POJ 2115 C Looooops [linear modulus equation]

Convert it to c * x = B-a mod (2 ^ k), and solve the minimum positive integer of the modulus linear equation.

Sample Input
3 3 2 16
3 7 2 16
7 3 2 16
3 4 2 16
0 0 0 0

Sample Output
0
2
32766
FOREVER

Equation solving: ax = B (mod n); [ax % n = B % n]
Set one solution of the linear modulus equation to x0.
Condition ①: d = gcd (a, n)
Condition ②: with d = ax1 + ny, the x1 value is obtained from the expanded Euclidean (Egcd ).
Condition ③: B % d = 0 (condition with solution)
Then x0 = x1 * (B/d );

Proof:
Because: it is easy to obtain d = gcd (a, n), then d = ax1 + ny ① (Extended Euclidean theorem)
Equation ① 2-side simultaneous modulus n: d % n = ax1 % n ②
B % d = 0, that is, B is a multiple of d;
Therefore, (B/d) must be an integer;
So from ②: B % n = d * (B/d) % n = ax1 * (B/d) % n = ax % n
So it is easy to see that x = x1 * (B/d) is an integer solution of the equation.

References:

C ++ code
# Include <iostream>
# Include <fstream>
# Include <algorithm>
# Include <string>
# Include <set>
// # Include <map>
# Include <queue>
# Include <utility>
# Include <iomanip>
# Include <stack>
# Include <list>
# Include <vector>
# Include <cstdio>
# Include <cstdlib>
# Include <cstring>
# Include <cmath>
# Include <ctime>
# Include <ctype. h>
Using namespace std;
# Define L long

L Egcd (L a, L B, L & x, L & y) // extends Euclidean
{
L d, tp;
If (B = 0)
{
X = 1, y = 0;
Return;
}
D = Egcd (B, a % B, x, y );
Tp = x;
X = y;
Y = tp-(a/B) * y;
Return d;
}
Void MLE (L a, L B, L n) // solves the modulus linear equation
{
L x, y, d;
D = Egcd (a, n, x, y );
If (B % d = 0)
{
L x0 = (B/d * x) % n + n;
Cout <x0% (n/d) <endl;
// For a group of remainder formed by countless solutions: the number of cycles is d, the length of the cycle is n/d, that is, the minimum positive integer is in n/d. This is what the teacher said, but I forgot why it involves the concept of group ......
}
Else puts ("FOREVER"); // no solution
}
Int main ()
{
L a, B, c;
Int k;
While (cin> a> B> c> k)
{
If (! A &&! B &&! C &&! K)
Break;
MLE (c, B-a, 1LL <k );
}
Return 0;
}