BSGS(基础篇，题目+详解)

基础篇

问题：

给出$a，b，p$，其中$gcd(a,p)=1$，求$x$满足
$$a^x\equiv b(modp)$$

思路：

设$x=A\sqrt{p}-B$其中$A\in [1,\sqrt{p}],B\in [0,\sqrt{p}]$，得到问题的变形
$$\begin{gathered} a^{A\sqrt{p}-B}\equiv b(modp) \\ {a}^{A\sqrt{p}}\equiv b{a}^B(modp) \end{gathered}$$
我们先枚举$B$，算出每个$b{a}^{B}modp$，用$unordered\_map$存起来，再枚举$A$，计算出$a^{A\sqrt{p}}$，在$unordered\_map$中找相同的值，这样的$A,B$就能恰好凑成一对答案。复杂度$O(\sqrt{p})$，如果用$map$的话，就多一个$log$。

模板：

unordered_map<ll, ll> mp;
ll bsgs(ll a, ll b, ll p) {
     
    if(a % p == 0) return -1;
    mp.clear();
    ll k = ceil(sqrt(p));
    for(int i=0; i<=k; i++) {
     
        mp[b] = i;
        b = b * a % p;
    }
    ll aa = qpow(a, k, p), A = aa;
    for(int i=1; i<=k; i++) {
     
        if(mp[aa]) {
     
            return 1ll * i * k - mp[aa] + 1;
        }
        aa = aa * A % p;
    }
    return -1;
}

题目：

[TJOI2007]可爱的质数/模板BSGS

#include
using namespace std;
typedef long long ll;
unordered_map<ll, ll> mp;
ll qpow(ll x, ll y, ll mod) {
     
	ll ans = 1;
	while(y) {
     
		if(y & 1) ans = ans * x % mod;
		x = x * x % mod;
		y >>= 1;
	}
	return ans;
}
ll bsgs(ll a, ll b, ll p) {
     
	if(a % p == 0 && b != 0) return -1;
	mp.clear(); 
	ll k = ceil(sqrt(p));
	for(int i=0; i<=k; i++) {
     
		mp[b] = i;
		b = b * a % p;
	}
	ll aa = qpow(a, k, p), A = aa;
	for(int i=1; i<=k; i++) {
     
		if(mp[A]) {
     
			return 1ll*i*k-mp[A];
		}
		A = A * aa % p;
	}
	return -1;
}
int main() {
     
#ifndef ONLINE_JUDGE 
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif  
    ll a, b, p;
    cin >> p >> a >> b;
    ll ans = bsgs(a, b, p);
    if(ans == -1) cout << "no solution";
    else cout << ans;
    return 0;
}

P2485 [SDOI2011]计算器
快速幂+扩展欧几里得+bsgs。

#include
using namespace std;
#define int long long
const int inf = 0x7fffffff;
unordered_map<int, int> mp;
int qpow(int x, int y, int mod) {
     
	int ans = 1;
	while(y) {
     
		if(y & 1) ans = 1ll * ans * x % mod;
		x = 1ll * x * x % mod;
		y >>= 1;
	}
	return ans;
}
int exgcd(int &x, int &y, int a, int b) {
     
	if(b == 0) {
     
		x = 1, y = 0;
		return a;
	}
	int d = exgcd(x, y, b, a%b);
	int tmp = y;
	y = x-a/b*y;
	x = tmp;
	return d;
}
signed main() {
     
#ifndef ONLINE_JUDGE 
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
	int n, k;
	cin >> n >> k;
	for(int i=1; i<=n; i++) {
     
		int y, z, p;
		cin >> y >> z >> p;
		if(k == 1) cout << qpow(y, z, p) << endl;
		else if(k == 2) {
     
			int u, v;
			int d = exgcd(u, v, y, p);
			if(z % d) puts("Orz, I cannot find x!");
			else {
     
				int l = p/d, r = z / d;
				cout << (1ll * (r % l) * (u % l) % l + l) % l << endl;
			}
		}
		else {
     
			mp.clear();
			if(y % p == 0 && z % p) {
     
				puts("Orz, I cannot find x!");
				continue;
			}
			int now = z%p, m = ceil(sqrt(p)), f = 0;
			mp[now] = 0;
			for(int i=1; i<m; i++) {
     
				now = now * y % p;
				mp[now] = i;
			}
			now = 1;
			int A = qpow(y, m, p);
			for(int i=1; i<=m; i++) {
     
				now = now * A % p;
				if(mp.count(now)) {
     
					cout << ((i*m - mp[now]) % p + p) % p << endl;
					f = 1;
					break;
				} 
			}
			if(!f) puts("Orz, I cannot find x!");
		}
	}
	return 0;
}