BSGS算法（模板）

BSGS （大步小步算法）

已知\(a、b、 c\)，求\(x\)。令\(a^x \equiv b \pmod c\)。

步骤

\[m = \lceil \sqrtｃ\ \rceil $$$$x = i*m-j\ \ (i\in[1, m], j\in[0, m])$$$$a^{i*m-j} \equiv b \pmod c$$$$a^{i*m}\equiv b*a^j \pmod c \]

枚举\(a^j(j\in[0, m])\)放入\(hash\)表里面，再枚举\(a^{i*m}\)，在\(hash\)表里面找有没有相同，如若有相同的那么就存在。

如果在询问较多的情况下，可以把\(bm = \lceil c^{\frac{2}{3}}\rceil\)、\(gm = \lceil c^{\frac{1}{3}}\rceil\)\(（i\in[1,gm],j\in[0, bm]）\)

实现代码

struct Hash{
    int head[maxm], cnt, mod;
    struct Node{
        int v, id, next;
    }node[maxn];
    void init(){
        mes(head, -1);
        cnt = 0;
        mod = 1333331;
    }
    void insert(int x, int id){
        int u = x%mod;
        node[++cnt].v = x;
        node[cnt].id = id;
        node[cnt].next = head[u];
        head[u] = cnt;
    }
    int find(int x){
        int u = x%mod;
        for(int i = head[u]; ~i; i = node[i].next){
            if(node[i].v == x)
                return node[i].id;
        }
        return -1;
    }
}hs;
  
ll qpow(ll a, ll b, int mod){
    ll ans = 1;
    while(b){
        if(b&1)
            ans = ans*a%mod;
        a = a*a%mod;
        b /= 2;
    }
    return ans;
}
struct BSGS{
    int bm, gm, a, b, x0, p;
      
    void init(){
        bm = (int)ceil(pow(p, 2.0/3));
        gm = (int)ceil(pow(p, 1.0/3));
        ll ans = 1;
        hs.init();
        for(int i = 0; i <= bm; i++){   //a^(i*bm - j)%p = b;
            hs.insert(ans, i);
            ans = ans*a%p;
        }
    }
  
    ll get(ll v){     
        v = qpow(v, p-2, p);
        ll num = qpow(a, bm, p);
        for(int i = 1; i <= gm; i++){
            v = v*num%p;
            int ans = hs.find(v);
            if(ans != -1)
                return 1ll*i*bm - ans;
        }
        return -1;
    }
}bsgs;
 

例题

牛客多校训练营第五场 - generator 2

\(x_n = (a*x_{n-1} + b)\%p\),给你\(n、x_0、a、b、p\)，\(Q\)个询问在\(x_0,x_1,x_2...,x_{n-1}\)找到最小符合\(x_i = v\)的\(i\),不存在输出\(-1\)。

思路

\[x_n = a*x_{n-1} + b$$$$x_n = a^n *x_0+b*\frac{1-a^n}{1-a}$$$$a^n = \frac{(a-1)_*x_n+b}{(a-1)*x_0+b}$$$$x_n = v, 令B = \frac{(a-1)_*v+b}{(a-1)*x_0+b}$$所以就是求$$a^n \equiv B\pmod p \]

\(a = 0、a = 1\)的情况要特殊考虑一下

AC代码

#include <map>
#include <set>
#include <list>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define  lowbit(x)  x & (-x)
#define  mes(a, b)  memset(a, b, sizeof a)
#define  fi         first
#define  se         second
#define  pb         push_back
#define  pii        pair<int, int>
  
typedef unsigned long long int ull;
typedef long long int ll;
const int    maxn = 1e6 + 10;
const int    maxm = 2e6 + 10;
const ll     INF  = 1e18 + 100;
const int    inf  = 0x3f3f3f3f;
const double pi   = acos(-1.0);
const double eps  = 1e-8;
using namespace std;
  
int n, m;
int cas, tol, T;
struct Hash{
    int head[maxm], cnt, mod;
    struct Node{
        int v, id, next;
    }node[maxn];
    void init(){
        mes(head, -1);
        cnt = 0;
        mod = 1333331;
    }
    void insert(int x, int id){
        int u = x%mod;
        node[++cnt].v = x;
        node[cnt].id = id;
        node[cnt].next = head[u];
        head[u] = cnt;
    }
    int find(int x){
        int u = x%mod;
        for(int i = head[u]; ~i; i = node[i].next){
            if(node[i].v == x)
                return node[i].id;
        }
        return -1;
    }
}hs;
  
ll qpow(ll a, ll b, int mod){
    ll ans = 1;
    while(b){
        if(b&1)
            ans = ans*a%mod;
        a = a*a%mod;
        b /= 2;
    }
    return ans;
}
struct BSGS{
    int bm, gm, a, b, x0, p;
      
    void init(){
        bm = (int)ceil(pow(p, 2.0/3));
        gm = (int)ceil(pow(p, 1.0/3));
        ll ans = 1;
        hs.init();
        for(int i = 0; i <= bm; i++){   //a^(i*bm - j)%p = b;
            hs.insert(ans, i);
            ans = ans*a%p;
        }
    }
  
    ll get(ll v){     
        v = qpow(v, p-2, p);
        ll num = qpow(a, bm, p);
        for(int i = 1; i <= gm; i++){
            v = v*num%p;
            int ans = hs.find(v);
            if(ans != -1)
                return 1ll*i*bm - ans;
        }
        return -1;
    }
}bsgs;
  
  
int main() {
    scanf("%d", &T);
    while(T--){
        ll n,x0, a, b, p;
        scanf("%lld%lld%lld%lld%lld", &n, &x0, &a, &b, &p);
        bsgs.x0 = x0; bsgs.a = a; bsgs.b = b; bsgs.p = p;
        bsgs.init();
        int q;ll v;
        scanf("%d", &q);
        ll inv = ((a-1)*x0%p+b+p)%p;
        if(inv == 0){
            while(q--){
                scanf("%lld", &v);
                if(x0 == v)
                    printf("0\n");
                else
                    printf("-1\n");
            }
            continue;
        }
        if(a == 0){
            while(q--){
                scanf("%lld", &v);
                if(v == x0)         //x = x0;
                    printf("0\n");
                else if(v == b) //xn = b
                    printf("1\n");
                else
                    printf("-1\n"); //无解
                continue;
            }
            continue;
        }
        if(a == 1){   //xn = x0 + nb;
            ll invb = qpow(b, p-2,p);
            while(q--){
                scanf("%lld", &v);
                if(b == 0){  
                    if(v == x0)
                        printf("0\n");
                    else
                        printf("-1\n");
                }
                else{   // n = (v - x0)/b;
                    ll ans = ((v-x0+p) % p * invb%p + p)%p;
                    if(ans < n)
                        printf("%lld\n", ans);
                    else
                        printf("-1\n");
                }
            }
            continue;
        }
        inv = qpow(inv, p-2, p);
        while(q--){
            scanf("%lld", &v);
            v = ((a-1)*v%p+b+p)%p*inv%p;
            ll ans = bsgs.get(v);
            if(ans == -1 || ans >= n)
                printf("-1\n");
            else
                printf("%lld\n", ans);
        }
    }
    return 0;
}