数论原根 及其求法

定义2：若G为n的原根，则当gcd(i,φ(n))==1，Gi也为n的原根
可以通过先求出n的最小原根，来枚举得到其他的合法原根。

#include 
#include 
#include 
#define ll long long
using namespace std;
const int maxn = 1e6 + 10;
bool vis[maxn] = {0};
int pri[maxn / 10] = {0};
int phi[maxn] = {0};
int n,k;
int temp[maxn] = {0};
int cnt = 0;
ll res[maxn] = {0};

void init () {
    for (int i = 2;i < maxn; ++ i) {
        if (!vis[i]) {
            pri[cnt ++] = i;
            for (int j = i + i;j < maxn;j += i) {
                vis[j] = 1;
            }
        }
    }
    phi[1] = 1;
    memset (vis,0,sizeof (vis));
    for (int i = 2;i < maxn; ++ i) {
        if (!vis[i]) {
            for (int j = i;j < maxn; j += i) {
                vis[j] = 1;
                if (phi[j] == 0) phi[j] = j;
                phi[j] = phi[j] / i * (i - 1);
            }
        }
    }
}

void div (int n) {
    k = 0;
    for (int i = 0;pri[i] * pri[i] <= n; ++ i) {
        if (n % pri[i] == 0) {
            temp[k ++] = pri[i];
            while (n % pri[i] == 0) n /= pri[i];
        }
    }
    if (n > 1) temp[k ++] = n;
}

ll lpow (ll a,ll b,ll m) {
    ll ans = 1;
    a %= m;
    while (b) {
        if (b & 1) {
            ans = ans * a % m;
            b --;
        }
        b >>= 1;
        a = a * a % m;
    }
    return ans % m;
}

bool check (int n) {
    if (n == 2 || n == 4) return true;
    if (n % 2 == 0) n /= 2;
    if (binary_search (pri + 1,pri + cnt,n)) return true;
    for (int i = 1;i < cnt && pri[i] * pri[i] <= n; ++ i) {
        if (n % pri[i] == 0) {
            while (n % pri[i] == 0) {
                n /= pri[i];
            }
            if (n == 1)return true;
            break;
        }
    }
    return false;
}

int main () {
    init ();
    while (scanf ("%d",&n) != EOF) {
        if (n == 2) {
            printf ("1\n");
            continue ;
        }
        if (check(n) == 0) {
            printf ("-1\n");
            continue;
        }
        int tot = 0;
        int p = phi[n];
        div (p);
        for (int q = 2;q < n; ++ q) {
            if (__gcd (q,n) != 1) continue;
            int flag = 1;
            for (int i = 0;i < k; ++ i) {
                int t = p / temp[i];
                if (lpow(q, t, n) % n == 1) {
                    flag = 0;
                    break;
                }
            }
            if(flag) {
                res[tot ++] = q;
                break;
            }
        }
        ll cc = res[0];
        tot = 0;
        for (int i = 1;i < p; ++ i) {
            if (__gcd (i,p) == 1) {
                res[tot ++] = cc;
            }
            cc = cc * res[0] % n;
        }
        sort (res,res + tot);
        if (tot) {
            printf ("%lld",res[0]);
            for (int i = 1;i < tot; ++ i) {
                printf (" %lld",res[i]);
            }
        }
        else {
            printf ("-1");
        }
        printf ("\n");
    }
    return 0;
}