BSGS&EXBSGS

BSGS

求最小的非负整数\(x\)满足\(a^x\equiv{b}\pmod{p}\)（\(\gcd(a,p)=1\)）

\[a^x\equiv{b}\pmod{p} \]

令\(x=ti-k(t=\sqrt{p},k\in[0,t-1])\)

\[a^{ti-k}\equiv{b}\pmod{p} \]

\[a^{ti}\equiv{b}*a^k\pmod{p} \]

先枚举\(k\)，将\(b*a^k\)的值存入\(hash\)表

再枚举\(i\in[1,t]\)，若\((a^t)^i\)的值出现了，则\(x=i*t-k\)

EXBSGS

求最小的非负整数\(x\)满足\(a^x\equiv{b}\pmod{p}\)

• \(b=1,x=0\)
• \(a=0,b=0,x=1\)
• \(a=0,b\ne{0}\)无解，否则

\[a*a^{x-1}\equiv{b}\pmod{p} \]

设\(d=(a,p)\)，若\(d\nmid{b}\)无解，否则

\[\frac{a}{d}*a^{x-1}\equiv{\frac{b}{d}}\pmod{\frac{p}{d}} \]

令\(A=a,B=\frac{b}{d},C=\frac{a}{d},P=\frac{p}{d}\)，则

\[CA^{x}\equiv{B}\pmod{P} \]

递归即可，当\(d=1\)时可直接\(BSGS\)

Code

int bsgs(int a, int b, int p, int c, int k)
{
	hs.clear();
	int q = 1, t = (int)(sqrt(p)) + 1;
	for (int i = 0; i < t; i ++ )
	{
		hs[1ll * q * b % p] = i;
		q = 1ll * q * a % p;
	}
	int cnt = q;
	q = 1ll * q * c % p;
	for (int i = 1; i <= t; i ++ )
	{
		if (hs.find(q) != hs.end()) return k + i * t - hs[q];
		q = 1ll * q * cnt % p;
	}
	return -1;
}

int exbsgs(int a, int b, int p)
{
	a %= p, b %= p;　// 特别注意1
	if (b == 1) return 0;
	if (!a)
	{
		if (!b) return 1;
		return -1;
	}
	int k = 0, A = a, B = b, C = 1, P = p, d = gcd(A, P);
	while (d != 1)
	{
		if (B % d) return -1;
		B /= d, P /= d, C = 1ll * C * (A / d) % P, k ++ ; // 特别注意2（除了d都是大写）
		d = gcd(A, P);
		if (C == B) return k; // 特别注意3
	}
	return bsgs(A, B, P, C, k);
}