bsgs+求数列通项——bzoj3122（进阶指南模板该进）

/*
已知递推数列 F[i]=a*F[i-1]+b (%c)
解方程F[x]=t

an+1 = b*an + c
an+1 + c/(b-1) = b(an + c/(b-1))
an+1 + c/(b-1) = b^(n-1) * (a1+c/(b-1))

根据这个数列可得
F[x] = (F[1] + b/(a-1))*a^(x-1) - b/(a-1) = t;

(F[1] + b/(a-1))*a^(x-1) = t+b/(a-1)
a^(x-1) = (t+b/(a-1)) / (F[1]+b/(a-1))
用逆元算出右边，再用bsgs算x即可
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll p,a,b,f1,t;

ll Pow(ll a,ll b,ll p){
    ll res=;
    while(b){
        if(b%)
            res=res*a%p;
        b>>=;a=a*a%p;
    }
    return res;
}
ll inv(ll x,ll p){
    return Pow(x,p-,p);
}

ll bsgs(ll a,ll b, ll p){
    //用来求a^x=b, 令x=i*t-j,(a^t)^i = b*a^j
    //先把右边的t个值存下来
    map<ll,ll>hash;
    hash.clear();
    b%=p;
    ll t=sqrt(p)+;
    ll tmp=b;
    for(ll j=;j<t;j++){
        ll val=b*Pow(a,j,p)%p;
        hash[val]=j;
    }

    a=Pow(a,t,p);
    if(a==)return b==?:-;
    for(int i=;i<=t;i++){//这个地方把循环起点改为1
        ll val=Pow(a,i,p);
        int j=hash.find(val)==hash.end()?-:hash[val];
        if(j>= && i*t-j>=)
            return i*t-j;
    }
    return -;
}
//F[i]=a*F[i-1]+b (%c)
int main(){
    int T;cin>>T;
    while(T--){
        cin>>p>>a>>b>>f1>>t;
        if(f1==t){puts("");continue;}
        if(a==){
            if(t==b)puts("");
            else puts("-1");
            continue;
        }
        if(a==){
            if(b==){puts("-1");continue;}
            ll  ans = (((t-f1)%p+p)%p*inv(b,p))%p;
            cout<<ans+<<'\n';
            continue;
        }

        a%=p;b%=p;f1%=p;t%=p;

        ll tmp=b*inv(a-,p)%p;
        t=(t+tmp)%p;
        t=t*inv(f1+tmp,p)%p;
        t=t*a%p;

        cout<<bsgs(a,t,p)<<'\n';
    }
}