division

I have a table software and columns in it as dev_cost , sell_cost . If dev_cost is 16000 and sell_cost is 7500. How do I find the quantity of software to be sold in order to recover the dev_cost ? I have queried as below: select dev_cost / sell_cost from software ; It is returning 2 as the answer. But we need to get 3, right? What would be the query for that? Thanks in advance. Ilmari Karonen Your columns have integer types, and integer division truncates the result towards zero . To get an accurate result, you'll need to cast at least one of the values to float or decimal : select cast(dev

I don't quite understand why I don't get a division by zero exception: int d = 0; d /= d; I expected to get a division by zero exception but instead d == 1 . Why doesn't d /= d throw a division by zero exception when d == 0 ? C++ does not have a "Division by Zero" Exception to catch. The behavior you're observing is the result of Compiler optimizations: The compiler assumes Undefined Behavior doesn't happen Division by Zero in C++ is undefined behavior Therefore, code which can cause a Division by Zero is presumed to not do so. And, code which must cause a Division by Zero is presumed to never

I have a dataframe: >>> dt COL000 COL001 QT STK_ID RPT_Date STK000 20120331 2.6151 2.1467 1 20120630 4.0589 2.3442 2 20120930 4.4547 3.9204 3 20121231 4.1360 3.8559 4 STK001 20120331 -0.2178 0.9184 1 20120630 -1.9639 0.7900 2 20120930 -2.9147 1.0189 3 20121231 -2.5648 2.3743 4 STK002 20120331 -0.6426 0.9543 1 20120630 -0.3575 1.6085 2 20120930 -2.3549 0.7174 3 20121231 -3.4860 1.6324 4 And I want the columns values divided by 'QT' column, somewhat like this: dt = dt/dt.QT # pandas does not accept this syntax The desired output is: STK_ID RPT_Date COL000 COL001 QT STK000 20120331 2.615110188 2

I have a function f(x)=(1^1)*(2^2)*(3^3)*.....(x^x) i have to calculate (f(x)/(f(x-r)*f(r)))modulo c i can calculate f(x) and (f(x-r)*f(r)). assume f(x) is a and f(x-r)*f(r) is b. c is some number that is very larger. `` so i how can calculate (a/b)%c your f(x) is just ᴨ (PI cumulative multiplication) squared it is hard to write it in here so i will deifine g(x0,x1) instead g(x0,x1)=x0*(x0+1)*(x0+2)*...*x1 so: f(x)=g(1,x)^2 computing h(x,r,c)=f(x)/(f(x-r)*f(r))%c when you rewrite it to g() you get: h(x,r,c)=((g(1,x)/(g(1,x-r)*g(1,r)))^2)%c now simplify (and lower magnitude) as much as you can

Possible Duplicate: What is the behavior of integer division in C? When I do a division calculation with an answer that's supposed to be between 1 and 0, it always returns 0. Try this: float a = 1; float b = 2; float c = a / b; //variable divide by variable gives 0.5 as it should float d = 1 / 2; //number divide by number gives 0 float e = 4 / 2; NSLog(@"%f %f %f %f %f", a,b,c, d, e); I get this from the console: 2013-01-17 14:24:17.512 MyProject[1798:c07] 1.000000 2.000000 0.500000 0.000000 0.500000 I have no idea what is going on. You're dividing integer with integer. The result will be an