BSGS&EXBSGS

BSGS

求最小的非负整数(x)满足(a^xequiv{b}pmod{p})（(pin{P})）

[a^xequiv{b}pmod{p} ]

令(x=ti-k(t=sqrt{p},kin[0,t-1]))

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

[a^{ti}equiv{b}*a^kpmod{p} ]

先枚举(k)，将(b*a^k)的值存入(hash)表

再枚举(iin[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^xequiv{b}pmod{p})

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

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

设(d=(a,p))，若(d mid{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);
}