
【算法分析】
将问题数字化。用１代表狼，２代表羊，３代表菜。则在河某一边物体的分布有以下
８种情况。
┏━━━━┯━┯━━━━━┯━━━━━━━━┯━━━┓
┃物体个数│０│ １ │ ２ │ ３ ┃
┠────┼─┼─┬─┬─┼──┬──┬──┼───┨
┃分布情况│０│１│２│３│1,2 │1,3 │2,3 │1,2,3 ┃
┠────┼─┼─┼─┼─┼──┼──┼──┼───┨
┃代码之和│０│１│２│３│３ │ ４ │ ５ │ ６ ┃
┠────┼─┼─┼─┼─┼──┼──┼──┼───┨
┃是否相克│ │ │ │ │相克│ │相克│ ┃
┗━━━━┷━┷━┷━┷━┷━━┷━━┷━━┷━━━┛
当（两物体在一起而且）代码和为３或５时，必然是相克物体在一起的情况。
【参考程序】
const
wt:array[0..3]of string[5]=(' ', 'WOLF ','SHEEP','LEAVE');
var left,right:array[1..3] of integer ;
what,i,total,left_rest,right_rest:integer;
procedure print_left; {输出左岸的物体}
var i:integer;
begin
total:=total+1;
write('(',total,')'); {第几次渡河}
for i:=1 to 3 do write(wt[left[i]]);
write('|',' ':4);
end;
procedure print_right;{输出右岸的物体}
var i:integer;
begin
write(' ':4,'|');
for i:=1 to 3 do if right[i]<>0 then write(wt[right[i]]);
writeln;
end;
procedure print_back(who:integer); {右岸矛盾时，需从右岸捎物体→左岸}
var i:integer;
begin
for i:=1 to 3 do begin
if not ((i=who) or (right[i]=0)) then begin
{要捎回左岸的物体不会时刚刚从左岸带来的物体，也不会是不在右岸的物体}
what:=right[i];
right[i]:=0;
print_left; {输出返回过程}
write('<-',wt[i]);
print_right;
left[i]:=what; {物体到达左岸}
end;
end;
end;
begin
total:=0;
for i:=1 to 3 do begin left[i]:=i; right[i]:=0;end;
repeat
for i:=1 to 3 do {共有３种物体}
if left[i]<>0 then {第Ｉ种物体在左岸}
begin
what:=left[i];left[i]:=0; {what:放置将要过河的物体编号}
left_rest:=left[1]+left[2]+left[3]; {求左岸剩余的物体编号总和}
if (left_rest=3) or (left_rest=5) then left[i]:=what
{假如左岸矛盾，则不能带第Ｉ种过河，尝试下一物体}
else {否则可带过河}
begin print_left; {输出过河过程}
write('->',wt[i]);
print_right;
right[i]:=what; {物体到达右岸}
if left_rest=0 then halt; {左岸物体已悉数过河}
right_rest:=right[1]+right[2]+right[3];
{求右岸剩余的物体编号总和}
if (right_rest=3)or(right_rest=5) then print_back(i)
{右岸有矛盾，要捎物体回左岸}
else begin print_left; {右岸有矛盾，空手回左岸}
write('<-',' ':5);
print_right;
end;
end;
end;
until false; {不断往返}
end.
【题目】七段数码管问题。从一个数字变化到其相邻的数字只需要通过某些段(数目不限)
1 或拿走某些段(数目不限)来实现.但不允许既增加段又拿起段.
┏━┓ 例如:3可以变到9,也可以变到1
6┃ 7┃2 ━┓ ┏━┓ ━┓ ┃
┣━┫ ┃ ┃ ┃ ┃ ┃
5┃ ┃3 ━┫ → ┗━┫ ━┫ → ┃
┗━┛ ┃ ┃ ┃ ┃
4 ━┛ ━┛ ━┛ ┃
要求:(1)判断从某一数字可以变到其它九个数字中的哪几个.
