If you are doing $R$'s complement, to get the negative of a number, change each digit $d$ to $R-1-d$, then add $1$ at the end, you do fine if $R$ is even. If $R$ is odd the divisions by $2$ don't come out even and you can't look at the leading digit to find the sign. For $R=4$, for example, the greatest three digit positive number is $133_4=31_{10}=\frac {4^3}2-1$ and the smallest negative number is $200_4=-32_{10}=-\frac {4^3}2$. For $R=3$ and three digit numbers, you would like the largest positive to be $111_3=13_{10}$ and the smallest negative to be $112_3=-13_{10}$, which fits your equation to within $1$, but you have to look at the whole number to know the sign.
There are a plethora of (poorly-tagged) questions on math.SE about twos-complement binary numbers. In summary, the usual definition of twos complement is, to convert a negative number $-N$ to twos complement, you write $N$ (the magnitude of the number) in binary, "flip" (take the complement of) all the bits, then add $1$ (a number with $1$ in the least significant place and $0$ everywhere else).
The conversion in the reverse direction is: "flip" all the bits
and add one, treat the result as a positive number, convert it to decimal,
and then flip the sign of the number so that you get back the original
negative number. So, for example, $1011\,1101$
in eight-bit twos-complement is $-0100\,0010_2 +1 = -0100\,0011_2 = -67_{10}.$
For ones complement you just flip all the bits. So, for example, $1011\,1101$
in eight-bit ones-complement is $-0100\,0010_2 = -66_{10}.$
Personally, I prefer to think of a twos-complement negative number $-N$ in $w$ bits as simply $2^w - N$. The ones-complement representation of $-N$ is then
$(2^w - 1) - N$.
Best Answer
I'll stick to 8-bit quantities, but the same applies in general.
The key to understanding two's complement is to note that we have a set of finitely many (in particular, $2^8$) values in which there is a sensible notion of addition by $1$ that allows us to cycle through all of the numbers. In particular, we have a system of modular arithmetic, in this case modulo $2^8 = 256$.
Intuitively, arithmetic modulo $n$ is a system of addition (and subtraction) in which overflow and underflow cause you to "cycle back" to a value from $0$ to $n-1$. A classic example is the usual "clock arithmetic", which is to say arithmetic modulo $12$.
For example, if it is $11\!:\!00$ now, then three hours later it will be $2\!:\!00$, since $$ 11 + 3 = 14 \equiv 2 \pmod {12} $$ and similarly, if it is $1\!:\!00$, then $4$ hours ago it was $9$ since $$ 1 - 4 = -3 \equiv 9 \pmod{12} $$ Notice that subtracting $4$ hours on the clock is the same as adding $12 - 4 = 8$ hours. In particular, we could have computed the above as follows: $$ 1 - 4 \equiv 1 + 8 = 9 \pmod{12} $$ That is: when performing arithmetic modulo $n$, we can subtract $x$ by adding $n-x$.
Now, let's apply this idea modulo $256$. How do you subtract $3$? Well, by the above logic, this is the same as adding $256 - 3 = 253$. In binary notation, we could say that subtracting $00000011$ is the same as adding $$ 1\overbrace{00000000}^8 - 00000011 = 1 + \overbrace{11111111}^8 - 00000011 = 11111101 $$ and there's your two's complement: the calculation $(11111111 - 00000011)$ "flips the bits" of $00000011$, and we add $1$ to this result.
Note 1: In the context of arithmetic with signed integers, we don't think of $11111101$ as being $253$ in our $8$-bit system, we instead consider it to represent the number $-3$. Rather than having our numbers go from $0$ to $255$ around a clock, we have them go from $-128$ to $127$, where $-x$ occupies the same spot that $n - x$ would occupy for values of $x$ from $1$ to $128$.
Succinctly, this amounts to saying that a number with 8 binary digits is deemed negative if and only if its leading digit (its "most significant" digit) is a $1$. For this reason, the leading digit is referred to as the "sign bit" in this context.
Note 2: An interesting infinite analog to the two's complement system of subtraction is that of infinite series 2-adic numbers. In particular, we can say something strange like $$ \dots 11111 = -1 $$ since $\dots 11111$ is the "infinite two's complement" of $1$.