洛谷P3846+P4195 BSGS及扩展BSGS模板题

1. BSGS

例题：P3846 【模板】BSGS / [TJOI2007] 可爱的质数

BSGS 的 oi wiki 介绍：链接

但是我感觉还是 chebs大佬的博客 更好理解。

示例程序：

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

ll p, b, n, s;
unordered_map<ll, ll> mp;

int main() {
    cin >> p >> b >> n;
    if (n == 1) {
        cout << 0 << endl;
        return 0;
    }
    s = ceill(sqrtl(p));

    for (ll i = 0, t = n; i < s; i++, t = t * b % p) {
        mp[t] = i;
    }
    ll bs = 1;
    for (int i = 0; i < s; i++) {
        (bs *= b) %= p;
    }
    for (ll i = 1, t = bs; i <= s; i++, t = t * bs % p) {
        auto it = mp.find(t);
        if (it != mp.end()) {
            ll l = i * s - (it->second);
            cout << l << endl;
            return 0;
        }
    }
    cout << "no solution\n";
    return 0;
}

双倍经验

2. 扩展BSGS（exBSGS）

例题：P4195 【模板】扩展 BSGS / exBSGS

exBSGS 的 oi wiki 介绍：链接

示例程序：

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

ll gcd(ll a, ll b) {
    return b ? gcd(b, a%b) : a;
}

// 返回 a^x = b(mod m) 的最小非负整数x，没有则返回-1
ll exBSGS(ll a, ll b, ll m) {
    a %= m, b %= m;
    if (b == 1 || m == 1)
        return 0;

    ll d, ax = 1, cnt = 0;
    while ((d = gcd(a, m)) > 1) {
        if (b % d)
            return -1;
        m /= d;
        b /= d;
        cnt++;
        (ax *= a / d) %= m;
        if (ax == b)
            return cnt;
    }

    unordered_map<ll, ll> mp;
    ll s = ceill(sqrtl(m));

    for (ll i = 0, t = b; i < s; i++, t = t * a % m) {
        mp[t] = i;
    }

    ll base = 1;
    for (int i = 0; i < s; i++) {
        (base *= a) %= m;
    }

    for (ll i = 1, t = ax; i <= s; i++) {
        (t *= base) %= m;
        auto it = mp.find(t);
        if (it != mp.end())
            return cnt + i * s - (it->second);
    }

    return -1;
}

ll a, p, b;

int main() {
    while (~scanf("%lld%lld%lld", &a, &p, &b) && a+p+b) {
        ll res = exBSGS(a, b, p);
        if (res == -1)
            puts("No Solution");
        else
            printf("%lld\n", res);
    }
    return 0;
}