(2)找出一种排列这十个数字的方案,便这样组成的十位数数值最小.
type kkk=set of 0..9;
const a:array[-1..9] of set of 1..7
=([5,6],[1,2,3,4,5,6],[2,3],[1,2,4,5,7],[1,2,3,4,7],[2,3,6,7],
[1,3,4,6,7],[1,3,4,5,6,7],[1,2,3],[1,2,3,4,5,6,7],[1,2,3,4,6,7]);
var
i,j:integer;
b:array[-2..9] of set of 0..9;
procedure number(p:string;s,l:integer;k:kkk);
{P:生成的数;s:用了几个数字;i:前一个是哪个数字;k:可用的数字}
var i:integer;
begin
for i:=0 to 9 do
if (i in k) and ( i in b[l]) then begin
{数字i未用过,且i可由前一个采用的数字变化而来}
if s=10 then begin writeln('Min:',p,i);readln;halt;end
else number(p+chr(48+i),s+1,i,k-[i]);
end;
end;
begin
for i:=1 to 9 do b[i]:=[];
b[-2]:=[0..9];
for i:=-1 to 8 do
for j:=i+1 to 9 do
if (a[i]<=a[j]) or (a[j]<=a[i]) then begin
b[i]:=b[i]+[j];
b[j]:=b[j]+[abs(i)];
end;
b[1]:=b[1]+b[-1];
for i:=0 to 9 do begin
write(i,' may turn to :');
for j:=0 to 9 do if j in b[i] then write(j,' ');
writeln;
end;
number('',1,-2,[0..9]);
end.
【题目】 把1-8这8个数放入下图8个格中,要求相邻的格(横,竖,对角线)上填的数不连续.
┌─┐
│①│
┌─┼─┼─┐
│②│③│④│
├─┼─┼─┤
│⑤│⑥│⑦│
└─┼─┼─┘
│⑧│
└─┘
【参考程序】
const lin:array[1..8] of set of 1..8 =
([3,2,4],[1,6,3,5],[5,7,1,2,4,6],[1,6,3,7],
[3,8,2,6],[2,4,3,5,7,8],[3,8,4,6],[5,7,6]);
var a:array[1..8] of integer;
total,i:integer; had:set of 1..8;
function ok(dep,i:integer):boolean; {判断是否能在第dep格放数字i}
var j:integer;
begin
ok:=true;
for j:=1 to 8 do {相邻且连续则不行}
if (j in lin[dep]) and (abs(i-a[j])=1) then ok:=false;
if i in had then ok:=false; {已用过的也不行}
end;
procedure output; {输出一种方案}
var j:integer;
begin
inc(total); write(total,':');
for j:=1 to 8 do write(a[j]:2);writeln;
end;
procedure find(dep:byte);
var i:byte;
begin
for i:=1 to 8 do begin {每一格可能放1-8这8个数字中的一个}
if ok(dep,i) then begin
a[dep]:=i; {把i放入格中}
had:=had+[i]; {设置已放过标志}
if (dep=8) then output
else find(dep+1);
a[dep]:=10; {回溯,恢复原状态}
had:=had-[i];
end;
end;
end;
begin
fillchar(a,sizeof(a),10);
total:=0; had:=[];
find(1);
writeln('End.');
end.
【题目】 在4×4的棋盘上放置8个棋,要求每一行,每一列上只能放置2个.
【参考程序1】
算法:8个棋子,填8次.深度为8.注意判断是否能放棋子时,两个两个为一行.
var a:array[1..8] of 0..4;
line,bz:array[1..4] of 0..2; {line数组:每行已放多少个的计数器}
{bz数组: 每列已放多少个的计数器}
total:integer;
procedure output; {输出}
var i:integer;
begin
inc(total); write(total,': ');
for i:=1 to 8 do write(a[i]); writeln;
end;
function ok(dep,i:integer):boolean;
begin
ok:=true;
if dep mod 2 =0 then {假如是某一行的第2个,其位置必定要在第1个之后}
if (i<=a[dep-1]) then ok:=false;
if (bz[i]=2) or(line[dep div 2]=2) then ok:=false;
{某行或某列已放满2个}
end;
procedure find(dep:integer);
var i:integer;
begin
for i:=1 to 4 do begin
if ok(dep,i) then begin
a[dep]:=i; {放在dep行i列}
inc(bz[i]); {某一列记数器加1}
inc(line[dep div 2]); {某一行记数器加1}
if dep=8 then output else find(dep+1);
dec(bz[i]); {回溯}
dec(line[dep div 2]);
a[dep]:=0;
end;
end;
end;
begin
total:=0; fillchar(a,sizeof(a),0); fillchar(bz,sizeof(bz),0);
find(1);
end.
