modulo

问题 I just read this topic (especially the last comments). Then I was wondering, why we actually need this was of giving the remainder. But it seems, that not many people "on google" were interested in that before... 回答1: If you're looking for reasons why you would want it, one is for what is known as "range reduction" Let's say you want sind function for computing the sine of an argument in degrees. A naive way to do this would be sind(x) = sin(x*pi/180) However pi here is not the true

问题 For example: int x = 35; if( x%2==1) { //do something } I just want to check the modulus value without assigning the result to x . Assume that value is in eax , so I can use DIV instruction, then put back the original value to eax etc.. but that seems inefficient. Can you suggest a better way? 回答1: To branch according to the modulo 2 of a value in al / ax / eax / rax : test al,1 jnz is_odd is_even: ; do something for even numbers. is_odd: ; do something for odd numbers. However, if all you

问题 Is there some cool algorithm with bit wise operations? 回答1: Often, the modulus and divide operations on a processor are the same thing. For instance, refer to http://jsimlo.sk/docs/cpu/index.php/div.html . This is the implementation of the divide instruction on Intel processors. 回答2: Most of the time, modulus is just computed by dividing the two numbers. The quotient is stored in one register, and the remainder is stored in the other register. You would go after the remainder. 回答3: If the

问题 Is there some cool algorithm with bit wise operations? 回答1: Often, the modulus and divide operations on a processor are the same thing. For instance, refer to http://jsimlo.sk/docs/cpu/index.php/div.html . This is the implementation of the divide instruction on Intel processors. 回答2: Most of the time, modulus is just computed by dividing the two numbers. The quotient is stored in one register, and the remainder is stored in the other register. You would go after the remainder. 回答3: If the

问题 Is it possible to use numpy's linalg.matrix_power with a modulo so the elements don't grow larger than a certain value? 回答1: In order to prevent overflow, you can use the fact that you get the same result if you first take the modulo of each of your input numbers; in fact: (M**k) mod p = ([M mod p]**k) mod p, for a matrix M . This comes from the following two fundamental identities, which are valid for integers x and y : (x+y) mod p = ([x mod p]+[y mod p]) mod p # All additions can be done on

问题 I've been trying to implement a modular exponentiator recently. I'm writing the code in VHDL, but I'm looking for advice of a more algorithmic nature. The main component of the modular exponentiator is a modular multiplier which I also have to implement myself. I haven't had any problems with the multiplication algorithm- it's just adding and shifting and I've done a good job of figuring out what all of my variables mean so that I can multiply in a pretty reasonable amount of time. The

问题 I'm brand new to Erlang. How do you do modulo (get the remainder of a division)? It's % in most C-like languages, but that designates a comment in Erlang. Several people answered with rem, which in most cases is fine. But I'm revisiting this because now I need to use negative numbers and rem gives you the remainder of a division, which is not the same as modulo for negative numbers. 回答1: In Erlang, 5 rem 3. gives 2 , and -5 rem 3. gives -2 . If I understand your question, you would want -5

问题 This question already has answers here : Why does C++ output negative numbers when using modulo? (3 answers) Closed 2 years ago . I have been writing a program for the following recurrence relation: An = 5An-1 - 2An-2 - An-3 + An-4 The output should be the Answer modulus 10^9 + 7.. I wrote a brute force approach for this one as follows... long long int t1=5, t2=9, t3=11, t4=13, sum; while(i--) { sum=((5*t4) - 2*t3 - t2 +t1)%MOD; t1=t2; t2=t3; t3=t4; t4=sum; } printf("%lld\n", sum); where MOD=

问题 I'm messing with the modulo operation in python and I understand that it will spit back what the remainder is. But what if the first number is smaller than the second? for instance 2 % 5 the answer is 2. How does that work? 2/5 = .4 回答1: Does this help 22 % 5 = 2 17 % 5 = 2 12 % 5 = 2 7 % 5 = 2 2 % 5 = 2 Maybe this 22 / 5 = 4 + 2/5 17 / 5 = 3 + 2/5 12 / 5 = 2 + 2/5 7 / 5 = 1 + 2/5 2 / 5 = 0 + 2/5 回答2: five goes into 2 zero times. 5*0 = 0 2-0 = 2. the answer is 2 回答3: 2 divided by 5 (integer

问题 This question already has answers here : Closed 9 years ago . Possible Duplicate: Recognizing when to use the mod operator What are the practical uses of modulus? I know what modulo division is. The first scenario which comes to my mind is to use it to find odd and even numbers, and clock arithmetic. But where else I could use it? 回答1: The most common use I've found is for "wrapping round" your array indices. For example, if you just want to cycle through an array repeatedly, you could use: