POJ 2115 单变元模线性方程

31 ll extgcd(ll a, ll b, ll &x, ll &y) {
32     ll d = a;
33     if (b) {
34         d = extgcd(b, a%b, y, x);
35         y -= (a / b)*x;
36     }
37     else x = 1, y = 0;
38     return d;
39 }
40 
41 ll line_mod_equation(ll a, ll b, ll n) {
42     ll x, y;
43     ll d = extgcd(a, n, x, y);
44     if (b%d == 0) {
45         x = x*(b / d) % (n / d);
46         if (x < 0) x += n / d;
47         return x;
48     }
49     return -1;
50 }
51 
52 int main() {
53     ios::sync_with_stdio(false), cin.tie(0);
54     ll a, b, c, k;
55     while (cin >> a >> b >> c >> k) {
56         if (!a && !b && !c && !k) break;
57         k = 1LL << k;
58         ll ans = line_mod_equation(c, b - a, k);
59         if (ans == -1) cout << "FOREVER" << endl;
60         else cout << ans << endl;
61     }
62     return 0;
63 }