模除

abs(i) - abs(j) < abs((i div j) * j) <= abs(i)

+10 / +3 == +3 +10 % +3 == +1 -10 / +3 == -3 -10 % +3 == -1
+10 / -3 == -3 +10 % -3 == +1 -10 / -3 == +3 -10 % -3 == -1

When integers are divided and the division is inexact, if both operands are positive the result of the / operator is the largest integer less than the algebraic quotient and the result of the % operator is positive. If either operand is negative, whether the result of the / operator is the largest integer less than or equal to the algebraic quotient or the smallest integer greater than or equal to the algebraic quotient is implementation-defined, as is the sign of the result of the % operator.

#include <stdio.h>

int main(void)
{
int x = -1,
y = +3;
if ((x%y > 0) ||
    ((x+y)%y == x%y))
    printf("This is a C90 translator behaving differently than C99\n");
}

The origin of this practice seems to have been a desire to map C’s division directly to the “natural” behavior of the target instruction set, whatever it may be, without requiring extra code overhead that might be necessary to check for special cases and enforce a particular behavior. However, the argument that Fortran programmers are unpleasantly surprised by this aspect of C and that there would be negligible impact on code efficiency was accepted by WG14, who agreed to require Fortran-like behavior in C99.

${\displaystyle x}$

${\displaystyle \left\lfloor x\right\rfloor }$ = the greatest integer less than or equal to ${\displaystyle x}$ (the floor of ${\displaystyle x}$);

${\displaystyle \left\lceil x\right\rceil }$ = the least integer greater than or equal to ${\displaystyle x}$ (the ceiling of ${\displaystyle x}$).

${\displaystyle \left\lceil x\right\rceil =\left\lfloor x\right\rfloor }$	if and only if ${\displaystyle x}$ is an integer,
${\displaystyle \left\lceil x\right\rceil =\left\lfloor x\right\rfloor +1}$	if and only if ${\displaystyle x}$ is not an integer;
${\displaystyle \left\lfloor -x\right\rfloor =-\left\lceil x\right\rceil }$;	${\displaystyle x-1<\left\lfloor x\right\rfloor \leq x\leq \left\lceil x\right\rceil <x+1}$.

……
If ${\displaystyle x}$ and ${\displaystyle y}$ are any real numbers, we define the following binary operation:

${\displaystyle x{\bmod {y}}=x-y\left\lfloor x/y\right\rfloor ,\ {\text{if }}y\neq 0}$;  ${\displaystyle x{\bmod {0}}=x}$.  (1)

From this definition we can see that, when ${\displaystyle y\neq 0}$,

${\displaystyle 0\leq {\frac {x}{y}}-\left\lfloor {\frac {x}{y}}\right\rfloor ={\frac {x{\bmod {y}}}{y}}<1}$.  (2)

Consequently
a) if ${\displaystyle y>0}$, then ${\displaystyle 0\leq x{\bmod {y}}<y}$;
b) if ${\displaystyle y<0}$, then ${\displaystyle 0\geq x{\bmod {y}}>y}$;
c) the quantity ${\displaystyle x-(x{\bmod {y}})}$ is an integral multiple of ${\displaystyle y}$.
We call ${\displaystyle x{\bmod {y}}}$ the remainder when ${\displaystyle x}$ is divided by ${\displaystyle y}$; similarly, we call ${\displaystyle \left\lfloor x/y\right\rfloor }$ the quotient.
When ${\displaystyle x}$ and ${\displaystyle y}$ are integers, “${\displaystyle {\text{mod}}}$” is therefore a familiar operation:

${\displaystyle 5{\bmod {3}}=2}$,  ${\displaystyle 18{\bmod {3}}=0}$,  ${\displaystyle -2{\bmod {3}}=1}$.  (3)

We have ${\displaystyle x{\bmod {y}}=0}$ if and only if ${\displaystyle x}$ is a multiple of ${\displaystyle y}$, that is, if and only if ${\displaystyle x}$ is divisible by ${\displaystyle y}$. The notation ${\displaystyle y\backslash x}$, read “${\displaystyle y}$ divides ${\displaystyle x}$,” means that ${\displaystyle y}$ is a positive integer and ${\displaystyle x{\bmod {y}}=0}$.
The “${\displaystyle {\text{mod}}}$” operation is useful also when ${\displaystyle x}$ and ${\displaystyle y}$ take arbitrary real values. For example, with trigonometric functions we can write

${\displaystyle \tan x=\tan(x{\bmod {\pi }})}$.

The quantity ${\displaystyle x{\bmod {1}}}$ is the fractional part of ${\displaystyle x}$; we have, by Eq. (1),

${\displaystyle x=\left\lfloor x\right\rfloor +(x{\bmod {1}})}$.

THEOREM. For any real numbers ${\displaystyle D}$ and ${\displaystyle d}$ with ${\displaystyle d\neq 0}$, there exists a unique pair of numbers ${\displaystyle q,r}$ satisfying the following conditions:

(a) ${\displaystyle q\in \mathbb {Z} }$;

(b) ${\displaystyle D=d\cdot q+r}$;

(c) ${\displaystyle 0\leq r<|d|}$.

${\displaystyle D\operatorname {div} d=q}$

${\displaystyle D{\bmod {d}}=r}$

${\displaystyle {\begin{aligned}D\operatorname {div} \,(-d)&=-(D\operatorname {div} d)\\D{\bmod {\left(-d\right)}}&=D{\bmod {d}}\\(-D)\operatorname {div} d&=-(D\operatorname {div} d)-I\cdot J\\(-D){\bmod {d}}&=-(D{\bmod {d}})+d\cdot I\cdot J\end{aligned}}}$

${\displaystyle J=\operatorname {sign} d}$

${\displaystyle I}$

${\displaystyle I={\text{if }}D/d\in \mathbb {Z} {\text{ then }}0{\text{ else }}1}$