BSGS 模板

模板如下：

扩展版本：
求解a^k=b  %p    求k,最小的k一定小于p,否则会重复，否则无解
***********************
gcd(a,p)=1时
设k=mi+v  m=sqrt(p);
i,v<=m

a^v=b*(a^-m)^i  %p

打表map for i=0~m-1 (a^i,i)
for i=0~m-1
check a^v存在

****************************
gcd(a,p)=t时
(A/t)^c*A^(x-c)=B/(t^c)  %C/(t^c)  c max
A^(x-c)=B/(t^c)*(A/t)^-c  %C/(t^c)



const int NN = 99991 ;          //sqrt(p)
int Hash[NN][2] ;
void insert(int id , LL vv){
    LL v = vv % NN ;
    while( Hash[v][0]!=-1 && Hash[v][1]!=vv){
        v++ ; if(v == NN) v-=NN ;
    }
    if(Hash[v][0]==-1 ){
        Hash[v][1] = vv ; Hash[v][0] = id ;
    }
}
int find(LL vv){
    LL v = vv % NN ;
    while( Hash[v][0]!=-1 && Hash[v][1]!=vv){
        v++ ;if(v == NN) v-=NN ;
    }
    if( Hash[v][0]==-1 )  return -1;
    return Hash[v][0] ;
}
void ex_gcd(LL a , LL b , LL& x , LL& y){
    if(b == 0){
        x = 1 ; y = 0 ;
        return ;
    }
    ex_gcd(b , a%b , x, y) ;
    LL t = x ;
    x = y;
    y = t - a/b*y ;
}
LL baby_step(LL A, LL B , LL C){        //A^x=B  %C  最小x，__gcd g++使用
    LL D=1 % C ,d=0;
    if(__gcd(A,C)!=1){
        LL ans = 1 ;
        for(LL i=0;i<=50;i++){
            if(ans == B)    return i ;
            ans = ans * A % C ;
        }
        LL tmp ;
        while( (tmp=__gcd(A,C)) != 1 ){
            if(B % tmp) return -1 ;
            d++ ;
            B/=tmp ;
            C/=tmp ;
            D = D*A/tmp%C ;
        }                       //D*A^(x-d)=B  %C
    }                   //printf("D=%lld A=%lld B=%lld C=%lld d=%lld
",D,A,B,C,d);
    memn(Hash);
    LL M = ceil( sqrt(C*1.0) ) ;
    LL rr = 1 ;
    for(int i=0;i<M;i++){
        insert(i, rr) ;
        rr = rr * A % C ;
    }                   //rr=A^M
    LL jj,x,y;
    for(int i=0;i<M;i++){
        ex_gcd(D, C , x, y) ;
        jj = find( (x * B % C+C)%C ) ;            //printf("f  %lld
",r);
        if(jj != -1){
            return  i*M+jj+d;
        }
        D = D * rr % C ;
    }
    return -1 ;
}


-1无解			sqrt(p)




=---------------------------------------


普通版本

//POJ 2417
//baby_step giant_step
// a^x = b (mod n) n为素数，a,b < n
// 求解上式 0 <= x < n的解
#include <cmath>
#include <cstdio>
#include <cstring>
#define MOD 76543
using namespace std;
int hs[MOD], head[MOD], next[MOD], id[MOD], top;
void insert(int x, int y)
{
    int k = x % MOD;
    hs[top] = x;
    id[top] = y;
    next[top] = head[k];
    head[k] = top++;
}
int find(int x)
{
    int k = x % MOD;
    for (int i = head[k]; i != -1; i = next[i])
        if (hs[i] == x)
            return id[i];
    return -1;
}
int BSGS(int a, int b, int n)
{
    memset(head, -1, sizeof(head));
    top = 1;
    if (b == 1)
        return 0;
    int m = sqrt(n * 1.0), j;
    long long x = 1, p = 1;
    for (int i = 0; i < m; i++, p = p * a % n)
        insert(p * b % n, i);
    for (long long i = m; ; i += m)
    {
        if ((j = find(x = x * p % n)) != -1)
            return i - j;
        if (i > n)
            break;
    }
    return -1;
}
int main()
{
    int a, b, n;
    while (~scanf("%d%d%d", &n, &a, &b))
    {
        int ans = BSGS(a, b, n);
        if (ans == -1)
            printf("no solution
");
        else
            printf("%d
", ans);
    }
}