How to use extended Euclidean algorithm to solve the modular linear Equation Group (detailed)

the a≡b (mod n) is called A and b about modulo n congruence, and the necessary and sufficient condition is that a-a is an integer multiple of n, that is, A-B=ZN (where z takes an integer).

while the modular linear equations ax≡b (mod n) can be written as AX-B=ZN (where z takes an integer), the moving term can be

Ax-zn=b, also known as the form of a two-yuan Ax+by=c equation, using the extended Euclidean algorithm (EXTGCD) can solve the equation whether there is a solution and a set of solutions, and can be based on the group to write a solution system, and then find a special solution, such as the smallest positive integer solution.

The procedure for expanding Euclid's algorithm is given below.

1typedefLong LongLL;2 voidEXTGCD (ll A, ll B, LL &d, LL &x0, LL &y0)3 {4     if(b = =0){5d=a;x=1; y=0;6     }7     Else{8EXTGCD (b,a%b,d,y,x);9Y-= x* (A/b); Ten     } One}

where A, b is a coefficient, D is a greatest common divisor of a, B, and X0,y0 represents a set of special solutions.

by gcd (A, b) = gcd (b,a%b)

ax0 + by0 = gcd (A, b) = gcd (b, a%b) = Bx1 + (a%b) y1

where a%b can also be written a-a/b*b,

so again =bx1 + (a-a/b*b) y1

finishing Ay1 + B (x1-a/b*y1)

X0=y1,y0= (x1-a/b*y1), which is equal to the coefficients , then the x0 and y0 are only required for X1 and Y1, so the x1,y1 is recursive to the next, so the recursion always seeks gcd (AN, 0) = an*xn + 0 * yn , so that xn=1,yn=0 can satisfy the formula, because AN*1+0*0=AN=GCD (an, 0). The x0,y0 is then handed back until it is obtained, and D is the greatest common divisor of a and B.

Here's how to tell if there is an integer solution.

Divide The ax+by=c both sides by D to get

A/D*X+B/D*Y=C/D, because D is a and b of the greatest common divisor, so A/d and b/d are integers, it is required x and y integer solution, C/D must be an integer, otherwise no integer solution.

in conclusion. : Set a,b,c as any integer, D=GCD (A, B), D=ax+by, if C is a multiple of D, there is an integer solution, otherwise there is no integer solution.

The following is how to find a solution system based on a set of solutions (x0,y0), and then find a set of special solutions.

First Use brought into The extension Euclid obtains a solution x0,y0 of AX+B*Y=GCD (A, b);

Observe two more formulas:

Ax+b*y=c 0)

A*x0+b*y0=d 1) ( make d=gcd (A, B))

Divide the 0) both sides by D to:

X* (A/D) +y* (b/d) =c/d

the 1) is multiplied by the b/d and the 0)-type comparison:

(X0*C/D) *a + (Y0*C/D) *b=c

A*x+b*y=c 0)

to x=x0*c/d,y=y0*c/d;

and then refine the above conclusions: Set a,b,c to be any integer, D=GCD (A, b), the equation D=ax+by a set of solutions is (X0,Y0), when C is a multiple of D, then the group of equations Ax+by=c a set of solutions is (X0*C/D,Y0*C/D), when C is not a multiple of D no integer solution.
connect down will A solution expands into a solution system:

       1) divided by D: x0* (A/D) +y0* (b/d) =1

because (A/D) and (b/d) coprime, write the equation:

       (x0+u* (B/D)) * (A/D) + (y0+v* (A/D)) * (b/d) =1 i.e. just u* (b/d) * (A/D)  + v* (A/d) * (B/D) =0 can be;
so x, y  x= (x0+u*) * (A/d)

Y= (y0+v* (A/D)) * (B/D)

Next 'll be Ask The smallest positive integer that satisfies the x of the equation .

so xmin= (x0* (C/D)) mod (b/d)

The specific example is POJ's 1061 frog dating

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How to use extended Euclidean algorithm to solve the modular linear Equation Group (detailed)