Number theory of the extended Euclidean related templates

Expand Euclid formula:

 

typedef long long LL;
LL exgcd(LL a,LL b,LL &x,LL &y){
     if(a==0&&b==0) return -1;
     if(b==0) { x=1;y=0; return a;  }
     LL d=exgcd(b,a%b,y,x);
     y-=a/b*x;
     return d;
}

 

Solutions ax + by = c Solution Set of:

 

typedef Long  Long LL; 
LL X0, yO, KX, KY; 
BOOL LE (LL A, B LL, LL C) {    // solution of linear equation ax + by = c 
    LL X1, Y1; 
    LL GCD = exgcd (A, B , X1, Y1);
     IF (% C GCD)
         return  to false ;   // no integer solution 
    X0 = X1 * C / GCD; // set of solutions 
    yO = Y1 * C / GCD; // set of solutions 
    KX = B / GCD; 
    KY = -a / GCD;
     return  to true ;    // solvable, the solution set is: (x0 + kx * t, y0 + ky * t) t is an integer 
}

 

The smallest positive integer x solving solutions:

 

/ * Solved smallest positive integer solutions of x * / 
typedef Long  Long LL;
 void minxx ( int A, int B, int C, int x, int Y) 
{ 
    int GCD = exgcd (A, B, x, Y); 
    LL X1, Y1, T; 
    T = B / GCD;
     IF (T < 0 ) 
    { 
        T = - T; 
    } 
    X1 = (X + T) * (C / GCD); 
    X1 = (X1% T + T)% T; 
    Y1 = (X1 * CA) / B; 
    the printf ( "x==%lld, y=%lld\n", x1, y1);
    return ;
}

 

void minxx(ll a, ll b, ll c, ll x, ll y)
{
    int gcd = exgcd(a, b, x, y);
    ll x1, y1, t;
    t = b / gcd;
    if( t<0 )
    {
        t = -t;
    }
    x1 = x;
    x1 = (c/gcd*x1%t+t)%t;
    y1 = (c-a*x1)/b;
    printf("%lld %lld\n", x1, y1);
    return ;
}

I ax≡b (mod n) with the equation for the unknown x solvable, if and only if gcd (a, n) | b (i.e. b% gcd (a, b) == 0), and the equation has a solution, Eq. there gcd (a, n) th solution.

Solving equation ax≡b (mod n) is equivalent to solving equation ax + ny = b, (x, y being an integer)

Solving linear congruence equation:

bool modular_linear_equation(int a,int b,int n)
{
    int x,y,x0,i;
    int d=exgcd(a,n,x,y);
    if(b%d)
        return false;
    x0=x*(b/d)%n;   //特解
    for(i=1;i<d;i++)
        printf("%d\n",(x0+i*(n/d))%n);
    return true;
}

Congruence equation ax≡b (mod n), if gcd (a, n) == 1, then the equation has only one solution.

In this case, if b == 1, congruence equation is ax = 1 (mod n), gcd (a, n) = 1.

In this case said determined modulo n x is a multiplication inverse element of a pair.

For congruence equation ax = 1 (mod n), gcd (a, n) = 1 is solved to solve the equation ax + ny = 1, x, y are integers.

ll inv(ll a, ll n)
{
    ll d, x, y;
    d = exgcd(a, n, x, y);
    return (d==1)?(x%n+n)%n:-1;
}