division

问题 I am trying to count the number of digits of an input. However, whenever I input 10 or 11 or any two digit number, the output is 325 . Why doesn't it work? inputnumber = int(input()) countnumber = inputnumber digitcount = 0 while countnumber > 0: digitcount += 1 countnumber = countnumber/10 print(digitcount) # result is 325 when input is 10 or 11 回答1: Your error mainly happened here: countnumber=countnumber/10 Note that you are intending to do integer division. Single-slash division in Python

问题 This question already has an answer here : Matlab element-wise division by zero (1 answer) Closed 3 years ago . I'm using matlab and I have two matrices : G = 1 1 1 1 1 1 1 1 and the scond: m = 4 4 4 4 0 0 0 0 I want this result : x = 1/4 1/4 1/4 1/4 0 0 0 0 What I did so far is this : x = G ./ m But it returns : x = 1/4 1/4 1/4 1/4 NaN NaN NaN NaN How can I avoid the divison by zero by placing a default value "0" if there is a division by zero ? 回答1: You can convert the NaNs back to zero: x

问题 This question already has an answer here : Matlab element-wise division by zero (1 answer) Closed 3 years ago . I'm using matlab and I have two matrices : G = 1 1 1 1 1 1 1 1 and the scond: m = 4 4 4 4 0 0 0 0 I want this result : x = 1/4 1/4 1/4 1/4 0 0 0 0 What I did so far is this : x = G ./ m But it returns : x = 1/4 1/4 1/4 1/4 NaN NaN NaN NaN How can I avoid the divison by zero by placing a default value "0" if there is a division by zero ? 回答1: You can convert the NaNs back to zero: x

问题 I need some division algorithm which can handle big integers (128-bit). I've already asked how to do it via bit shifting operators. However, my current implementation seems to ask for a better approach Basically, I store numbers as two long long unsigned int 's in the format A * 2 ^ 64 + B with B < 2 ^ 64 . This number is divisible by 24 and I want to divide it by 24 . My current approach is to transform it like A * 2 ^ 64 + B A B -------------- = ---- * 2^64 + ---- 24 24 24 A A mod 24 B B

问题 mov eax, 0 mov ebx, 0 mov edx, 0 mov ax, 31 mul cx mov bx, 12 div bx add bp, ax mov eax, 0 mov ebx, 0 mov bp, bp mov al, 7 div al can anyone tell me whats wrong with the div al instruction in this block of code, so as I'm debugging every number of bp i calculated, when i divide by al it give me 1 as the remainder, why is this happen? the remainder should be store back to ah register thank in advance edited code : mov eax, 0 mov ebx, 0 mov edx, 0 mov ax, 31 mul cx mov bx, 12 div bx add bp, ax

问题 I am currently using a formula to calculate a variable, which looks like: =(Fields!MontantInitialRegime.Value/Fields!MontantInitial.Value)*Fields!SoldeFacture.Value It works well, except for the cases where both numerator and denominator equal zero. I've done some research and I've tried to solve it by using the formula below, which doesn't work. =IIF(Fields!MontantInitialRegime.Value = 0, Fields!SoldeFacture.Value, Fields!MontantInitialRegime.Value / IIF(Fields!MontantInitial.Value = 0,

问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: implement division with bit wise operator I recently got into more depth by bitwise functions, and started to implement basic arithmetic functions with bitwise operators. So far I have gotten (+, -, *) . However I'm not really sure how to approach division. I know that I could somehow use multiplication instead, but not sure how to approach this using that method either. So how would I implement division using

问题 Maybe I can't see obvious thing but: int x1 = 2; int y1 = 4; int x2 = 11; int y2 = 7; double res = (y2-y1)/(x2-x1); System.out.println(res); Output: 0.0 Why? 回答1: you need to initially define those variables as doubles and it should work. 回答2: The problem is you're doing integer arithmetic. You need a typecast in order to convert the numerator or denominator to floating point first (e.g.): int x1 = 2; int y1 = 4; int x2 = 11; int y2 = 7; double res = (double)(y2-y1)/(x2-x1); System.out

问题 Accidentally I stumbled upon the following difference. The division 50/0.02 returns a float: 2500.0 However a floor division 50//0.02 returns - as it seems to me - a wrong answer: 2499.0 Can anybody explain how this difference is caused? 回答1: This is due to python's floating point precision problems. 0.02 is actually as Martijn Pieters suggested 0.02000000000000000041633363423443370265886187553405762 and so this when divided by 50 gives a value like 2499.99999999999994795875 , and with floor

问题 Accidentally I stumbled upon the following difference. The division 50/0.02 returns a float: 2500.0 However a floor division 50//0.02 returns - as it seems to me - a wrong answer: 2499.0 Can anybody explain how this difference is caused? 回答1: This is due to python's floating point precision problems. 0.02 is actually as Martijn Pieters suggested 0.02000000000000000041633363423443370265886187553405762 and so this when divided by 50 gives a value like 2499.99999999999994795875 , and with floor