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Anonymous users2024-02-07varn,i,j:integer;
beginreadln(n);
for i:=1 to n do beginfor j:=1 to i do beginwrite（‘*
end；writeln；
end；readln；
end.Upstairs is wrong, what about the n variable.
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Anonymous users2024-02-06That's a bit of a problem.
Why 4 rows are 2 stars and 5 rows are one star.
You don't make it clear.
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Anonymous users2024-02-05This reading is wrong.
Follow your way, and he will become.
p[1].name='jason 1 1'
p[1].x='herry'So it doesn't match the type, and you should do that.
Put your for i:=1 to n do
with p[i] do
readln(name,x,y);
Change to for i:=1 to n do
with p[i] do
beginreadln(st);
name:=copy(st,1,pos(' ',st)-1);
delete(st,1,pos(' ',st));
val(copy(st,1,pos(' ',st)-1),x,code);
delete(st,1,pos(' ',st));
val(st,y,code);
end;It's just a normal string process.
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Anonymous users2024-02-04vari,k,c:char;
n:longint;
beginreadln(n);
c:=chr(n+64);Calculate the last letter for i:= in the last row'a'to c do enumerates the last letter of each row begin
for k:=c downto succ(i) do write(' ');The first letter is also the last letter in the row, and the letter before this letter is a space, so the output space, succ() is used to take the successor, and you can get a letter after i.
for k:=i downto 'a' do write(k);Print from the last letter of the row to A
for k:='b' to i do write(k);Print from B to the last letter of the row, because A has already been printed and only one A is in the center of each row, so don't print repeatedly.
writeln;After printing a row, wrap the line.
end;end.
The specific steps are as follows:
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