$X_{i +1} \equiv a \times x_i + b(mod p)$
求最小的$n$，使得$x_n=t$
枚举找通项可得 $X_n=a^{n - 1}\times X_1+b\times \dfrac{a^{n - 1} - 1}{a - 1}$
设$c$为$a - 1$的逆元则有答案$n$ , $(x_1+b \times c) \times a^{n-1} \equiv b \times c + t (mod\ p)$
则有$a^{n-1} \equiv (b \times c + t) * (x_1 + b \times c)^{-1} (mod\ p)$
$BSGS$即可
1 #include <iostream>
2 #include <cstdio>
3 #include <cmath>
4 #include <map>
5 #include <algorithm>
6 #define inc(i) (++ (i))
7 #define dec(i) (-- (i))
8 #define int long long
9 using namespace std;
10
11 int T , a , b , t , x , P;
12
13 inline int KSM(int x , int p)
14 {
15 int Ans = 1;
16 while(p)
17 {
18 if(p & 1) Ans = Ans * x % P;
19 x = x * x % P , p >>= 1;
20 }
21 return Ans;
22 }
23
24 map <int , int> M;
25 inline int BSGS(int A , int B , int C)
26 {
27 if(A % C == 0 && B != 0) return -2;
28 if(A % C == 0 && B == 0) return 1;
29 M.clear();
30 int m = (int)ceil(sqrt(1.0 * C)) , Now;
31 Now = B % C , M[Now] = 0;
32 for(int i = 1 ; i <= m ; inc(i))
33 Now = Now * A % C , M[Now] = i;
34 A = KSM(A , m) , Now = 1;
35 for(int i = 1 ; i <= m ; inc(i))
36 {
37 Now = Now * A % C;
38 if(M[Now]) return i * m - M[Now];
39 }
40 return -2;
41 }
42
43 signed main()
44 {
45 scanf("%lld" , &T);
46 while(T --)
47 {
48
49 scanf("%lld%lld%lld%lld%lld" , &P , &a , &b , &x , &t);
50 if(x == t)
51 {
52 puts("1");
53 continue;
54 }
55 if(a == 0)
56 {
57 if(b == t) puts("2");
58 else puts("-1");
59 continue;
60 }
61 if(a == 1)
62 {
63 if(b == 0) puts("-1");
64 else
65 {
66 int B = (t - x + P) % P;
67 B = (B * KSM(b , P - 2)) % P;
68 printf("%lld\n", B + 1);
69 }
70 continue;
71 }
72 int c = KSM(a - 1 , P - 2) * b % P;
73 // printf("c = %d\n" , c);
74 b = (t + c) * KSM(x + c , P - 2) % P;
75 // printf("B = %d\n" , b);
76 printf("%lld\n" , BSGS(a , b , P) + 1);
77 }
78 return 0;
79 }