generator 2

题目链接

题意

给一个一阶递推式,计算在$modp$意义下第几个数字等于$v$,一共有$Q$次询问。

思路

首先由给的递推式计算出一个特征方程。然后就变成了一个离散对数的问题。

预处理出小步,然后枚举大步。

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using namespace std;
typedef long long ll;


const ll maxn = 1e6;
ll mod;
struct {
static const int mod = 12000005;
int hs[mod], head[mod], nxt[mod], id[mod], top;

void init(){
memset(head, -1, sizeof(head));
top = 1;
};

void insert(ll x, int y) {
int k = x % mod;
hs[top] = x, id[top] = y, nxt[top] = head[k], head[k] = top++;
}

int find(int x) {
int k = x % mod;
for (int i = head[k]; i != -1; i = nxt[i]) if (hs[i] == x) return id[i];
return -1;
}
}mp;
ll bin(ll x, ll n, ll MOD)
{
ll ret = MOD != 1;
for (x %= MOD; n; n >>= 1, x = x * x % MOD)
if (n & 1)
ret = ret * x % MOD;
return ret;
}
inline ll get_inv(ll x, ll p) { return bin(x, p - 2, p); }


ll m, ma;
ll Init(ll a, ll p)
{
mp.init();
ll v = 1;
for(int i=1;i<=m;i++)
{
v = v * a % p;
mp.insert(v,i);
}
return v;
}

ll BSGS(ll a, ll b, ll p, ll init_d)
{
a %= p;
if (!a && !b)
return 1;
if (!a)
return -1;
ll v = init_d;
ll vv = init_d;
ll inv_b = get_inv(b, p);

for(ll i=1;i<=ma;i++)
{
auto it = mp.find(vv * inv_b % p);
if (it != -1)
return i * m - it;
vv = vv * v % p;
}
return -1;
}

int main()
{
ios::sync_with_stdio(false);
ll T;
cin>>T;
while (T--)
{
ll n, x0, a, b, p;
ll Q;
cin>>n>>x0>>a>>b>>p>>Q;
if (a == 0)
{
while (Q--)
{
ll v;
cin>>v;
if (v == x0)
{
cout<<0<<endl;
}
else if (v == b)
{
cout<<1<<endl;
}
else
{
cout<<-1<<endl;
}
}
}
else if (a == 1)
{
ll invb = get_inv(b, p);
while (Q--)
{
ll v;
cin>>v;
ll ans = (((v - x0 + p) % p) * invb) % p;
if (ans >= n)
{
cout<<-1<<endl;
}
else
{
cout<<ans<<endl;
}
}
}
else
{
ll bais = b * get_inv(a - 1, p) % p;
ll y0 = (x0 + bais) % p;
ll inv_y0 = get_inv(y0, p);
m = min(maxn, n);
ma = p / maxn + 3;
ll init_d = Init(a, p);
while (Q--)
{
ll v;
cin>>v;
v = (v + bais) % p;
v = (v * inv_y0) % p;
ll res = BSGS(a, v, p, init_d);
if (res >= n)
res = -1;
cout<<res<<endl;
}
}
}
}