Задача 15 (стр.54)

CLS

INPUT n, m

DIM A(n, m), B(m)

FOR i = 1 TO n

FOR j = 1 TO m

INPUT A(i, j)

NEXT j, i

FOR j = 1 TO m

FOR i = 1 TO n

IF A(i, j) MOD 3 = 0 THEN K = A(i, j): EXIT FOR

NEXT i

B(j) = K

IF B(j) <> 0 THEN

FOR i = 1 TO n

A(i, j) = A(i, j) / B(j)

NEXT i

END IF

NEXT j

FOR i = 1 TO n

FOR j = 1 TO m

PRINT A(i, j);

NEXT j

PRINT

NEXT i

END

Задача 10 (стр.57)

CLS

INPUT n

DIM A(n, n)

FOR i = 1 TO n

FOR j = 1 TO n

INPUT A(i, j)

NEXT j, i

max = A(1, 1)

K1 = 1

K2 = 1

FOR i = 1 TO n

FOR j = 1 TO n

IF (A(i, j) > max AND i = j) OR (A(i, j) > max AND i + j = n + 1) THEN

max = A(i, j)

K1 = i

K2 = j

END IF

NEXT j, i

C = n MOD 2 + 1

SWAP A(K1, K2), A(C, C)

FOR i = 1 TO n

FOR j = 1 TO n

PRINT A(i, j);

NEXT j

PRINT

NEXT i

END

Задача 3 (стр.62)

CLS

INPUT n, m

DIM A(n, m)

FOR i = 1 TO n

FOR j = 1 TO m

INPUT A(i, j)

NEXT j, i

INPUT C

FOR i = 1 TO n

S = 0

FOR j = 1 TO m

S = S + A(i, j)

NEXT j

IF S <= C THEN K = i: EXIT FOR

NEXT i

FOR i = K TO n - 1

FOR j = 1 TO m

A(i, j) = A(i + 1, j)

NEXT j, i

n = n - 1

FOR i = 1 TO n

FOR j = 1 TO m

PRINT A(i, j);

NEXT j

PRINT

NEXT i

END

Задача 4 (стр. 67)

CLS

INPUT n, m

DIM A(n + 1, m + 1)

FOR i = 1 TO n

FOR j = 1 TO m

INPUT A(i, j)

NEXT j, i

min = A(1, 1)

K1 = 1

K2 = 1

FOR i = 1 TO n

FOR j = 1 TO m

IF A(i, j) < min THEN min = A(i, j): K1 = i: K2 = j

NEXT j, i

FOR i = n TO K1 STEP -1

FOR j = 1 TO m

A(i + 1, j) = A(i, j)

NEXT j, i

n = n + 1

FOR j = 1 TO m

A(K1, j) = 0

NEXT j

FOR j = m TO K2 STEP -1

FOR i = 1 TO n

A(i, j + 1) = A(i, j)

NEXT i, j

m = m + 1

FOR i = 1 TO n

A(i, K2) = 0

NEXT i

FOR i = 1 TO n

FOR j = 1 TO m

PRINT A(i, j);

NEXT j

PRINT

NEXT i

END

ЛИТЕРАТУРА

1. Г. Зельднер. Программируем на языке QBASIC 4.5. –М.: ABF, 1996.

2. О.И. Мельникова, А.Ю. Бонюшкина. Начала программирования на языке QBASIC. – М.: ЭКОМ, 1997.

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Примеры решения типовых задач с использованием одномерных массивов......