快速离散对数
用法
给定原根 \(g\) 以及模数 \(\mathrm{mod}\)，\(T\) 次询问 \(x\) 的离散对数。
复杂度 \(\mathcal O(\mathrm{mod}^{2/3} + T \log \mathrm{mod})\)。
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79 | #include<bits/stdc++.h>
using namespace std;
int power(int a, int b, int p){
int r = 1;
while(b){
if(b & 1) r = 1ll * r * a % p;
b >>= 1, a = 1ll * a * a % p;
}
return r;
}
namespace BSGS {
unordered_map <int, int> M;
int B, U, P, g;
void init(int g, int P0, int B0){
M.clear();
B = B0;
P = P0;
U = power(power(g, B, P), P - 2, P);
int w = 1;
for(int i = 0;i < B;++ i){
M[w] = i;
w = 1ll * w * g % P;
}
}
int solve(int y){
int w = y;
for(int i = 0;i < P / B;++ i){
if(M.count(w)){
return i * B + M[w];
}
w = 1ll * w * U % P;
}
return -1;
}
}
const int MAXN = 1e5 + 3;
int H[MAXN], P[MAXN], H0, p, h, g, mod;
bool V[MAXN];
int solve(int x){
if(x <= h){
return H[x];
}
int v = mod / x, r = mod % x;
if(r < x - r){
return ((H0 + solve(r)) % (mod - 1) - H[v] + mod - 1) % (mod - 1);
} else {
return (solve(x - r) - H[v + 1] + mod - 1) % (mod - 1);
}
}
int main(){
ios :: sync_with_stdio(false);
cin.tie(nullptr);
int T;
cin >> g >> mod;
h = sqrt(mod) + 1;
BSGS :: init(g, mod, sqrt(1ll * mod * sqrt(mod) / log10(mod)));
H0 = BSGS :: solve(mod - 1);
H[1] = 0;
for(int i = 2;i <= h;++ i){
if(!V[i]){
P[++ p] = i;
H[i] = BSGS :: solve(i);
}
for(int j = 1;j <= p && P[j] <= h / i;++ j){
int &p = P[j];
H[i * p] = (H[i] + H[p]) % (mod - 1);
V[i * p] = true;
if(i % p == 0)
break;
}
}
cin >> T;
while(T --){
int x, tmp = 0;
cin >> x;
cout << solve(x) << "\n";
}
return 0;
}
|