Divisor

About divisor

Definitions :

When the remainder of the integer n by an integer d is 0, i.e., d divisible by n, is called d n, about the number, n is a multiple of d, denoted by d | n

In the Fundamental Theorem of Arithmetic \ (N \) can be decomposed into the following way

\[N=\prod_{i=1}^m p_i^ {c_i}, \ p_1<p_2<…<p_m , \ c_i ∈ N^*\]

Then \ (N \) number is a positive number from about:

\[(c_i+1)*(c_i+2)*…(c_m+1)=\prod_{i=1}^{m}(c^i+1)\]

\ (N \) of all positive divisors are:

\[\prod_{i=1}^{m}{(\sum_{j=0}^{c^i}(p_i)^j)}\]

Solution \ (N \) is set about several positive

Trial division

$ \ $ Qquad if a number \ (X \) is the \ (N \) divisors, then \ (N / d≤ \ sqrt N \) is also \ (N \) divisors of.

$ \ Qquad $ because the divisor always come in pairs, so scanning \ (the X-= 1- \ sqrt N \ ∈ Z \) , try whether \ (the X-| N \) . But we have to special sentenced perfect square, because for a perfect square \ (\ sqrt N \ ∈ Z \) .

    int factor[1600], num = 0;
    for(int i = 1; i * i <= n; i++) {
        if(n % i == 0) {
            factor[++num] = i;
            if(i != n/i) 
                factor[++num] = n / i;
        }
    }

$ \ $ Qquad request \ (1- \ sqrt N \) n-number divisor of each set - COLORED

The basic idea:

Different from the trial division, we can consider each in turn number \ (x \) , places \ (x \) is a divisor number is \ (x, 2x, 3x ... \)

    vector<int> factor[SIZE];
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n / i; j++)
            factor[i*j].push_back(i);       
    //输出
    for(int i = 1; i <= n ;i++) {
        for(int j = 0; j < factor[i].size; j++)
            printf("%d ",factor[i][j]);
        putchar('\n'); 
    }

$ \ $ Qquad in a small data ( \ (n∈ [4,16] \) ), is faster than the trial division multiple methods, complexity \ (O (N \ sqrt N ) \)

Divisor related content:

1. The greatest common divisor of | \ (gcd \)

Definitions :

If the natural numbers \ (x \) satisfies \ (x | a \) and \ (the X-| b \) , called \ (x \) is \ (a \) is (b \) \ common divisor, then \ ( max (x) \) is \ (a \) and \ (B \) is the greatest common divisor, referred to as \ (GCD (a, B) \) .
Similarly, if while meeting \ (a | x \) and \ (B | X \) , then \ (X \) is the \ (A \) and (B \) \ common multiple, in such \ (X \ ) in a minimum, for the \ (a, b \) the least common multiple, referred to as \ (LCM (a, B) \) .

theorem:

\[∀a,b∈N, \qquad \qquad gcd(a,b)*lcm(a,b)=a*b \]

Proof reader as defined above can try yourself, you can also "reference algorithm contest Step-up Guide"

Solution

In On prime factor decomposition we have already mentioned solving \ (gcd \) algorithm: 1. Decreases Technique 2. Euclidean algorithm.

"Nine Chapters on Arithmetic" are:

\[∀a,b∈N,a≥b,\quad gcd(a,b)=gcd(b,a-b)=gcd(a,a-b)\]

\[∀a,b∈N,\quad gcd(2a,2b)=2gcd(a,b)\]

It is important above two theorems, which involve binary Optimized behind us \ (gcd \)

Euclidean algorithm:

\[∀a,b∈N,b≠0,\qquad gcd(a,b)=gcd(b,a \quad mod \quad b)\]

This gives the familiar Euclidean algorithm :

    //递归形式
    int gcd(int a, int b) {
        return b ? gcd(b, a % b) : a;
    } 

    //非递归形式
    int gcd(int a, int b) {
        int temp;
        while(b) {
            temp=a % b;
            a = b;
            b = temp;
        }   
        return a;
    }

$ \ Qquad \ qquad \ (Euclidean algorithm complexity is \) O (log (A + B)) $. For high-precision arithmetic, modulo not easy to achieve, it is recommended Decreases Technique instead.

$ \ Qquad \ qquad \ (Of course, \) gcd $ can also be optimized (optimization is binary mentioned earlier).