【参考程序2】
算法:某一行的放法可能性是(1,2格),(1,3格),(1,4格)....共6种放法
const
fa:array[1..6] of array[1..2]of 1..4=((1,2),(1,3),(1,4),(2,3),(2,4),(3,4));
{六种可能放法的行坐标}
var
a:array[1..8] of 0..4;
bz:array[1..4] of 0..2; {列放了多少个的记数器}
total:integer;
procedure output;
var i:integer;
begin
inc(total);
write(total,': ');
for i:=1 to 8 do write(a[i]);
writeln;
end;
function ok(dep,i:integer):boolean;
begin
ok:=true; {判断现在的放法中,相应的两列是否已放够2个}
if (bz[fa[i,1]]=2) or (bz[fa[i,2]]=2) then ok:=false;
end;
procedure find(dep:integer);
var i:integer;
begin
for i:=1 to 6 do begin {共有6种可能放法}
if ok(dep,i) then begin
a[(dep-1)*2+1]:=fa[i,1];{一次连续放置2个}
a[(dep-1)*2+2]:=fa[i,2];
inc(bz[fa[i,1]]); {相应的两列,记数器均加1}
inc(bz[fa[i,2]]);
if dep=4 then output else find(dep+1);
dec(bz[fa[i,1]]); {回溯}
dec(bz[fa[i,2]]);
a[(dep-1)*2+1]:=0;
a[(dep-1)*2+2]:=0;
end;
end;
end;
begin
total:=0; fillchar(a,sizeof(a),0); fillchar(bz,sizeof(bz),0);
find(1);
end.
【题目】迷宫问题.求迷宫的路径.(深度优先搜索法)
【参考程序1】
const
Road:array[1..8,1..8]of 0..3=((1,0,0,0,0,0,0,0),
(0,1,1,1,1,0,1,0),
(0,0,0,0,1,0,1,0),
(0,1,0,0,0,0,1,0),
(0,1,0,1,1,0,1,0),
(0,1,0,0,0,0,1,1),
(0,1,0,0,1,0,0,0),
(0,1,1,1,1,1,1,0)); {迷宫数组}
FangXiang:array[1..4,1..2]of -1..1=((1,0),(0,1),(-1,0),(0,-1));{四个移动方向}
WayIn:array[1..2]of byte=(1,1); {入口坐标}
WayOut:array[1..2]of byte=(8,8); {出口坐标}
Var i,j,Total:integer;
Procedure Output;
var i,j:integer;
Begin
For i:=1 to 8 do begin
for j:=1 to 8 do begin
if Road[i,j]=1 then write(#219); {1:墙}
if Road[i,j]=2 then write(' '); {2:曾走过但不通的路}
if Road[i,j]=3 then write(#03) ; {3:沿途走过的畅通的路}
if Road[i,j]=0 then write(' ') ; {0:原本就可行的路}
end; writeln;
end; inc(total); {统计总数} readln;
end;
Function Ok(x,y,i:byte):boolean; {判断坐标(X,Y)在第I个方向上是否可行}
Var NewX,NewY:shortint;
Begin
Ok:=True;
Newx:=x+FangXiang[i,1];
Newy:=y+FangXiang[i,2];
If not((NewX in [1..8]) and (NewY in [1..8])) then Ok:=False; {超界?}
If Road[NewX,NewY]=3 then ok:=false; {是否已走过的路?}
If Road[NewX,NewY]=1 then ok:=false; {是否墙?}
End;
Procedure Howgo(x,y:integer);
Var i,NewX,NewY:integer;
Begin
For i:=1 to 4 do Begin {每一步均有4个方向可选择}
If Ok(x,y,i) then Begin {判断某一方向是否可前进}
Newx:=x+FangXiang[i,1]; {前进,产生新的坐标}
Newy:=y+FangXiang[i,2];
Road[Newx,Newy]:=3; {来到新位置后,设置已走过标志}
If (NewX=WayOut[1]) and(NewY=WayOut[2]) Then Output
Else Howgo(Newx,NewY); {如到出口则输出,否则下一步递归}
Road[Newx,Newy]:=2; {堵死某一方向,不让再走,以免打转}
end;
end;
End;
Begin
total:=0;
Road[wayin[1],wayin[2]]:=3; {入口坐标设置已走标志}
Howgo(wayin[1],wayin[2]); {从入口处开始搜索}
writeln('Total is ',total); {统计总数}
end.
