Integer division that rounds up: Difference between revisions
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<math> | <math> | ||
\begin{align} | \begin{align} | ||
\left \ | \left \lfloor \frac{x+y-1}{y} \right \rfloor | ||
&= | &= | ||
\left \ | \left \lfloor \frac{x}{y} + \frac{y-1}y \right \rfloor \\ | ||
&= | &= | ||
\left \ | \left \lfloor \frac{ay}{y} + \frac{y-1}y \right \rfloor \\ | ||
&= | &= | ||
\frac{ay}{y} \\ | \frac{ay}{y} \\ |
Revision as of 11:11, 8 July 2024
Introduction
Usual integer division rounds down: for . To round up (if overflow is not an issue), you can use following algorithm with the usual roundig down division:
Proof
Proof is in two parts; 1st if divides , and if not. Note that usual integer division rounds down.
Part 1. If divides we have for some . Thus we have
because . This part is ok.
Part 2. If does not divide we have for some and . Thus we have