Returns the same result as the remainder of an ax running C (mod m)

 

Convert the form

Ax-c = My

It can be seen that the original formula has a solution when and only when the linear equation ax-My = C has a solution

Set G = gcd (a, m)

Then the number of all shapes, such as ax-my, is a multiple of G.

Therefore, if G does not divide C, the original equation has no solution.

 

Assume that G is divisible by C:

Use the Extended Euclidean algorithm to calculate a special Au + mv = g (U0, V0)

Therefore, an integer c/g can be used to multiply the upper limit.

Au0 * (C/G) + mv0 * (C/G) = C

Obtain the original solution X0 = U0 * (C/g)

 

Number of solutions:

Assume that X1 is another solution of ax sans C (mod m ).

Ax1 ax2 (mod m), so m division ax1-ax2

So (M/G) Division (A/G) (x1-x2)

Because (M/G) and (A/G) mutual quality, so (M/G) Division (x1-x2)

The equation x = x0 + K * (M/G) (k = 0, 1, 2 ,...... G-1)

Total g

 1 void solve(int a, int c, int m) 2 { 3     int u0, v0; 4     int g = ex_gcd(a, m, u0, v0); 5     if(c%g != 0) 6     { 7         printf("The equation has no solution!\n"); 8         return; 9     }10     int i, x;11     for(i=0; i<g; ++i)12     {13         x = c/g*u0 + m/g*i;14         x = x % m;15         if(x<0)16             x+=m;17         printf("%d\n", x);18     }19 }

Code Jun