【题目】一笔画问题
从某一点出发,经过每条边一次且仅一次.(具体图见高级本P160)
【参考程序】
const max=6;{顶点数为6}
type shuzu=array[1..max,1..max]of 0..max;
const a:shuzu {图的描述与定义 1:连通;0:不通}
=((0,1,0,1,1,1),
(1,0,1,0,1,0),
(0,1,0,1,1,1),
(1,0,1,0,1,1),
(1,1,1,1,0,0),
(1,0,1,1,0,0));
var
bianshu:array[1..max]of 0..max; {与每一条边相连的边数}
path:array[0..1000]of integer; {记录画法,只记录顶点}
zongbianshu,ii,first,i,total:integer;
procedure output(dep:integer); {输出各个顶点的画法顺序}
var sum,i,j:integer;
begin
inc(total);
writeln('total:',total);
for i:=0 to dep do write(Path[i]);writeln;
end;
function ok(now,i:integer;var next:integer):boolean;{判断第I条连接边是否已行过}
var j,jj:integer;
begin
j:=0; jj:=0;
while jj<>i do begin inc(j);if a[now,j]<>0 then inc(jj);end;
next:=j;
{判断当前顶点的第I条连接边的另一端是哪个顶点,找出后赋给NEXT传回}
ok:=true;
if (a[now,j]<>1) then ok:=false; {A[I,J]=0:原本不通}
end; { =2:曾走过}
procedure init; {初始化}
var i,j :integer;
begin
total:=0; {方案总数}
zongbianshu:=0; {总边数}
for i:=1 to max do
for j:=1 to max do
if a[i,j]<>0 then begin inc(bianshu[i]);inc(zongbianshu);end;
{求与每一边连接的边数bianshu[i]}
zongbianshu:=zongbianshu div 2; {图中的总边数}
end;
procedure find(dep,nowpoint:integer); {dep:画第几条边;nowpoint:现在所处的顶点}
var i,next,j:integer;
begin
for i:=1 to bianshu[nowpoint] do {与当前顶点有多少条相接,则有多少种走法}
if ok(nowpoint,i,next) then begin {与当前顶点相接的第I条边可行吗?}
{如果可行,其求出另一端点是NEXT}
a[nowpoint,next]:=2; a[next,nowpoint]:=2; {置成已走过标志}
path[dep]:=next; {记录顶点,方便输出}
if dep < zongbianshu then find(dep+1,next) {未搜索完每一条边}
else output(dep);
path[dep]:=0; {回溯}
a[nowpoint,next]:=1; a[next,nowpoint]:=1;
end;
begin
init; {初始化,求边数等}
for first:=1 to max do {分别从各个顶点出发,尝试一笔画}
fillchar(path,sizeof(path),0);
path[0]:=first; {记录其起始的顶点}
writeln('from point ',first,':');readln;
find(1,first); {从起始点first,一条边一条边地画下去}
end.
【题目】城市遍历问题.
给出六个城市的道路连接图,找出从某一城市出发,遍历每个城市一次且仅一次的最短路径
及其路程长度.(图见高级本P147}
【参考程序】
const
a:array[1..6,1..6]of 0..10 {城市间连接图.数字表示两城市间的路程}
=((0,4,8,0,0,0),
(4,0,3,4,6,0),
(8,3,0,2,2,0),
(0,4,2,0,4,9),
(0,6,2,4,0,4),
(0,0,0,9,4,0));
var
had:array[1..6]of boolean; {某个城市是否已到过}
pathmin,path:array[1..6]of integer; {记录遍历顺序}
ii,first,i,summin,total:integer;
procedure output(dep:integer); sum,i,j:integer;
sum:=0; i:=2 6 {求这条路的路程总长}
if sum><6 then find(dep+1)
else output(dep);
had[i]:=false; {回溯}
path[dep]:=0;
end;
end;
begin
for first:=1 to 6 do begin {轮流从每一个城市出发,寻找各自的最短路}
fillchar(had,sizeof(had),false);
fillchar(path,sizeof(path),0);
total:=0;
SumMin:=maxint; {最短路程}
path[1]:=first;had[first]:=true;{处理出发点的城市信息,记录在册并置到过标志}
find(2); {到下一城市}
writeln('from city ',first,' start,total is:',total,' the min sum:',summin);
for i:=1 to 6 do write(PathMin[i]);writeln; {输出某个城市出发的最短方案}
end;
end.