Why can optimize? Because use of the Euclidean algorithm modulo operation, so that constant slow running large, so with optimized binary Decreases + \ (GCD \)

$a=0, \quad return \quad b; \quad \mid \quad b=0, \quad return \quad a;$
\ (A <B, A \ Quad XOR = \ Quad B, \ Quad B \ Quad XOR = \ A Quad, \ Quad A \ Quad XOR = \ Quad B; \) (unique binary exchange)
$a$ & $1$ 且 $b$ & $1$，$ \quad ans=2gcd(a >> 1,b>>1);$
$a$ & $1$ 且！$b$ & $1$，$ \quad ans=gcd(a >> 1,b);$
！$a$ & $1$ 且 $b$ & $1$，$ \quad ans=gcd(a,b >> 1);$
！$a$ & $1$ 且！$b$ & $1$，$ \quad ans=gcd(a - b,b)$。

    //下面代码返回gcd(a,b)的值同时把b赋予这个值，不需要可以把&去掉
    int gcd(int a,int &b) {
        if(a == 0 || b == 0) return b = a + b; 
        int n = 0, m = 0;
        for(; !(a & 1); a >>= 1, n++); 
        for(; !(b & 1); b >>= 1, m++);
        n = m < n ? m : n;
        while(a) {
            if(a < b) { a ^= b, b ^= a, a ^= b;}
            if(!(a -= b)) return b <<= n;
            while(!(a & 1)) a >>= 1;
        }
    }

However, note that when the above code is wrong negative, because the negative right \ (1 \) bit and other \ (2 \) not the same as, the above negative encountered normal use bit operation instead of division.

2. The prime function of Euler

definition:

\ (∀a, b∈N \) , if the \ (GCD (A, B) =. 1 \) , called \ (a, b \) prime.

For three numbers and above analogy to the situation, not repeat them here, readers can access relevant information on their own.

Euler function

$ \ $ Qquad \ (. 1-N \) with \ (N \) prime number is called the Euler function, referred to as \ ([Phi] (N) \) .

In the basic arithmetic theorem, \ (N = \ prod_. 1} = {I ^ m ^ {P_i} C_i \) , then:

\[φ(N)=N*\frac{1-p_1}{p_1}*\frac{1-p_2}{p_2}*\frac{1-p_3}{p_3}*…*\frac{1-p_m}{p_m}=N*\prod_{prime \ p|N}(1-\frac{1}{p})\]

The calculation formula of Euler function, we need only decomposition of the quality factor, to obtain the Euler function:

    //参考代码(源于《算法竞赛进阶指南》) 
    int phi(int n) {
        int ans = n;
        for(int i = 2; i <= sqrt(n); i++) 
            if(n % i == 0) {
                ans = ans / i * (i - 1);
                while(n % i == 0) n /= i;
            }
        if(n > 1) ans = ans / n * (n - 1);
        return ans;
    }

The nature of the Euler function:

\ (∀n> 1,1-n \ ) with \ (n-\) prime number and is \ (n-* [Phi] (n-) / 2 \) .
If \ (GCD (A, B) =. 1 \) , i.e. \ (a, b \) prime, then \ ([Phi] (ab &) = [Phi] (A) [Phi] (B) \) .
Set \ (P \) is a prime number, if \ (p \ mid n \) and \ (P ^ 2 \ MID n-\) , then \ ([Phi] (n-) = [Phi] (n-/ P) * P \) .
Set \ (P \) is a prime number, if \ (p \ mid n \) and \ (P ^ 2 \ the nmid n-\) , then \ (φ (n) = φ (n / p) * (p-1) \) .
$\sum_{d \mid n}φ(d)=n$。

Multiplicative function:

$ \ Qquad $ if \ (gcd (A, b) = 1 \) , there \ (f (ab) = f (A) f (b) \) , then known as the function \ (f \) is a multiplicative function

If \ (F \) is the product of the function and the Fundamental Theorem of Arithmetic in \ (n-= \ prod_ {I =. 1} ^ m P_i ^ {C_i} \) , then $ f (n) = \ prod_ {i = 1} ^ mf (p_i ^ { c_i}) $.

(Sixth with the nature of the Euler function)

\ (* \) Is completely multiplicative function:

\[∀a,b∈Z, \qquad f(ab)=f(a)f(b)\]

Expand on multiplicative function is very large, the content is more esoteric, following a brief introduction:

Common multiplicative function as

\ ([Phi] (n-) \) - Euler function
\ ([sigma] (n-) \) - divisors function
\ ([mu] (n-) \) - Mobius function
\ (σ_0 (n-) \) - about a few number of functions
\ (σ_k (n-) \) - about the number and function number
\ (GCD (n-, K) \) - the greatest common divisor function, when \ (K \) the fixed time
\ (1 (the n-) = 1 \) - I do not know what this is
\ (f (the n-) = the n-\) - I still do not know what is

$ \ Qquad $ Another point is multiplicative function are linearly screen

Dirichlet convolution

First, start by supplementing the next number theoretic function definition:

Codomain: arbitrarily set value range comprising
Arithmetical functions: domain is a positive integer, accompany domain is a complex function

Well, we can begin to introduce Dirichlet convolution of.

