
用pascal的语言 输入3 5 输出 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15
可以的话在做道输入35输出167121325811143491015第一道题目是输入35输出123451098761112131415...
可以的话在做道
输入3 5
输出1 6 7 12 13
2 5 8 11 14
3 4 9 10 15
第一道题目是输入3 5
输出1 2 3 4 5
10 9 8 7 6
11 12 13 14 15 展开
输入3 5
输出1 6 7 12 13
2 5 8 11 14
3 4 9 10 15
第一道题目是输入3 5
输出1 2 3 4 5
10 9 8 7 6
11 12 13 14 15 展开
展开全部
第一道
var a,b,i,j:longint;
begin
readln(a,b);
for i:=1 to a*b div 5 do
begin
if i mod 2=1 then
for j:=1+(i-1)*5 to 5+(i-1)*5 do
write(j,' ')
else
for j:=5+(i-1)*5 downto 1+(i-1)*5 do
write(j,' ');
writeln;
end;
end.
第二道:
var a,b,i,j,k:longint;
aa:array[0..10,0..10] of longint;
begin
k:=0;
readln(a,b);
for i:=1 to b do
begin
if i mod 2=1 then
for j:=1 to a do
begin
k:=k+1;
aa[j,i]:=k;
end
else for j:=a downto 1 do
begin
k:=k+1;
aa[j,i]:=k;
end;
end;
for i:=1 to a do
begin
for j:=1 to b do
write(aa[i,j],' ');
writeln;
end;
end.
辛辛苦苦打了半小时,求满意答案
var a,b,i,j:longint;
begin
readln(a,b);
for i:=1 to a*b div 5 do
begin
if i mod 2=1 then
for j:=1+(i-1)*5 to 5+(i-1)*5 do
write(j,' ')
else
for j:=5+(i-1)*5 downto 1+(i-1)*5 do
write(j,' ');
writeln;
end;
end.
第二道:
var a,b,i,j,k:longint;
aa:array[0..10,0..10] of longint;
begin
k:=0;
readln(a,b);
for i:=1 to b do
begin
if i mod 2=1 then
for j:=1 to a do
begin
k:=k+1;
aa[j,i]:=k;
end
else for j:=a downto 1 do
begin
k:=k+1;
aa[j,i]:=k;
end;
end;
for i:=1 to a do
begin
for j:=1 to b do
write(aa[i,j],' ');
writeln;
end;
end.
辛辛苦苦打了半小时,求满意答案
展开全部
螺旋方阵
const z=5;
var a:array[1..z,1..z] of integer;
i,j,d,k,l,m,n:integer;
begin
k:=z;
i:=1;j:=1; l:=0;m:=1;
for d:=1 to z*z do
begin
a[i,j]:=d;
l:=l+1;
if (k=l) and (m=1) then begin dec(k);m:=0;n:=1;l:=0;end;
if (k=l) and (n=1) then begin m:=-1;n:=0;l:=0;end;
if (k=l) and (m=-1) then begin m:=0;n:=-1;l:=0;dec(k);end;
if (k=l) and (n=-1) then begin m:=1;n:=0;;l:=0;end;
i:=i+n;
j:=j+m;
end;
for i:=1 to z do
begin
for j:=1 to z do
write(a[i,j]:4);
writeln;
end;
readln;
end.
const z=5;
var a:array[1..z,1..z] of integer;
i,j,d,k,l,m,n:integer;
begin
k:=z;
i:=1;j:=1; l:=0;m:=1;
for d:=1 to z*z do
begin
a[i,j]:=d;
l:=l+1;
if (k=l) and (m=1) then begin dec(k);m:=0;n:=1;l:=0;end;
if (k=l) and (n=1) then begin m:=-1;n:=0;l:=0;end;
if (k=l) and (m=-1) then begin m:=0;n:=-1;l:=0;dec(k);end;
if (k=l) and (n=-1) then begin m:=1;n:=0;;l:=0;end;
i:=i+n;
j:=j+m;
end;
for i:=1 to z do
begin
for j:=1 to z do
write(a[i,j]:4);
writeln;
end;
readln;
end.
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