题意：

传送门
已知递推公式\(x_i = a*x_{i - 1} + b\mod p\)，\(p\)是素数，已知\(x_0，a，b，p\)，给出一个\(n\)和\(v\)，问你满足\(x_i = v\)且\(i < n\)的最小的\(i\)是多少。

思路：

经过一些举例我们可以发现\(x_i = a^ix_0 + b\frac{a^i-1}{a-1}\mod p\)，当\(a \neq 1\)可得\(a^i = \frac{(a-1)v+b}{(a-1)x_0+b}\mod p\)。再当\(a \geq 1\)时可用\(BSGS\)求解，其他直接特判。
但是因为\(q\)的存在，如果直接套模板复杂度\(O(\sqrt p + q\sqrt p)\)，\(3e7*T\)必\(TLE\)。因为每组询问的预处理都相同，那么我们不妨把预处理扩大，这样每次查询所要暴力的次数就变小了。预处理大小设为\(1e6\)，手写\(Hash\)更快。

代码：

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<cmath>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 300 + 5;
const int INF = 0x3f3f3f3f;
const ull seed = 131;
const ll MOD = 1e9;
using namespace std;//BSGS
const int M = 5e6;
int hs[M], head[M], nex[M], id[M], top;
void _insert(ll x, int y){int k = x % M;hs[top] = x, id[top] = y, nex[top] = head[k], head[k] = top++;
}
int _find(ll x){int k = x % M;for(int i = head[k]; i != -1; i = nex[i]){if(hs[i] == x) return id[i];}return -1;
}
ll ppow(ll a, ll b, ll mod){ll ret = 1;while(b){if(b & 1) ret = ret * a % mod;a = a * a % mod;b >>= 1;}return ret;
}
ll loop, up;
void preBSGS(ll p, ll a){   // a^x = b mod pmemset(head, -1, sizeof(head));top = 1;up = ceil(1e6);ll t = 1;for(int i = 0; i <= up; i++){if(i == up) loop = t;_insert(t, i);t = 1LL * t * a % p;}
}
ll BSGS(ll A, ll B, ll P){  // a^x = b mod pll m = ceil(P * 1.0 / 1e6);ll obj = ppow(B, P - 2, P), x;for(int i = 1; i <= m; i++){obj = 1LL * obj * loop % P;if((x = _find(obj)) != -1){return 1LL * i * up - x;}}return -1;
}
ll n, x0, a, b, p, Q;
int main(){int T;scanf("%d", &T);while(T--){scanf("%lld%lld%lld%lld%lld", &n, &x0, &a, &b, &p);scanf("%lld", &Q);preBSGS(p, a);while(Q--){ll v;scanf("%lld", &v);if(a == 0){if(v == x0) printf("0\n");else if(v == b && n - 1 >= 1) printf("1\n");else printf("-1\n");}else if(a == 1){v = ((v - x0) % p + p) % p;if(b == 0){printf("%d\n", v == 0? 0 : -1);continue;}ll ans = 1LL * v * ppow(b, p - 2, p) % p;printf("%lld\n", ans >= n? -1 : ans);}else{ll ret = (a * v % p - v + b) % p;ret = ret * ppow((a * x0 - x0 + b) % p, p - 2, p) % p;ret = (ret + p) % p;ll ans = BSGS(a, ret, p);printf("%lld\n", ans >= n? -1 : ans);}}}return 0;
}