BSGS算法及擴展

BSGS算法

\(Baby Step Giant Step\)算法，即大步小步算法，縮寫爲\(BSGS\)
拔山蓋世算法

它是用來解決這樣一類問題
\(y^x = z (mod\ p)\)，給定\(y,z,p>=1\)求解\(x\)

普通的\(BSGS\)只能用來解決\(gcd(y,p)=1\)的狀況

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

怎麼求解，爲了方便，設\(x=a*m-b\)
那麼\(y^{a*m}=z*y^b(mod \ p), a\in(0,m+1]\)

直接暴力辣，把右邊的\(b\)枚舉\([0,m)\)，算出\(z*y^b(mod \ p)\)，哈希存起來
而後左邊\(a\)枚舉\((0, m+1]\)，算出\(y^{a*m}(mod \ p)\)查表就好了

而後不知道爲何要用\(exgcd\)，只會\(map\)...

代碼

[SDOI2011]計算器

# 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;
}

擴展BSGS

對於\(gcd(y, p)\ne1\)怎麼辦？

咱們把它寫成\(y*y^{x-1}+k*p=z, k\in Z\)的形式

根據\(exgcd\)的理論
那麼若是\(y,p\)的\(gcd\)不是\(z\)的約數就不會有解

設\(d=gcd(y,p)\)
那麼
\[\frac{y}{d}*y^{x-1}+k*\frac{p}{d}=\frac{z}{d}\]

遞歸到\(d=1\)
設之間的全部的\(d\)的乘積爲\(g\)，遞歸\(c\)次
令\(x'=x-c, p'=\frac{p}{g},z'=\frac{z}{g}\)
那麼
\[y^{x'}*\frac{y^c}{g}=z'(mod \ p')\]

那麼\(BSGS\)求解就行了

代碼

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;
}