Задание 6.
Выполнить умножение методом анализа двух разрядов множителя одновременно, начиная с младших разрядов.
Использовать дополнительный код.
Решение
1
| 1
| 0
| 1
| 1
| 0
| 0
| 1
| 0
| 1
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
| 1
| 0
| 1
| 0
| 1
| 1
| 1
| 1
| 0
| 1
| 1
| 0
|
| +
|
|
|
|
|
|
|
|
|
|
|
|
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
|
|
|
|
|
|
|
|
|
|
|
| 1
| 1
| 0
| 1
| 1
| 0
| 0
| 1
| 0
| 1
|
|
|
|
|
|
|
|
|
|
|
| 1
| 1
| 0
| 1
| 1
| 0
| 0
| 1
| 0
| 1
|
|
|
|
|
|
|
|
|
|
|
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
|
|
|
|
|
|
|
|
|
|
|
| 1
| 1
| 0
| 1
| 1
| 0
| 0
| 1
| 0
| 1
|
|
|
|
|
|
|
|
|
|
|
| 1
| 1
| 0
| 1
| 1
| 0
| 0
| 1
| 0
| 1
|
|
|
|
|
|
|
|
|
|
|
| 1
| 1
| 0
| 1
| 1
| 0
| 0
| 1
| 0
| 1
|
|
|
|
|
|
|
|
|
|
|
| 1
| 1
| 0
| 1
| 1
| 0
| 0
| 1
| 0
| 1
|
|
|
|
|
|
|
|
|
|
|
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
|
|
|
|
|
|
|
|
|
|
|
| 1
| 1
| 0
| 1
| 1
| 0
| 0
| 1
| 0
| 1
|
|
|
|
|
|
|
|
|
|
|
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
|
|
|
|
|
|
|
|
|
|
|
| 1
| 1
| 0
| 1
| 1
| 0
| 0
| 1
| 0
| 1
|
|
|
|
|
|
|
|
|
|
|
|
| 1
| 0
| 0
| 1
| 0
| 1
| 0
| 0
| 1
| 1
| 0
| 1
| 0
| 1
| 0
| 0
| 0
| 0
| 1
| 1
| 1
| 0
| Ответ: 10010100110101000011102 |