HDu4794 斐波那契循环节

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
",fibmod(n)>>1ll);
    }
    return 0;
}
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