Introduction

If you're new to the world of competitive programming, you may have noticed that some questions, typically combinatorial and probability questions, have this funny habit of asking you to calculate a huge number, then tell you that "because this number can be huge, please output it modulo $$$10^9 + 7$$$". Like, it's not enough that they ask you to calculate a number they know will overflow basic integer data types, but now you need to apply the modulo operation after that? Even worse are those that say you need to calculate a fraction $$$\frac pq$$$ and ask you to output $$$r$$$ where $$$r \cdot q = p \text{ mod } n$$$... not only do you have to calculate a fraction with huge numbers, how in the world are you going to find $$$r$$$?

Actually, the modulo is there to make the calculation easier, not harder. This may sound counterintuitive, but once you know how modular arithmetic works, you'll see why too. Soon you'll be solving these problems like second nature.

Terminology and notation

For convenience, I will define the notation $$$n \text{ mod } m$$$ to mean $$$n - \lfloor \dfrac nm \rfloor \cdot m$$$, where $$$\lfloor x \rfloor$$$ is the largest integer that doesn't exceed $$$x$$$. This may or may not correspond to the expression n % m in your programming language (% is often called the "modulo operator" but in some instances, it's more correct to call it the "remainder operator"). If -8 % 7 == 6, you're fine, but if it is -1, you'll need to adjust it by adding $$$m$$$ to any negative results. If you use Java/Kotlin, the standard library function Math.floorMod(n, m) does what we need.

Also for convenience, I will also define the $$$\text{ mod }$$$ operator to have lower precedence than addition or subtraction, thus $$$ax + b\ \text{ mod } m \Rightarrow (ax + b) \text{ mod } m$$$. This probably does not correspond with the precedence of the % operator.

The value $$$m$$$ after the modulo operator is known as the modulus

"Basic" arithmetic

First off, some important identities about the modulo operator:

$$$(a \text{ mod } m) + (b \text{ mod } m)\ \text{ mod } m = a + b\ \text{ mod } m$$$

$$$(a \text{ mod } m) - (b \text{ mod } m)\ \text{ mod } m = a - b\ \text{ mod } m$$$

$$$(a \text{ mod } m) \cdot (b \text{ mod } m)\ \text{ mod } m = a \cdot b\ \text{ mod } m$$$

These identities have the very important consequence in that you generally don't need to ever store the "true" values of the large numbers you're working with, only their residue $$$\text{ mod } m$$$. You can then add, subtract, and multiply with them as much as you need for your problem, taking the modulo as often as needed to avoid integer overflow. You may even decide to wrap them into their own object class with overloaded operators if your language supports them, though you may have to be careful of any object allocation overhead. If you use Kotlin like I do, consider using the inline class feature.

But what about division and fractions? That's slightly more complicated, and requires a concept called the "modular multiplicative inverse". The modular multiplicative inverse of a number $$$a$$$ is the number $$$a^{-1}$$$ such that $$$a \cdot a^{-1}\ \text{ mod } m = 1$$$

But how do you actually find such a number? Bruteforcing all numbers to a prime number close to a billion will usually cause you to exceed the time limit. There are two faster ways to calculate the inverse: the extended GCD algorithm, and Fermat's little theorem. Though the extended GCD algorithm is more versatile and sometimes slightly faster, the Fermat's little theorem method is often more popular, simply because it's almost "free" once you implement exponentiation, another useful operation to do, so that's what we'll cover here.

Fermat's theorem says that as long as the modulus $$$m$$$ is a prime number ($$$10^9 + 7$$$ is prime, and so is $$$998 244 353$$$, another common modulus in these problems) and $$$a \text{ mod } m \neq 0$$$, then $$$a^m \text{ mod } m = a \text{ mod } m$$$. Working backwards, $$$a^{m-1} \text{ mod } m = a \cdot a^{m-2}\ \text{ mod } m = 1$$$, therefore the number we need is $$$a^{m-2} \text{ mod } m$$$.

Multiplying $$$m-2$$$ times would still take too long; therefore a trick known as exponentiation by squaring is needed.