Extended Euclid Euclid &&

Euclidean algorithm:

Euclidean algorithm, also known as flounder division, referred to gcd, used to calculate the greatest common divisor of two integers

Principle: Euclidean algorithm to solve the greatest common divisor (common factor)

Provided two atoms a, b (a> b), with the GCD (a, b) represents a, b greatest common divisor

r = a (mod b) () is the remainder of a divided by b, q is a quotient divided by b, i.e. a ÷ b = q ....... r. Euclidean i.e. to prove gcd (a, b) = gcd (b, r).

Proved that: a / b = q; a = q * b + r;

<1> when r = 0 when, b is the greatest common divisor of a and b

<2> when r = 0, the! Set a, b is the greatest common divisor c, a = m * c; b = n * c; ( c is because a, b greatest common divisor, the m, n mutually quality)

r = aq * b; r = (m * c) -q * b (n * c) = (mq * b * n) * c, a divisor can be seen also c of r,

So we can continue to b / r 0 r until we come to a, b is the greatest common divisor

Recursion:

#include<iostream>
using namespace std;
int gcd(int a,int b){
    if(b==0)
        return a;
    return 
        gcd(b,a%b);
}
int main ()
{
    int a, b, years;
    cin>>a>>b;
    years = gcd (a, b);
    cout << years;
    return  0 ;
}

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optimization:

#include<iostream>
using namespace std;
int gcd(int a,int b){
    return b?gcd(b,a%b):a;
}
int main ()
{
    int a, b, years;
    cin>>a>>b;
    years = gcd (a, b);
    cout << years;
    return  0 ;
}

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Extended Euclidean Algorithm:

Extended Euclidean Algorithm, referred exgcd, generally used to solve the equation, solving the linear congruence equations, solving the inverse mode, and the like

Extended Euclidean Algorithm: find a * x + b * y = c in the general solution.

Pei Shu theorem: for integers a, b, they are about x, y linear indeterminate equation

prove:

. Provided a * x + b * y = t,

<1> When b = 0, t = a (because gcd algorithm, if (b == 0) return a;), there are a * x = a, easy to get x = 1.

<2> When b = 0, the! Set a * x1 + b * y1 = gcd (a, b), b * x2 + (a% b) * y2 = gcd (b, a% b); Because gcd (a , b) = gcd (b, a% b), there is simultaneous: a * x1 + b * y1 = b * x2 + (a% b) * y2.

a% b = a- (a / b) * b; (a / b direction trade integer).

The ax1 + by1 = bx2 + (aa / b * b) y2

　　　　 ax1+by1=bx2+ay2-a/b*by2

　　　　 ax1+by1=ay2+bx2-b*a/b*y2

　　　　 ax1+by1=ay2+b(x2-a/b*y2)

　　　　 Solve for x1 = y2, y1 = x2-a / b * y2

The proof of the above, the recursive approach in the realization of the recursive into the first layer, i.e., until reach the last layer when b = 0 returns x = 1, y = 0

Then according to x = y ', y = x'-a / b / y' (x 'and y' and the next layer in the x y) obtained when a layer of solution. When the solution is calculated continuously and return layer, eventually returned to the first layer, to obtain the original solution

int exgcd(int a,int b,int &x,int &y){
    if(b==0){
        x=1;
        y=0;
        return a;
    }
    int r=exgcd(b,a%b,x,y);
    int t=x;
    x = y;
    y=t-a/b*y;
    return r;
}

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Euclid expand inversion yuan

PS: a ≡ b (mod m) 读作a同余于b模m，或读作a与b关于模m同余。 比如 26 ≡ 14 (mod 12)。

什么是逆元？

为什没求逆元跟欧几里得算法联系起来了呢？根据上面裴蜀定理，我们知道gcd是a,b两个数线性组合的最小值，其他组合值都是gcd的倍数，

当gcd为1时，a,b互质，满足

int cal(int a,int m)
{
    int x,y;
    int gcd = ex_gcd(a,m,x,y);
    //cout << "a " << a << " m " << m << " x " << x << " y " << y << endl;
    if(1%gcd!=0) return -1;
    x*=1/gcd;
    m = abs(m);
    int ans = x%m;
    if(ans<=0) ans += m;
    return ans;
}

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这里1%gcd是看gcd是不是1，前面说了，d（这里是1）应该是gcd的倍数，而且不互质的两个数没有逆元。