bzoj1319

数论

这个幂指数很难搞，那么我们取个log

去取log得有底数，那么自然这个底数能表示出所有的数

原根满足这个性质

那么我们求出原根，再去log

变成k*ind(x)=ind(a) (mod phi(p))

phi(p)=p-1

又因为g^ind(a)=a (mod p)

那么我们用bsgs求出ind(a)

那么我们就要解出ind(x)的所有解，用exgcd

然后求出最小自然数数解，然后通解是x+b*t

那么每次加b就行了

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + ;
int p, k, a, top;
ll Ans[N], st[N];
ll power(ll x, ll t)
{
    ll ret = ;
    for(; t; t >>= , x = x * x % p) if(t & ) ret = ret * x % p;
    return ret;
}
int find_root(int p)
{
    if(p == ) return ;
    int tmp = p - ;
    for(int i = ; i * i <= p; ++i) if(tmp % i == )
    {
        st[++top] = i;
        while(tmp % i == ) tmp /= i;
    }
    if(tmp != ) st[++top] = tmp;
    for(int g = ; g < p; ++g)
    {
        bool flag = true;
        for(int j = ; j <= top; ++j)
        {
            if(power(g, (p - ) / st[j]) == )
            {
                flag = false;
                break;
            }
        }
        if(flag) return g;
    }
    return -;
}
int bsgs(int a, int b, int p)
{
    map<ll, int> mp;
    int m = sqrt(p);
    ll x = , pw = power(a, m);
    for(int i = ; i <= m; ++i)
    {
        mp[x] = i * m;
        x = x * pw % p;
    }
    x = ;
    for(int i = ; i < m; ++i)
    {
        ll tmp = power(x, p - ) * b % p;
        if(mp.find(tmp) != mp.end()) return mp[tmp] + i;
        x = x * a % p;
    }
}
ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if(b == )
    {
        x = ;
        y = ;
        return a;
    }
    int ret = exgcd(b, a % b, y, x);
    y -= (a / b) * x;
    return ret;
}
int main()
{
    cin >> p >> k >> a;
    ll g = find_root(p), c = bsgs(g, a, p), b = p - , x, y, gcd = exgcd(k, b, x, y);
    if(c % gcd != ) return puts(""), ;
    c /= gcd;
    b /= gcd;
    k /= gcd;
    x = (x * c % b + b) % b;
    while(x < p)
    {
        Ans[++Ans[]] = power(g, x);
        x += b;
    }
    sort(Ans + , Ans + Ans[] + );
    printf("%d\n", Ans[]);
    for(int i = ; i <= Ans[]; ++i) printf("%lld\n", Ans[i]);
    return ;
}