poj2417 Discrete Logging BSGS裸题

给a^x == b (mod c)求满足的最小正整数x，

用BSGS求，令m=ceil(sqrt(m)),x=im-j，那么a^(i*m)=b*aj%p;,

我们先枚举j求出所有的ba^{j%p,1<=j<m复杂度O（sqrt(c)）,然后枚举1<=i<=m，求出a}(im)在ba^j找满足条件的答案，最后的答案就是第一个满足条件的i*m-j，复杂度O（sqrt（c））

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define fio ios::sync_with_stdio(false);cin.tie(0)

using namespace std;

const double eps=1e-6;
const int N=1000000+10,maxn=5000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f;

map<ll,ll>ma;
inline ll quick(ll a,ll b,ll p)
{
    ll ans=1;
    while(b)
    {
        if(b&1)ans=ans*a%p;
        a=a*a%p;b>>=1;
    }
    return ans;
}
int main()
{
    ll p,a,b;
    while(~scanf("%lld%lld%lld",&p,&a,&b))
    {
        if(a%p==0)
        {
            puts("no solution");
            continue;
        }
        ma.clear();
        ll m=ceil(sqrt(p));
        ll now=b%p;
        ma[now]=0;
        for(int i=1;i<=m;i++)
        {
            now=(now*a)%p;
            ma[now]=i;
        }
        ll ans=-1,t=quick(a,m,p);now=1;
        for(int i=1;i<=m;i++)
        {
            now=now*t%p;
            if(ma[now])
            {
                ans=i*m-ma[now];
                break;
            }
        }
        if(ans!=-1)printf("%lld\n",(ans%p+p)%p);
        else puts("no solution");
    }
    return 0;
}
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