【题目】棋子移动问题
[参考程序]
const
n=3; {n<5}
type
ss=string[2*n+1];
ar=array[1..630]of ss;
var
a:ar;
f,z:array[1..630] of integer;
i,j,k,m,h,t,k1:integer;
s,d:ss;
q:boolean;
procedure print (x:integer);
var t:array[1..100] of integer;
y:integer;
begin
y:=0;
repeat
y:=y+1;
t[y]:=x;
x:=f[x];
until x=0;
writeln(a[t[y]]:2*n+4);
writeln(copy('-------------------------',1,2*n+5));
for x:=2 to y do writeln(x-1:2,':',a[t[y+1-x]]);
end;
begin
s:='_';d:='_';
for i:=1 to n do begin
s:='o'+s+'*';
d:='*'+d+'o';
end;
a[1]:=s;f[1]:=0;z[1]:=n+1;
q:=false;
i:=1;j:=2; t:=0;
repeat
for h:=1 to 4 do begin
k:=z[i];k1:=k;s:=a[i];
case h of
1:if k>1 then k1:=k-1;
2:if k<(2*n+1) then k1:=k+1;
3:if (k>2) and (s[k-1]<>s[k-2]) then k1:=k-2;
4:if (k<(2*n)) and(s[k+1]<>s[k+2]) then k1:=k+2;
end;
if k<>k1 then begin
s[k]:=s[k1];s[k1]:='_';
m:=1;
while (a[m]<>s) and (m< j-1) do m:=m+1;
if a[m] >>s then begin
a[j]:=s;f[j]:=i;z[j]:=k1;
if s=d then begin
print(j);
q:=true;
end;
j:=j+1;
end;
end;
end; {end for}
i:=i+1;
until q or (i=j);
readln;
end.
【题目】求集合元素问题（1,2x+1,3X+1类)
某集合Ａ中的元素有以下特征：
(1)数1是A中的元素
(2)如果X是A中的元素,则2x+1,3x+1也是A中的元素
(3)除了条件(1),(2)以外的所有元素均不是A中的元素
[参考程序1]
uses crt,dos;
var a:array[1..10000]of longint;
b:array[1..10000]of boolean;
times,n,m,long,i:longint;
hour1,minute1,second1,sec1001:word;
hour2,minute2,second2,sec1002:word;
begin
write('N=');readln(n);
{ gettime(hour1,minute1,second1,sec1001);
times:=minute1*60+second1;
writeln(minute1,':',second1);}
fillchar(b,sizeof(b),0);
a[1]:=1;m:=2;long:=1;
while long<=n do begin
for i:=1 to long do
if (a[i]*2=m-1) or (a[i]*3=m-1) then
if not b[m] then begin
inc(long);a[long]:=m;b[m]:=true;break;
end;
inc(m);
end;
{ gettime(hour2,minute2,second2,sec1002);
times:=minute2*60+second2-times;
writeln(minute2,':',second2);
writeln('Ok! Uses Time: ',times);}
for i:=1 to n do write(a[i],' ');
readln;
end.
[参考程序2]
uses crt;
const n=10000;
var a:array[1..n] of longint;
i,j,k,l,y:longint;
begin
clrscr;
fillchar(a,sizeof(a),0);
i:=1;j:=1;
a[i]:=1;
repeat
y:=2*a[i]+1;
k:=j;
while y〈a[k] do begin
a[k+1]:=a[k];
k:=k-1;
end;
if y>a[k] then begin
a[k+1]:=y;j:=j+1;
end
else for l:=k+1 to j do a[l]:=a[l+1];
j:=j+1;
a[j]:=3*a[i]+1;
inc(i);
until k>=n;
for i:=1 to n do begin
write(a[i],' ');
if (i mod 10 =0 ) or (i=n) then writeln
end;
end.
[参考程序3]
uses crt;
var a:array[1..10000]of longint;
n,i,one,another,long,s,x,y:longint;
begin
write('n=');readln(n);
a[1]:=1;long:=1;one:=1;another:=1;
while longy then begin s:=y;inc(another);end
else begin s:=x;inc(one);inc(another);end;
inc(long);a[long]:=s;
end;
for i:=1 to n do write(a[i],' ');
end.
[参考程序4]
var n:integer;
top,x:longint;
function init(x:longint):boolean;
begin
if x=1 then init:=true
else if((x-1)mod 2=0)and(init((x-1)div 2))
or((x-1)mod 3=0)and(init((x-1)div 3))then
init:=true
else init:=false;
end;
begin
write('input n:');
readln(n);
x:=0;
top:=0;
while top< n do begin
x:=x+1;
if init(x) then
top:=top+1;
write(x:8);
end;
write('output end.');
readln
end.
|
网友评论：（只显示最新10条。评论内容只代表网友观点，与本站立场无关！）