51Nod-1038-X^A Mod P

ACM模版

描述

题解

第一次接触原根这个玩意儿，感觉真恶心，一脸懵逼啊……

leader_win’s blog 讲得倒是十分详细，可是我依然懵逼着，数论真的恶心，太多太多╮(╯﹏╰)╭……

代码

#include <cmath>#include <vector>#include <cstdio>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;const int MAXN = 100100;ll qk_pow(ll a, ll b, ll mod){    ll ret = 1;    while (b)    {        if (b & 1)        {            ret = ret * a % mod;        }        b >>= 1;        a = a * a % mod;    }    return ret;}ll ex_gcd(ll a, ll b, ll &x, ll &y){    if (b == 0)    {        x = 1;        y = 0;        return a;    }    else    {        ll r = ex_gcd(b, a % b, y, x);        y -= x * (a / b);        return r;    }}vector<ll> a;bool check(ll g, ll p){    for (int i = 0; i < a.size(); i++)    {        if (qk_pow(g, (p - 1) / a[i], p) == 1)        {            return 0;        }    }    return 1;}//  求解原根ll primitive_root(ll p){    ll tmp = p - 1;    for (int i = 2; i <= tmp / i; i++)    {        if (tmp % i == 0)        {            a.push_back(i);            while (tmp % i == 0)            {                tmp /= i;            }        }    }    if (tmp != 1)    {        a.push_back(tmp);    }    ll g = 1;    while (true)    {        if (check(g, p))        {            return g;        }        ++g;    }}struct sa{    ll x;    int id;    bool operator < (const sa &b) const    {        if (x == b.x)        {            return id < b.id;        }        return x < b.x;    }} rec[MAXN];//  求解离散对数ll discerte_log(ll x, ll n, ll m){    int s = (int)(sqrt((double)m + 0.5));    while ((ll)s * s <= m)    {        s++;    }    ll cur = 1;    sa tmp;    for (int i = 0; i < s; i++)    {        tmp.x = cur;        tmp.id = i;        rec[i] = tmp;        cur = cur * x % m;    }    sort(rec, rec + s);    ll mul = qk_pow(cur, m - 2, m) % m;    cur = 1;    for (int i = 0; i < s; i++)    {        ll more = n * cur % m;        tmp.x = more;        tmp.id = -1;        int j = (int)(lower_bound(rec, rec + s, tmp) - rec);        if (rec[j].x == more)        {            return i * s + rec[j].id;        }        cur = cur * mul % m;    }    return -1;}//  求解n次剩余vector<ll> residue(ll p, ll n, ll a){    vector<ll> ret;    if (a == 0)    {        ret.push_back(0);        return ret;    }    ll g = primitive_root(p);    ll m = discerte_log(g, a, p);    if (m == -1)    {        return ret;    }    ll A = n, B = p - 1, C = m, x, y;    ll G = ex_gcd(A, B, x, y);    if (C % G != 0)    {        return ret;    }    x = x * (C / G) % B;    ll delta = B / G;    for (int i = 0; i < G; i++)    {        x = ((x + delta) % B + B) % B;        ret.push_back(qk_pow(g, x, p));    }    sort(ret.begin(), ret.end());    ret.erase(unique(ret.begin(), ret.end()), ret.end());    return ret;}ll P, A, B;int main(){    int T;    scanf("%d", &T);    while (T--)    {        a.clear();        scanf("%lld%lld%lld", &P, &A, &B);        vector<ll> ans;        ans = residue(P, A, B);        if (ans.empty())        {            puts("No Solution");        }        else        {            for (int i = 0; i < ans.size(); i++)            {                printf("%lld ", ans[i]);            }            putchar(10);        }    }    return 0;}