4.1 KiB
| id | title | author |
|---|---|---|
| intro-nt | Introductory Number Theory | Darren Yao |
- Prime Factorization
- GCD & LCM
- Modular Arithmetic
Additional Resources
- CSES 21.1, 21.2
- CF CodeNCode
Prime Factorization
A number a is called a divisor or a factor of a number b if b is divisible by a, which means that there exists some integer k such that b = ka. Conventionally, 1 and n are considered divisors of n. A number n > 1 is prime if its only divisors are 1 and n. Numbers greater than 1 that are not prime are composite.
Every number has a unique prime factorization: a way of decomposing it into a product of primes, as follows:
n = {p_1}^{a_1} {p_2}^{a_2} \cdots {p_k}^{a_k} where the p_i are distinct primes and the a_i are positive integers.
Now, we will discuss how to find the prime factorization of an integer.
(pseudocode)
This algorithm runs in O(\sqrt{n}) time, because the for loop checks divisibility for at most \sqrt{n} values. Even though there is a while loop inside the for loop, dividing n by i quickly reduces the value of n, which means that the outer for loop runs less iterations, which actually speeds up the code.
Let's look at an example of how this algorithm works, for n = 252.
(table)
At this point, the for loop terminates, because i is already 3 which is greater than \lfloor \sqrt{7} \rfloor. In the last step, we add 7 to the list of factors v, because it otherwise won't be added, for a final prime factorization of \{2, 2, 3, 3, 7\}.
GCD & LCM
The greatest common divisor (GCD) of two integers a and b is the largest integer that is a factor of both a and b. In order to find the GCD of two numbers, we use the Euclidean Algorithm, which is as follows:
$$
\gcd(a, b) = \begin{cases}
a & b = 0 \
\gcd(b, a \bmod b) & b \neq 0 \
\end{cases}
This algorithm is very easy to implement using a recursive function in Java, as follows:
public int gcd(int a, int b){
if(b == 0) return a;
return gcd(b, a % b);
}
For C++, use the built-in __gcd(a,b). Finding the GCD of two numbers can be done in O(\log n) time, where n = \min(a, b).
The least common multiple (LCM) of two integers a and b is the smallest integer divisible by both a and b.
The LCM can easily be calculated from the following property with the GCD:
\operatorname{lcm}(a, b) = \frac{a \cdot b}{\gcd(a, b)}If we want to take the GCD or LCM of more than two elements, we can do so two at a time, in any order. For example,
\gcd(a_1, a_2, a_3, a_4) = \gcd(a_1, \gcd(a_2, \gcd(a_3, a_4)))Modular Arithmetic
In modular arithmetic, instead of working with integers themselves, we work with their remainders when divided by m. We call this taking modulo m. For example, if we take m = 23, then instead of working with x = 247, we use x \bmod 23 = 17. Usually, m will be a large prime, given in the problem; the two most common values are 10^9 + 7, and 998\,244\,353. Modular arithmetic is used to avoid dealing with numbers that overflow built-in data types, because we can take remainders, according to the following formulas:
todo no support for gather
$$
\begin{gather*}
(a+b) \bmod m = (a \bmod m + b \bmod m) \bmod m \\
(a-b) \bmod m = (a \bmod m - b \bmod m) \bmod m \\
(a \cdot b) \pmod{m} = ((a \bmod m) \cdot (b \bmod m)) \bmod m \\
a^b \bmod {m} = (a \bmod m)^b \bmod m
\end{gather*}
$$
Modular Exponentiation
?
Modular Inverse
Under a prime moduli, division exists.
Problems
- USACO
- Both are also Knapsack DP problems.
- Cow Poetry
- First consider the case where there are only two lines with the same class.
- Requires fast modular exponentiation for full credit.
- Exercise
- Prime factorization
- Other