19.07.2007, 15:16
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это ответ
Здравствуйте, Ласточка!
Решение задачи 3 в приложении. Действия А - Д сделаны в виде процедур.
Приложение:
program Project2;const n = 4; m = 5;var a: array[1..n, 1..m] of integer; i, j: integer;procedure maxabs;var maxabs, jmaxabs: integer;begin for i := 1 to n do begin maxabs := abs(a[i, 1]); jmaxabs := 1; for j := 1 to m do if abs(a[i, j]) > maxabs then begin maxabs := a[i, j]; jmaxabs := j; end; a[i, jmaxabs] := a[i, jmaxabs] * (-1); end;end;procedure last;var ilast: integer;begin for j := 1 to m do begin ilast := 0; for i := 1 to n do if a[i, j] < 0 then ilast := i; if ilast > 0 then a[ilast, j] := 0; end;end;procedure mul;var first, last: integer;begin for i := 1 to n do begin first := a[i, 1]; last := a[i, m]; for j := 1 to m do begin if a[i, j] > 0 then a[i, j] := a[i, j] * first else if a[i, j] < 0 then a[i, j] := a[i, j] * last; end; end;end;procedure replace;var k, l: integer;begin writeln(‘Vvedite k‘); readln(k); writeln(‘Vvedite l‘); readln(l); for j := 1 to m do a[k, j] := a[k, j] * (-1); for i := 1 to n do a[i, l] := a[i, l] * (-1);end;procedure add;var k1, k2: integer;begin writeln(‘Vvedite k1‘); readln(k1); writeln(‘Vvedite k2‘); readln(k2); for i := 1 to n do inc(a[i, k1], a[i, k2]);end;begin //randomize; {инициализация} for i := 1 to n do for j := 1 to m do a[i, j] := random(12) - 3; {вывод} for i := 1 to n do begin for j := 1 to m do write(a[i, j], ‘ ‘); writeln; end; writeln; maxabs;{А} last;{Б} mul;{В} replace;{Г} add;{Д} {вывод} for i := 1 to n do begin for j := 1 to m do write(a[i, j], ‘ ‘); writeln; end; readln;end.