Найти максимальный элемент побочной диагонали

var
  a: array [1..5, 1..5] of integer;
  i, j, max, maxi: integer;
 
begin
  randomize;
  for i := 1 to 5
  do 
  begin
    for j := 1 to 5
    do 
    begin
      A[i, j] := random(50);
      write(A[i, j]:3);
    end;
    writeln;
  end;
  max := A[1, 1];
  for i := 1 to 5
    do
    for j := 1 to 5
    do 
    begin
      if i = j
        then if A[i, j] > max then begin max := A[i, j];maxi := i end;
    end;
  writeln;
  writeln(max);
  for i := 1 to 5
    do swap(A[maxi, i], A[i, maxi]);
  for i := 1 to 5
  do 
  begin
    for j := 1 to 5
      do
      write(A[i, j]:3);
    writeln;
  end;
end.

  max:=a[1,5]; for i=2 to 5 do if max<a[i,6-i] then max:=a[i,6-i];

  for i=1 to 5 do if max=a[i,6-i] then
    for j:=1 to 5 do swap(a[i,j],a[j,i]);

var
  b: array [1..5,1..5] of Boolean;
 
  for i:=1 to 5 do for j:=1 to 5 do b[i,j]:=max=a[i,j];
  for i:=1 to 5 do for j:=1 to 5 do if b[i,j] then
    for k:=1 to 5 do swap(a[i,k],a[k,j]);

procedure swap(var a, b: Integer);
var t: Integer;
begin
  t:=a; a:=b; b:=t;
end;

const m = 5;
var a: array[1..m, 1..m] of integer;
    i, j, t, max: integer;
begin
  randomize;
  writeln('Source array:');
  for i := 1 to m do
    begin
      for j := 1 to m do
        begin
          a[i, j] := -99 + random(199);
          write(a[i, j]:4)
        end;
      writeln
    end;
  max := 1;
  for i := 2 to m do
    if a[i, m - i + 1] > a[max, m - max + 1]
      then max := i;
  writeln('Max element of secondary diagonal: a[', max, ', ', m - max + 1, '] = ', a[max, m - max + 1]);
  for i := 1 to m do
    begin
      t := a[max, m - i + 1];
      a[max, m - i + 1] := a[i, m - max + 1];
      a[i, m - max + 1] := t
    end;
  writeln('Result:');
  for i := 1 to m do
    begin
      for j := 1 to m do write(a[i, j]:4);
      writeln
    end;
  readln
end.