pascal enters N numbers and displays them on the screen in the reverse order from when they were ent

Just use an array.

var a:array[1..1000]of longint;

b:array[1..1000]of longint;

i,x,n:longint;

beginread(n);

for i:=1 to n do read(a[i]);

for i:=n downto 1 do b[i]:=a[n-i+1];

for i:=1 to n do write(b[i],' ');

end.This method is a little bit more violent, and you can also swap directly on an array.

var a:array[1..1000]of longint;

i,x,n,j:longint;

beginread(n);

for i:=1 to n do read(a[i]);

for i:=1 to (n div 2)dobeginx:=a[n-i+1];

a[n-i+1]:=a[i];

a[i]:=x;

end;for i:=1 to n do write(a[i],' ');

end.This method will work a little faster.

Pure hand begging.

The crux of the matter is to deal with the 0 at the end, and the procedure is as follows:

Enter a positive integer and output it in reverse order, with a space after each digit stop.

Enter a positive integer n, and you can assume that n is in the int range.

The output outputs n in reverse order, with a space after each number, and the output occupies a row. /#include

int main()

int i;

int n;

scanf("%d",&n);

int num = 0;

for(i = 0; ;i ++

num = n % 10;

n /=10;

printf("%d",num);

if( n < 1)

break;

return 0;

The crux of the matter is to deal with the 0 at the end, and the procedure is as follows:

Enter a positive integer and output it in reverse order, with a space after each number.

Input a positive integer n, and you can assume that n is output in the int range and output n in reverse order, with a space after each number, and the output takes up a row. /#include

int main()

int i;

int n;

scanf("%d",&n);

int num = 0;

for(i = 0; ;i ++

num = n % 10;

n /= 10;

printf("%d",num);

if( n < 1)

break;

return 0;

It's super easy to handle it as a string

vari,j:longint;

n:string;

beginreadln(n);

for i:=length(n) downto 1 doif n[i]<>'0' then break;This step is to remove the precedent 0

for j:=i downto 1 do write(n[j]);

writeln;

end.Of course, you can use arrays as well.

Define n as an array of characters n:array [1..9] of char;

Do this by also noting the length.

Of course, it is possible to do it directly with plastic surgery!

varn:longint;

beginreadln(n);

while (n mod 10=0) do n:=n div 10;This step is to remove the precedent 0

while n<>0 do

beginwrite(n mod 10);output last digit n:=n div 10; Delete the last digit end;

writeln;

end.Choose from three methods

The crux of the matter is to deal with the 0

The procedure is as follows: var s:string;

i,j:integer;

beginreadln(s);Integers are stored as strings;

for i:=length(s) downto 1 doif s[i]<>'0' then break;This step is to find the position of the non-0 number on the far right of the S string and write it down with i;

for j:=i downto 1 do write(s[j]);

readln;

end.