1樓：聽不清啊
var n,i,j,k:byte;
begin
write('n i j='); readln(n,i,j);
if abs(j-n)writeln(n-abs(k-n));
writeln;
for i:=1 to n+n-1 do
begin
for j:=1 to n+n-1 do
begin
if abs(j-n)write(n-abs(k-n):2);
end;
writeln;
end;
end.
執行截圖中沒有輸出指定的一個數，但是**中已經加入。可以直接複製執行。望採納。
2樓：匿名使用者
vara:array[1..9,1..9] of integer;
i,j,k,fj,fk:integer;
begin
fj:=0;
fk:=1;
for i:=1 to 4 do
begin
j:=i;
k:=i;
fj:=0;
fk:=0;
repeat
a[j,k]:=i;
if (j=i) and (k=i) thenbegin
fj:=0;
fk:=1;
end;
if (j=i) and (k+i>=10) thenbegin
fj:=1;
fk:=0;
end;
if (j+i>=10) and (k+i>=10) thenbegin
fj:=0;
fk:=-1;
end;
if (j+i>=10) and (k=i) thenbegin
fj:=-1;
fk:=0;
end;
j:=j+fj;
k:=k+fk;
until (j=i) and (k=i) ;
end;
a[5,5]:=5;
for j:=1 to 9 do
begin
for k:=1 to 9 do
write(a[j,k]:2);
writeln;
end;
writeln;
read(j,k);
writeln(a[j,k]);
end.
3樓：發動吧命運之骰
uses math;
const f:array[1..8,1..
2] of integer=((0,1),(0,-1),(1,0),(1,1),(1,-1),(-1,0),(-1,1),(-1,-1));
vardl:array[0..10000,1..2] of longint;
bz:array[0..100,0..100] of boolean;
a:array[0..100,0..100] of longint;
i,j,k,l,t,n,m,head,tail:longint;
begin
readln(n);
for i:=0 to n*2 do
begin
bz[0,i]:=true;
bz[i,0]:=true;
bz[n+1,i]:=true;
bz[i,n+1]:=true;
end;
l:=n;
t:=n;
bz[l,t]:=true;
a[l,t]:=0;
dl[1,1]:=l;
dl[1,2]:=t;
head:=1;
tail:=1;
while head<=tail do
begin
inc(k);
t:=tail+1;
for i:=head to tail do
begin
for j:=1 to 8 do
begin
if not bz[dl[i,1]+f[j,1],dl[i,2]+f[j,2]] then
begin
inc(tail);
bz[dl[i,1]+f[j,1],dl[i,2]+f[j,2]]:=true;
dl[tail,1]:=dl[i,1]+f[j,1];
dl[tail,2]:=dl[i,2]+f[j,2];
a[dl[tail,1],dl[tail,2]]:=k;
end;
end;
end;
head:=t;
end;
readln(i,j);
writeln(k-a[i,j]);
end.
大致思路是從中間開始向四周寬搜，有問題追問
4樓：匿名使用者
這是個9行9列的方陣，第5行第5列為中心，其值為5，由裡往外是中心對稱的。
vara:array[1..9,1..9] of integer;
i,j:integer;
begin
readln(i,j);
if (i=5)and(j=5) then a[i,j]:=5;
if ((abs(i-5)=4)and(abs(j-5)<=4))or((abs(j-5)=4)and(abs(i-5)<=4)) then a[i,j]:=1;
if ((abs(i-5)=3)and(abs(j-5)<=3))or((abs(j-5)=3)and(abs(i-5)<=3)) then a[i,j]:=2;
if ((abs(i-5)=2)and(abs(j-5)<=2))or((abs(j-5)=2)and(abs(i-5)<=2)) then a[i,j]:=3;
if ((abs(i-5)=1)and(abs(j-5)<=1))or((abs(j-5)=1)and(abs(i-5)<=1)) then a[i,j]:=4;
writeln(a[i,j]);
end.
按題目的意思，不需要用陣列，只需用i，j兩個變數，在上面程式的then後直接輸出即可。
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