Definition of \ (f, g \) number theoretic function, their Dirichlet convolution can be expressed as \ (F * G \) , provided \ (F * G = H \) :

\[h(n)=\sum _{d|n}f(d)g\Big(\frac{n}{d}\Big)\]

If \ (f, g \) is a multiplicative function, obviously, \ (H \) is also a multiplicative function.

prove

$ \ $ Disposed qquad \ (n = a * b \ ) and \ (a, b \) coprime, i.e. \ (GCD (A, B) =. 1 \) :

\[h(n)=\sum _{d_1|a, d_2|b}f(d_1d_2)g\Big(\frac{a}{d_1}\frac{b}{d_2}\Big)\]

\[=\sum_{d_1|a, d_2|b}f(d_1)f(d_2)g\Big(\frac{a}{d_1}\Big)g\Big(\frac{b}{d_2}\Big)\]
\[=\sum_{d_1|a}f(d_1)g\Big(\frac{a}{d_1}\Big)\sum_{d_2|b}f(d_2)g\Big(\frac{b}{d_2}\Big)\]
\[=h(a)*h(b)\]

$ \ Qquad $ proof.

Algorithms

Dirichlet convolution operations to meet:

\ (F * F * G = G \) (commutative)
\ ((F * G) = F * H * (F * G) \) (associative law)
\ (F * (G + H) + = F * F * G H \) (distributive property)
If \ (f, g \) is the product of a function, the \ (f * g \) is multiplicative function. (nature)

Dirichlet convolution related

Below or return to the Euler function:

$ \ $ Qquad we can use the \ (of Eratosthenes \) sieve, is calculated according to the Euler function, the \ (O (NlogN) \) within solved \ (2-N \) in each of the Euler number function.

    int phi[SIZE];
    void eluer(int n) {
        for(int i = 2; i <= n; i++) phi[i] = i;
        for(int i = 2; i <= n; i++) 
          if(phi[i] == i) 
              for(int j = i; j <= n; j += i;) 
                phi[j] = phi[j] / i * (i - 1);
    }

$ \ Qquad $ but since that is a multiplicative function can be linear sieve, then how to optimize a linear function of the Euler it? Let's review a prime number sieve linear thinking ( prime number sieve Detailed ), linear sieve, each composite number \ (n \) will be its prime factors sieve once, use the following several properties:

Theorem three: Let \ (P \) is a prime number, if \ (p \ mid n \\) and $ P ^ 2 \ MID n- \ (, then \) [Phi] (n-) = [Phi] (n-/ P) * P $ .
Theorem four: Let \ (P \) is a prime number, if \ (p \ mid n \\) and $ P ^ 2 \ the nmid n- \ (, then \) [Phi] (n-) = [Phi] (n-/ P) * (P -1) $

We can use in screening these two composite number theorem, from (φ (n / p) \ ) \ recursive to \ ([phi] (the n-) \) .

About prime number sieve, linear sieve because there are two kinds of writing, in fact, is much the same, but for the convenience of the reader, here are given by:

    //法一 ： 
    int v[SIZE], pri[SIZE], phi[SIZE], num;
    void promoted_eluer(int n) {
        for(int i = 2; i <= n; i++) {
            if(v[i] == 0) 
                pri[++num] = i, phi[i] = i - 1;
            for(int j = 1; j <= num && i * pri[j] <=n; j++) {
                v[i * pri[j]] = 1;
                phi[i * pri[j]]=
                  phi[i] * (i % pri[j] ? pri[j] - 1 : pri[j]);
                if(i % pri[j] == 0) break;
            }
        }
    }

    //法二： 
    int v[SIZE], pri[SIZE], phi[SIZE], num;
    void promoted_eluer(int n) {
        for(int i = 2; i <= n; i++) {
            if(v[i] == 0) 
                pri[++num] = i, phi[i] = i - 1, v[i] = i;
            for(int j = 1; j <= num; j++) {
                if(pri[j] > v[i] || pri[j] > n / i) break;
                v[i * pri[j]] = pri[j];
                phi[i * pri[j]]=
                  phi[i] * (i % pri[j] ? pri[j] - 1 : pri[j]);
            }
        }
    }

※ end of chapter exercises

$END$

$PS:$

More than explain the sequence and contents Reference: Li Yudong "algorithm contest Step-up Guide"

About the number of relevant