题意:Arnold变换把矩阵(x,y)变成((x+y)%n,(x+2*y)%n),问最小循环节

题解:仔细算前几项能看出是斐波那契数论modn,然后套个斐波那契循环节板子即可

//#pragma GCC optimize(2)

//#pragma GCC optimize(3)

//#pragma GCC optimize(4)

//#pragma GCC optimize("unroll-loops")

//#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")

#include<bits/stdc++.h>

#define fi first

#define se second

#define db double

#define mp make_pair

#define pb push_back

#define pi acos(-1.0)

#define ll unsigned long long

#define vi vector<int>

#define mod 1000000007

#define ld long double

#define C 0.5772156649

//#define ls l,m,rt<<1

//#define rs m+1,r,rt<<1|1

#define pll pair<ll,ll>

#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 Max(a,b) ((a)>(b)?(a):(b))

#define Min(a,b) ((a)<(b)?(a):(b))

#define fin freopen("a.txt","r",stdin)

#define fout freopen("a.txt","w",stdout)

#define fio ios::sync_with_stdio(false);cin.tie(0)

template<typename T>

inline T const& MAX(T const &a,T const &b){return a>b?a:b;}

template<typename T>

inline T const& MIN(T const &a,T const &b){return a<b?a:b;}

inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}

inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}

inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}

inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;

const ll INF=0x3f3f3f3f3f3f3f3f;

const int N=100000+10,maxn=300000+10,inf=0x3f3f3f3f;

struct Node{

    ll row,col;

    ll a[3][3];

};

Node mul(Node x,Node y,ll c)

{

    Node ans;

    ans.row=x.row,ans.col=y.col;

    memset(ans.a,0,sizeof ans.a);

    for(int i=0;i<x.row;i++)

        for(int j=0;j<x.col;j++)

           for(int k=0;k<y.col;k++)

               ans.a[i][k]=(ans.a[i][k]+x.a[i][j]*y.a[j][k]+c)%c;

    return ans;

}

Node quick_mul(Node x,ll n,ll c)

{

    Node ans;

    ans.row=x.row,ans.col=x.col;

    memset(ans.a,0,sizeof ans.a);

    for(int i=0;i<ans.col;i++)ans.a[i][i]=1;

    while(n){

        if(n&1)ans=mul(ans,x,c);

        x=mul(x,x,c);

        n>>=1;

    }

    return ans;

}

bool ok(ll x,ll n)

{

    Node A;A.row=A.col=2;

    A.a[0][0]=A.a[0][1]=A.a[1][0]=1;A.a[1][1]=0;

    A=quick_mul(A,x-1,n);

    return ((A.a[0][0]+A.a[0][1])%n==1)&&((A.a[1][0]+A.a[1][1])%n==0);

}

bool erci(ll x,ll p){return qp(x,(p-1)>>1,p)==1;}

ll n,fac[N],cnt;

vector<pli>v;

ll fibmod(ll n)

{

    v.clear();

    for(ll i=2;i*i<=n;i++)

    {

        if(n%i==0)

        {

            int co=0;

            while(n%i==0)n/=i,co++;

            v.pb(mp(i,co));

        }

    }

    if(n!=1)v.pb(mp(n,1));

    ll ans=1;

    for(int i=0;i<v.size();i++)

    {

        ll p=v[i].fi,m=v[i].se;

        ll te,gp,x;

        if(p==2)gp=3;

        else if(p==3)gp=8;

        else if(p==5)gp=20;

        else

        {

            if(erci(5,p))te=p-1;

            else te=2*(p+1);

            cnt=0;

            for(ll j=1;j*j<=te;j++)

            {

                if(te%j==0)

                {

                    fac[++cnt]=j;

                    if(j*j!=te)fac[++cnt]=te/j;

                }

            }

            sort(fac+1,fac+1+cnt);

            for(int j=1;j<=cnt;j++)

            {

                if(ok(fac[j],p))

                {

                    gp=fac[j];

                    break;

                }

            }

        }

        x=gp*(ll)pow(p,m-1);

        ans=ans/gcd(ans,x)*x;

    }

    return ans;

}

int main()

{

    while(~scanf("%llu",&n))

    {

        if(n==2)puts("3");

        else printf("%llu\n",fibmod(n)>>1ll);

    }

    return 0;

}

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