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How did they come up with the equation for calculating module of 2 numbers “a%b = a-b*(a//b)”
Asking the user for input until they give a valid responsehow come aSet.add(something) updates the original set even its done in a different function?Boolean testing: Python prints '1' or 'True'How can I use lists in an equation?Calculating multiple exponential curve equations in dataframe with groupbyHow do I plot a parametric equation in Python?How to solve an equation which has a solution variable in form of function at both side of equationCoding Problem using Python - Laplace equation, electrostatics. How to run a function only in a masked part of an arrayHow do I tell which module a function comes from?Calculate the number of times a combination is used in a log
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
if we try different example you can see it satisfy the equation
print(13 % -4)
print(-10 % 9)
the question is where did they get the equation
a%b = b*(a//b)-a from
python-3.x
add a comment |
if we try different example you can see it satisfy the equation
print(13 % -4)
print(-10 % 9)
the question is where did they get the equation
a%b = b*(a//b)-a from
python-3.x
The equation is actuallya%b = a-b*(a//b)
– Tamas Hegedus
Mar 24 at 1:22
add a comment |
if we try different example you can see it satisfy the equation
print(13 % -4)
print(-10 % 9)
the question is where did they get the equation
a%b = b*(a//b)-a from
python-3.x
if we try different example you can see it satisfy the equation
print(13 % -4)
print(-10 % 9)
the question is where did they get the equation
a%b = b*(a//b)-a from
python-3.x
python-3.x
edited Mar 24 at 1:24
jamil
asked Mar 24 at 1:10
jamiljamil
215
215
The equation is actuallya%b = a-b*(a//b)
– Tamas Hegedus
Mar 24 at 1:22
add a comment |
The equation is actuallya%b = a-b*(a//b)
– Tamas Hegedus
Mar 24 at 1:22
The equation is actually
a%b = a-b*(a//b)
– Tamas Hegedus
Mar 24 at 1:22
The equation is actually
a%b = a-b*(a//b)
– Tamas Hegedus
Mar 24 at 1:22
add a comment |
1 Answer
1
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oldest
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It is the intuitive way to define a "remainder" operation, and also it's actually in the python documentation, you just have to tweak the equation a littlebit.
The floor division and modulo operators are connected by the following
identity: x == (x//y)*y + (x%y). Floor division and modulo are also
connected with the built-in function divmod(): divmod(x, y) == (x//y,
x%y).
So subtracting (x//y)*y
from x == (x//y)*y + (x%y)
gives you x%y == x - (x//y)*y
.
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is the intuitive way to define a "remainder" operation, and also it's actually in the python documentation, you just have to tweak the equation a littlebit.
The floor division and modulo operators are connected by the following
identity: x == (x//y)*y + (x%y). Floor division and modulo are also
connected with the built-in function divmod(): divmod(x, y) == (x//y,
x%y).
So subtracting (x//y)*y
from x == (x//y)*y + (x%y)
gives you x%y == x - (x//y)*y
.
add a comment |
It is the intuitive way to define a "remainder" operation, and also it's actually in the python documentation, you just have to tweak the equation a littlebit.
The floor division and modulo operators are connected by the following
identity: x == (x//y)*y + (x%y). Floor division and modulo are also
connected with the built-in function divmod(): divmod(x, y) == (x//y,
x%y).
So subtracting (x//y)*y
from x == (x//y)*y + (x%y)
gives you x%y == x - (x//y)*y
.
add a comment |
It is the intuitive way to define a "remainder" operation, and also it's actually in the python documentation, you just have to tweak the equation a littlebit.
The floor division and modulo operators are connected by the following
identity: x == (x//y)*y + (x%y). Floor division and modulo are also
connected with the built-in function divmod(): divmod(x, y) == (x//y,
x%y).
So subtracting (x//y)*y
from x == (x//y)*y + (x%y)
gives you x%y == x - (x//y)*y
.
It is the intuitive way to define a "remainder" operation, and also it's actually in the python documentation, you just have to tweak the equation a littlebit.
The floor division and modulo operators are connected by the following
identity: x == (x//y)*y + (x%y). Floor division and modulo are also
connected with the built-in function divmod(): divmod(x, y) == (x//y,
x%y).
So subtracting (x//y)*y
from x == (x//y)*y + (x%y)
gives you x%y == x - (x//y)*y
.
answered Mar 24 at 1:21
Tamas HegedusTamas Hegedus
19.4k43370
19.4k43370
add a comment |
add a comment |
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The equation is actually
a%b = a-b*(a//b)
– Tamas Hegedus
Mar 24 at 1:22