luoguP3306 [SDOI2013]随机数生成器

题意

将\(x_1,x_2,x_3...x_n\)写出来可以发现通项为\(a^{i-1}*x_1+b*\sum\limits_{j=0}^{i-2}a^j=a^{i-1}*x_1+b*\frac{1-a^{i-1}}{1-a}=(x_1-\frac{b}{1-a})a^{i-1}+\frac{b}{1-a}\)

所求变为求一个\(i\)满足：
\(t\equiv (x_1-\frac{b}{1-a})a^{i-1}+\frac{b}{1-a}\pmod{p}\)
\(a^{i-1}\equiv \frac{t-(\frac{b}{1-a})}{x_1-\frac{b}{1-a}}\pmod{p}\)

上BSGS即可，注意特判\(a=0/1\)。

\(a=0\)：对于任意\(i,x_i=b\)
\(a=1\)：\(x_i=x_1+(i-1)*b\)

解下对应的方程即可。

code:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int T;
ll a,b,x1,mod,goal;
inline ll power(ll x,ll k,ll mod)
{
    ll res=1;
    while(k)
    {
        if(k&1)res=res*x%mod;
        x=x*x%mod;k>>=1;
    }
    return res;
}
inline ll inv(ll x,ll mod){return power(x,mod-2,mod);}
inline ll BSGS(ll a,ll b,ll mod)
{
    if(b==1)return 0;
    unordered_map<ll,int>mp;mp.clear();
    a%=mod,b%=mod;
    ll t=ceil(sqrt(mod));
    ll now=1;
    for(int i=0;i<=t;i++)mp[b*now%mod]=i,now=now*a%mod;
    a=power(a,t,mod);
    if(!a)return !b?1:-1;
    now=a;
    for(int i=1;i<=t;i++)
    {
        if(mp.count(now))return i*t-mp[now];
        now=now*a%mod;
    }
    return -1;
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld%lld%lld%lld%lld",&mod,&a,&b,&x1,&goal);
        if(x1==goal){puts("1");continue;}
        if(!a){puts(b==goal?"2":"-1");continue;}
        if(a==1)
        {
            goal=((goal-x1)%mod+mod)%mod;
            ll d=__gcd(b,mod);
            if(goal%d){puts("-1");continue;}
            ll res=(goal*inv(b,mod)+1)%mod;
            if(!res)res=mod;
            printf("%lld\n",res);
            continue;
        }
        ll res=BSGS(a,((goal-b*inv(1-a,mod)%mod)%mod+mod)%mod*inv(((x1-b*inv(1-a,mod)%mod)%mod+mod)%mod,mod)%mod,mod);
        printf("%lld\n",res==-1?res:res+1);
    }
    return 0;
}

标签：luoguP3306,frac,res,ll,生成器,SDOI2013,inline,return,mod
来源： https://www.cnblogs.com/nofind/p/11933148.html