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NT Part 3: Playing with Modulars, and Euler Phi Function

 Quick stuff first, fast exponentiation in logarithm time.


Let us calculate $a^b$ in modular $m$ in $O(\log b)$.
It uses binary expansion of $b$, and is very very straightforward.

ll exp(ll x, ll n)  
{  
    if(n==0) return 1;  
    if(n==1) return x;  
    if(n%2==0) return exp((x*x)%mod,n/2);  
    if(n%2==1) return (x*exp((x*x)%mod,n/2))%mod;  
}


Now, let us talk about modular inverses.

By using Extended Euclidean Algorithm, we can get the inverse of $a$ modulo $m$.

#include <iostream>
int inv(int a, int m)
{
    int temp=m, q, t, u=0, v=1;
    if(m==1) return 0;
    while(a>1)
    {
        q=a/m;
        t=m;
        m=a%m;
        a=t;
        t=u;
         u=v-q*u;
         v=t;
    }
    if(v<0) v+=temp;
    return v;
}
int main(void)
{
    int a, m;
    std::cin>>a>>m;
    std::cout<<inv(a,m);
}


Of course, logarithm time. If $m$ is prime, we can do a lot of different things.

Fermat's Little Theorem gives $a^{p-1} \equiv 1 \pmod{p}$ if $(a,p)=1$, where $p$ is a prime.
Therefore, we can calculate the modular inverse of $a$ as $a^{p-2}$, by fast exponentiation.
Time Complexity is $O(\log p)$.

Also, you can get the modular inverse of all numbers in $[1,n]$ in $O(n)$.
The code for this is shown below. The proof for the correctness is left to the reader (not difficult)

#include <iostream>
typedef long long ll;
using namespace std;
int inv[111111], n;
ll mod=1e9+7;
int main(void)
{
    cin>>n;
    int i;
    inv[1]=1;
    for(i=2 ; i<=n ; i++)
    {
        inv[i]=((mod-mod/i)*inv[mod%i])%mod;
        cout<<inv[i]<<endl;
    }
}


We can also calculate $\binom{n}{m}$ in modulo $p$ ($p$ is a prime) very fast using Lucas' Theorem.

Lucas' Theorem basically states that $\binom{n}{m} \equiv \binom{n_0}{m_0} \cdot \binom{\lfloor \frac{n}{p} \rfloor}{\lfloor \frac{m}{p} \rfloor}$, where $n_0$ is $n$ modulo $p$ and $m_0$ is $m$ modulo $p$.

This is very efficient when $p$ is small and $n, m$ is huge. We can precalculate the factorials and inverse of factorials modulo $p$ by using the above code, and solve each queries in $O(\log_p \text{max} (n,m))$.

Also, we can use Chinese Remainder Theorem to solve a system of modular equations.

Let us solve $x \equiv r_i \pmod {m_i}$, where $m_i$ are pairwise coprime.
(If they are not coprime, break them into prime powers, and if some are contradictory, there are no solutions.)

The CRT itself gives an algorithm to get our answer.
Set $M= \prod_{i=1}^n m_i$, and $u_i = \frac{M}{m_i}$. Also, set $s_i$ as the modular inverse of $u_i$ in modulo $m_i$. Then our answer is $\sum_{i=1}^n r_is_iu_i \pmod{M}$

We learned how to calculate modular inverse in logarithm time above. So the time complexity is $O(n \log MAX)$.

long long int r[111111]; // remainders
long long int m[111111]; // modulars
long long int M=1; // product
int n; // number of equation
int res(void)
{
    int i;
    for(i=1 ; i<=n ; i++)
    {
        M=M*m[i];
    }
    long long int ret=0;
    for(i=1 ; i<=n ; i++)
    {
        ret+=r[i]*inv(M/m[i],m[i])*(M/m[i]);
        ret=ret%M;
    }
    return ret;
}


$\phi (n)$ is the number of positive integers no more than $n$ which is coprime with $n$.
Formula is $\phi (n) = n \prod_{p|n} (1-\frac{1}{p})$. Proof is Inclusion-Exclusion.
Also, we have the formula $\sum_{d|n} \phi (d) = n$.

Of course, for the calculation of Euler Phi numbers, we can tweak the Eratosthenes's Sieve algorithm a little bit.

void preprocess(void)
{
    int i, j;
    eulerphi[1]=1;
    for(i=2 ; i<=122000 ; i++)
    {
        eulerphi[i]=i;
        primechk[i]=1;
    }
    for(i=2 ; i<=122000 ; i++)
    {
        if(primechk[i]==1)
        {
            eulerphi[i]-=eulerphi[i]/i;
            for(j=2 ; i*j<=122000 ; j++)
            {
                primechk[i*j]=0;
                eulerphi[i*j]-=eulerphi[i*j]/i;
            }
        }
    }
}


You could also calculate $\phi (n)$ by using prime factorization of $n$.

Now let's get to the fun stuff. The mobius function.

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