Ответы Mail.ru: помогите решить уравнение для 3-го класса 32-16...

Артикул: VVF53.32-16. Фото VVF53.32-16 Регулирующий клапан , фланцевый 2-х ходовой клапан DN32, kvs 16 Siemens.Италия / Серия А. 16:00.edit: your post about "16/16 division and 32/16 division which both take 18 cycles." -- dsPICs have a conditional subtract operation in assembly. Consider using this as your computational primitive.

Клуб итальянской Серии А заинтересован в услугах игрока ЦСКА

Answered 1 year ago. Originally Answered: What number comes next in the series: 2, 4, 8, 16, 32, 64?Решение 49-8*3-16=9. Вычитание 11100110001-00100101110 в двоичной.32-16÷n=32 16÷n= 32-30 16÷n=2 n=16÷2 n=8 Ответ:n=8.

algorithm - 64/32-bit division on a processor with 32/16-bit division

2,16. 0,64. #5. #8. 4,16. 32. 4,17. 3,65.b1=-64;b2=32. q=b2/b1=32/(-64)=-1/2 |q|<1.32-16:n=30 16:n=30 n=30*16 n=480 Ответ: n=480.

My copy of Knuth (The Art of Computer Programming) is at work, so I will be able to't check it till Monday, however that may be my first source. It has a complete phase on arithmetic.

edit: your submit about "16/16 division and 32/16 division which both take 18 cycles." -- dsPICs have a conditional subtract operation in assembly. Consider the usage of this as your computational primitive.

Also observe that if X = XH * 232 + XL and D = DH * 216 + DL, then if you're searching for

(Q,R) = X/D the place X = Q * D + R

where Q = QH * 216 + QL, R = RH * 216 + RL, then

XH * 232 + XL = DH * QH * 232 + (DL * QH + DH * QL) * 216 + (DL * QL) + RH * 216 + RL

This suggests (by means of taking a look at phrases which can be the prime 32 bits) to make use of the next procedure, similar to long department:

(QH, R0) = XH / (DH+1) -> XH = QH * (DH+1) + R0 [32/16 divide] R1 = X - (QH * 216) * D [requires a 16*32 multiply, a shift-left by 16, and a 64-bit subtract] calculate R1' = R1 - D * 216 whilst R1' >= 0, regulate QH upwards by way of 1, set R1 = R1', and goto step 3 (QL, R2) = (R1 >> 16) / (DH+1) -> R1 = QL * (DH+1) + R2 [32/16 divide] R3 = R1 - (QL * D) [calls for a 16*32 multiply and a 48-bit subtract] calculate R3' = R3 - D whilst R3' >= 0, regulate QL upwards through 1, set R3 = R3', and goto step 7

Your 32-bit quotient is the pair (QH,QL), and 32-bit the rest is R3.

(This assumes that the quotient is not higher than 32-bit, which you want to grasp ahead of time, and will simply test forward of time.)