Цитата(Lesenok @ 5.12.2008, 19:14  ) | | как делаются задачи такого рода? |
(модуль - MYUNIT.PAS)
| Код | unit myunit;
interface
const
size= 3;
type
E = integer;
matrix = array[1 .. size, 1 .. size] of E;
function min(const m: matrix): E;
function max(const m: matrix): E;
procedure add(const a, b: matrix; var res: matrix);
procedure sub(const a, b: matrix; var res: matrix);
procedure scale(const a: matrix; f: E; var res: matrix);
procedure print(const m: matrix);
implementation
function min(const m: matrix): E;
var
i, j: integer;
T: E;
begin
T := m[1, 1];
for i := 1 to size do
for j := 1 to size do
if m[i, j] < T then T := m[i, j];
min := T;
end;
function max(const m: matrix): E;
var
i, j: integer;
T: E;
begin
T := m[1, 1];
for i := 1 to size do
for j := 1 to size do
if m[i, j] > T then T := m[i, j];
max := T;
end;
procedure add(const a, b: matrix; var res: matrix);
var i, j: integer;
begin
for i := 1 to size do
for j := 1 to size do
res[i, j] := a[i, j] + b[i, j];
end;
procedure sub(const a, b: matrix; var res: matrix);
var i, j: integer;
begin
for i := 1 to size do
for j := 1 to size do
res[i, j] := a[i, j] - b[i, j];
end;
procedure scale(const a: matrix; f: E; var res: matrix);
var i, j: integer;
begin
for i := 1 to size do
for j := 1 to size do
res[i, j] := a[i, j] * f;
end;
procedure print(const m: matrix);
var i, j: integer;
begin
for i := 1 to size do begin
for j := 1 to size do
write(m[i, j]:5);
writeln
end;
end;
end. |
(основная программа)
| Код | uses myunit;
const
a: matrix = (
(1, 2, 3),
(4, 5, 6),
(6, 3, 4)
);
b: matrix = (
(4, 7, 9),
(2, 6, 5),
(5, 3, 8)
);
c: matrix = (
(4, 9, 2),
(5, 7, 1),
(3, 5, 2)
);
var
z, t1, t2: matrix;
begin
scale(b, max(a), t1); { t1 = b*max(a) }
scale(c, min(b), t2); { t2 = c*min(b) }
add(t1, t2, t1); { t1 = t1 + t2 = b*max(a) + c*min(b) }
scale(t1, max(b), t1); { t1 = t1 * max(b) }
scale(a, max(c), t2); { t2 = a * max(c) }
sub(t1, t2, t1); { t1 = t1 - t2 }
print(t1);
end. |
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