ZOJ 3254

zoj 3254 

传送门 : ZOJ 3254

题解 : 扩展BSGS + 判断循环节

#include <cstdio>#include <cmath>#include <algorithm>using namespace std;typedef long long ll;typedef unsigned long long ull;const int N = (1 << 16) + 10;ll A, P, D;ull M;/**注意答案范围*/struct Bj{    ll v;    int id;    bool operator < (const Bj &rhs) const{        return v == rhs.v ? id < rhs.id : v < rhs.v;    }}R[N];ll FastPowMod(ll a, ll b, ll p){    ll ret = 1 % p;    while(b){        if(b & 1) ret = ret * a % p;        a = a * a % p;        b >>= 1;    }    return ret;}int findL(ll v, int r){    int l = 0, h = r - 1;    while(l <= h){        int m = (l + h) >> 1;        if(R[m].v == v) return R[m].id;        if(R[m].v > v) h = m - 1;        else l = m + 1;    }    return -1;}void extend_gcd(ll a, ll b, ll &x, ll &y){    if(b == 0){        x = 1;        y = 0;        return;    }    extend_gcd(b, a % b, y, x);    y -= a / b * x;}ll inv (ll a, ll b, ll p){    ll x, y;    extend_gcd(a, p, x, y);    return ((b * x) % p + p) % p;}ll BSGS(ll b, ll p, ll n, ll phi){/**BSGS*/    int m = ceil(sqrt((double)p));    for(int i = 0; i < m; ++i) {        R[i].id = i;        R[i].v = i == 0 ? 1 % p : b * R[i - 1].v % p;    }    sort(R, R + m);    int cnt = 1;    for(int i = 1; i < m; ++i) if(R[i].v != R[i - 1].v) R[cnt++] = R[i];    ll tmp = n;    ll bm = inv(FastPowMod(b, m, p), 1, p);    for(int i = 0; i < m; ++i){        int pos = findL(tmp, cnt);/**二分*/        if(~pos) return i * m + pos;        tmp = tmp * bm % p;    }    return M + 1;}ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);}ll getPhi(ll x){/**欧拉*/    ll ret = x;    for(ll i = 2; i * i <= x; ++i){        if(x % i == 0){            while(x % i == 0) x /= i;            ret = ret / i * (i - 1);        }    }    if(x > 1) ret = ret / x * (x - 1);    return ret;}ll cal(ll b, ll p, ll phi){/**循环节*/    ll m = phi;    for(ll x = 2; x * x <= phi; ++x){        if(phi % x == 0){            //ll tmp = 1;            while(phi % x == 0) phi /= x;            while(m % x == 0 && FastPowMod(b, m / x, p) == 1) m /= x;            /**枚举phi的质因子（循环节必定是phi的约数）*/        }    }    if(phi > 1) {        while( m % phi == 0 && FastPowMod(b, m / phi, p) == 1) m /= phi;    }    return m;}ull solve(ll b, ll p, ll n){    ll d;    ll cnt = 0, ret = 0, tmp = 1 % p, tx = 1 % p, tn = n, tp = p;    ll Ret = 1 % p;    while((d = gcd(b, p)) != 1){        if(tx == tn) return 1;/**去约数完成之前找到*/        if(n % d){            return 0;        }        n /= d;        p /= d;        tmp = b / d * tmp % p;        tx = tx * b % tp;        ++cnt;    }    ll phi = getPhi(p);    ll R =  inv(tmp, n, p); /**逆元*/                                                                                                                           inv(tmp, n, p);    ll pos = BSGS(b, p, R, phi);    if((pos += cnt) <= M){        ++ret;        ret += (M - pos) / cal(b, p, phi);    }    return ret;}int main(){    //freopen("in.txt", "r", stdin);    while(~scanf("%lld%lld%lld%llu", &A, &P, &D, &M)){        ull ans = solve(A % P, P, D);        printf("%llu\n", ans);    }    return 0;}