bzoj 1420 Discrete Root

题目传送门

　　戳我来传送

题目大意

　　给定$k, p, a$，求$x^{k}equiv a pmod{p}$在模$p$意义下的所有根。

　　考虑模$p$下的某个原根$g$。

　　那么$x  = g^{ind_{g}x}, a = g^{ind_{g}a}$。

　　所以原方程转化为$g^{kcdot ind_{g}x}equiv g^{ind_{g}a} pmod{p}$。

　　所以方程等价于$kcdot ind_{g}x equiv ind_{g}a pmod{varphi(p)}$。

　　用exgcd解出$ind_{g}x$的所有可能的解，再用快速幂算出$x$即可。

Code

/**
 * bzoj     
 * Problem#1420
 * Accepted
 * Time: 44ms
 * Memory: 2032k 
 */
#include <bits/stdc++.h>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef bool boolean;

typedef class HashMap {
    private:
        static const int M = 46666; 
    public:
        int ce;
        int h[M], key[M], val[M], next[M];
        
        HashMap() {    }
        
        void insert(int k, int v) {
            int ha = k % M;
            for (int i = h[ha]; ~i; i = next[i])
                if (key[i] == k) {
                    val[i] = v;
                    return;
                }
            ++ce, key[ce] = k, val[ce] = v, next[ce] = h[ha], h[ha] = ce;
        }
        
        int operator [] (int k) {
            int ha = k % M;
            for (int i = h[ha]; ~i; i = next[i])
                if (key[i] == k)
                    return val[i];
            return -1;
        }
        
        void clear() {
            ce = -1;
            memset(h, -1, sizeof(h));
        }
}HashMap; 

void exgcd(int a, int b, int& d, int& x, int& y) {
    if (!b)
        d = a, x = 1, y = 0;
    else {
        exgcd(b, a % b, d, y, x);
        y -= (a / b) * x;
    }
}

int qpow(int a, int pos, int p) {
    int pa = a, rt = 1;
    for ( ; pos; pos >>= 1, pa = pa * 1ll * pa % p)
        if (pos & 1)
            rt = rt * 1ll * pa % p;
    return rt;
}

int inv(int a, int n) {
    int d, x, y;
    exgcd(a, n, d, x, y);
    return (x < 0) ? (x + n) : (x);
}

int p, k, a;
int g;

inline void init() {
    scanf("%d%d%d", &p, &k, &a);
}

HashMap mp;
int ind(int a, int b, int p) {
    mp.clear();
    int cs = sqrt(p - 0.5), ac = qpow(a, cs, p), iac = b * 1ll * inv(ac, p) % p, pw = 1;
    for (int i = cs - 1; ~i; i--)
        iac = iac * 1ll * a % p, mp.insert(iac, i);
    for (int i = 0; i < p; i += cs, pw = pw * 1ll * ac % p)
        if (~mp[pw])
            return mp[pw] + i;
    return -1;
}

int top = 0;
int fac[50];
void getfactors(int x) {
    for (int i = 2; i * i <= x; i++) {
        if (!(x % i)) {
            fac[++top] = i;
            while (!(x % i))    x /= i; 
        }
    }
    if (x != 1)    fac[++top] = x;
}

boolean isroot(int x) {
    int pos = p - 1;
    for (int i = 1; i <= top; i++)
        if (qpow(x, pos / fac[i], p) == 1)
            return false;
    return true;
}

inline void solve() {
    if (p == 1 || !a) {
        printf("1
0");
        return;
    }
    if (p == 2)
        g = 1;
    else {
        getfactors(p - 1);
        for (g = 2; !isroot(g); g++);
    }

    int inda = ind(g, a, p), d, x, y, cir;
    exgcd(k, p - 1, d, x, y);
    if (inda % d) {
        puts("0");
        return;
    }
    cir = (p - 1) / d;
    x = x * 1ll * (inda / d) % cir;
    (x <= 0) ? (x += cir) : (0);
    vector<int> res;
    for (; x < p; x += cir)
        res.push_back(qpow(g, x, p));
    sort(res.begin(), res.end());
    printf("%d
", (signed) res.size());
    for (int i = 0; i < (signed) res.size(); i++)
        printf("%d
", res[i]);
}

int main() {
    init();
    solve();
    return 0;
}