【算法乱讲】BSGS

Description

Given a prime P, 2 <= P < 231, an integer B, 2 <= B < P, and an integer N, 1 <= N < P, compute the discrete logarithm of N, base B, modulo P. That is, find an integer L such that BL== N (mod P)

Input

Read several lines of input, each containing P,B,N separated by a space.

Output

For each line print the logarithm on a separate line. If there are several, print the smallest; if there is none, print "no solution".

Sample Input

5 2 1
5 2 2
5 2 3
5 2 4
5 3 1
5 3 2
5 3 3
5 3 4
5 4 1
5 4 2
5 4 3
5 4 4
12345701 2 1111111
1111111121 65537 1111111111

Sample Output

0
1
3
2
0
3
1
2
0
no solution
no solution
1
9584351
462803587



显然，这是一道bsgs的裸题
那么bsgs是什么玩意呢，
我们先玩一玩式子
令 m=ceil(sqrt(p))设a^(m*i+j)=b(mod p)

显然 a^j*a^(m*i)=b(mod p)　
　    a^j=b*a^(-m*i) (mod p)

因此，我们预处理所有可能的a^j丢进哈希表中然后我们枚举i，
看看有没有可能对应的j所以我们的算法时间复杂度为O(n^0.5)

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<map>
#include<vector>
#include<set>
#define il inline
#define re register
using namespace std;
typedef long long ll;
struct hash_set{
    ll v[111111];
    int next[111111],g[111111],w[111111],tot;
    il void clear(){
        memset(g,false,sizeof(g));tot=0;
    }
    il void insert(ll h,int f){
        v[++tot]=h;
        w[tot]=f;
        next[tot]=g[h%100007];
        g[h%100007]=tot;
    }
    il int find(ll h){
        for(re int i=g[h%100007];i;i=next[i])
            if(h==v[i]) return w[i];
        return -1;
    }
} p;
ll A,B,P,m,t,s;
il ll ksm(re ll base,re ll pow){
    if(pow<0){
        cout<<"-1";exit(0);
    }
    ll ans=1;
    for(;pow;pow>>=1){
        if(pow&1) ans=ans*base%P;
        base=base*base%P; 
    }
    return ans;
}
il ll rev(re ll a){
    return ksm(a,P-2);
}
il void init(){
    p.clear();
    m=ceil(sqrt(P));t=1;
    for(int i=0;i<m;i++){
        if(p.find(t)<0) p.insert(t,i);
        t=t*A%P;
    }
    //cout<<endl;
    for(int i=0,l;i<=P/m;i++){
        t=rev(ksm(A,m*i));
    //    cout<<t<<" "<<m*i<<" ";
        s=t*B%P;
    //    cout<<s<<endl;
        l=p.find(s);
        if(l>=0){
            printf("%lld\n",m*i+l);
            return;
        }
    }
    printf("no solution\n");
}
int main(){
    while(scanf("%lld%lld%lld",&P,&A,&B)!=EOF){
        init();
    }
    return 0;
}