BSGS算法及扩展

BSGS算法

$Baby Step Giant Step$算法，即大步小步算法，缩写为$BSGS$
拔山盖世算法

它是用来解决这样一类问题
$y^x = z (mod\ p)$，给定$y,z,p>=1$求解$x$

普通的$BSGS$只能用来解决$gcd(y,p)=1$的情况

设$x=a*m+b, m=\lceil \sqrt p \rceil, a\in[0,m), b\in[0,m)$
那么$y^{a*m}=z*y^{-b} (mod\ p)$

怎么求解，为了方便，设$x=a*m-b$
那么$y^{a*m}=z*y^b(mod \ p), a\in(0,m+1]$

直接暴力辣，把右边的$b$枚举$[0,m)$，算出$z*y^b(mod \ p)$，哈希存起来
然后左边$a$枚举$(0, m+1]$，算出$y^{a*m}(mod \ p)$查表就行了

然后不知道为什么要用$exgcd$，只会$map$...

代码

[SDOI2011]计算器

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;

template <class Int>
IL void Input(RG Int &x){
    RG int z = 1; RG char c = getchar(); x = 0;
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    x *= z;
}

IL void None(){
	puts("Orz, I cannot find x!");
}

int p;

IL int Pow(RG ll x, RG ll y){
	RG ll ret = 1;
	for(; y; x = x * x % p, y >>= 1)
		if(y & 1) ret = ret * x % p;
	return ret;
}

map <int, int> pw;

IL void BSGS(RG int x, RG int y){
	pw.clear();
	if(y == 1){
		puts("0");
		return;
	}
	RG int ans = -1, m = sqrt(p) + 1, xx, s = y;
	for(RG int i = 0; i < m; ++i){
		pw[s] = i;
		s = 1LL * s * x % p;
	}
	xx = Pow(x, m), s = 1;
	for(RG int i = 1; i <= m + 1; ++i){
		s = 1LL * s * xx % p;
		if(pw.count(s)){
			ans = i * m - pw[s];
			break;
		}
	}
	if(ans < 0) None();
	else printf("%d\n", ans);
}

int T, k, y, z;

int main(RG int argc, RG char* argv[]){
	for(Input(T), Input(k); T; --T){
		Input(y), Input(z), Input(p);
		if(k == 1) printf("%d\n", Pow(y, z));
		else if(k == 2){
			RG int d = (y % p) ? 1 : p;
			if(z % d) None();
			else printf("%lld\n", 1LL * Pow(y, p - 2) * z % p);
		}
		else{
			if(y % p) BSGS(y % p, z % p);
			else None();
		}
	}
    return 0;
}

扩展BSGS

对于$gcd(y, p)\ne1$怎么办？

我们把它写成$y*y^{x-1}+k*p=z, k\in Z$的形式

根据$exgcd$的理论
那么如果$y,p$的$gcd$不是$z$的约数就不会有解

设$d=gcd(y,p)$
那么

$$\frac{y}{d}*y^{x-1}+k*\frac{p}{d}=\frac{z}{d}$$

递归到$d=1$
设之间的所有的$d$的乘积为$g$，递归$c$次
令$x'=x-c, p'=\frac{p}{g},z'=\frac{z}{g}$
那么

$$y^{x'}*\frac{y^c}{g}=z'(mod \ p')$$

那么$BSGS$求解就好了

代码

SPOJMOD Power Modulo Inverted

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
# define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
using namespace std;
typedef long long ll;

template <class Int>
IL void Input(RG Int &x){
    RG int z = 1; RG char c = getchar(); x = 0;
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    x *= z;
}

map <int, int> pw;

IL int Gcd(RG int x, RG int y){
	return !y ? x : Gcd(y, x % y);
}

IL int Pow(RG ll x, RG ll y, RG int p){
	RG ll ret = 1;
	for(; y; x = x * x % p, y >>= 1)
		if(y & 1) ret = ret * x % p;
	return ret;
}

int a, b, p;

IL int EX_BSGS(){
	if(b == 1) return 0;
	pw.clear();
	RG int cnt = 0, t = 1, s, x, m;
	for(RG int d = Gcd(a, p); d != 1; d = Gcd(a, p)){
		if(b % d) return -1;
		++cnt, b /= d, p /= d, t = 1LL * t * a / d % p;
		if(b == t) return cnt;
	}
	s = b, m = sqrt(p) + 1;
	for(RG int i = 0; i < m; ++i){
		pw[s] = i;
		s = 1LL * s * a % p;
	}
	x = Pow(a, m, p), s = t;
	for(RG int i = 1; i <= m; ++i){
		s = 1LL * s * x % p;
		if(pw.count(s)) return i * m - pw[s] + cnt;
	}
	return -1;
}

int ans;

int main(RG int argc, RG char* argv[]){
	for(Input(a), Input(p), Input(b); a + b + p;){
		a %= p, b %= p, ans = EX_BSGS();
		if(ans < 0) puts("No Solution");
		else printf("%d\n", ans);
		Input(a), Input(p), Input(b);
	}
    return 0;
}