斐波那契数列寻找mod n 循环节 模板

该代码来自ACdreamer

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <math.h>

using namespace std;
typedef unsigned long long LL;

const int M = 2;

struct Matrix
{
    LL m[M][M];
};

Matrix A;
Matrix I = {1,0,0,1};

Matrix multi(Matrix a,Matrix b,LL MOD)
{
    Matrix c;
    for(int i=0; i<M; i++)
    {
        for(int j=0; j<M; j++)
        {
            c.m[i][j] = 0;
            for(int k=0; k<M; k++)
                c.m[i][j] = (c.m[i][j]%MOD + (a.m[i][k]%MOD)*(b.m[k][j]%MOD)%MOD)%MOD;
            c.m[i][j] %= MOD;
        }
    }
    return c;
}

Matrix power(Matrix a,LL k,LL MOD)
{
    Matrix ans = I,p = a;
    while(k)
    {
        if(k & 1)
        {
            ans = multi(ans,p,MOD);
            k--;
        }
        k >>= 1;
        p = multi(p,p,MOD);
    }
    return ans;
}

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

const int N = 400005;
const int NN = 5005;

LL num[NN],pri[NN];
LL fac[NN];
int cnt,c;

bool prime[N];
int p[N];
int k;

void isprime()
{
    k = 0;
    memset(prime,true,sizeof(prime));
    for(int i=2; i<N; i++)
    {
        if(prime[i])
        {
            p[k++] = i;
            for(int j=i+i; j<N; j+=i)
                prime[j] = false;
        }
    }
}

LL quick_mod(LL a,LL b,LL m)
{
    LL ans = 1;
    a %= m;
    while(b)
    {
        if(b & 1)
        {
            ans = ans * a % m;
            b--;
        }
        b >>= 1;
        a = a * a % m;
    }
    return ans;
}

LL legendre(LL a,LL p)
{
    if(quick_mod(a,(p-1)>>1,p)==1) return 1;
    else                           return -1;
}

void Solve(LL n,LL pri[],LL num[])
{
    cnt = 0;
    LL t = (LL)sqrt(1.0*n);
    for(int i=0; p[i]<=t; i++)
    {
        if(n%p[i]==0)
        {
            int a = 0;
            pri[cnt] = p[i];
            while(n%p[i]==0)
            {
                a++;
                n /= p[i];
            }
            num[cnt] = a;
            cnt++;
        }
    }
    if(n > 1)
    {
        pri[cnt] = n;
        num[cnt] = 1;
        cnt++;
    }
}

void Work(LL n)
{
    c = 0;
    LL t = (LL)sqrt(1.0*n);
    for(int i=1; i<=t; i++)
    {
        if(n % i == 0)
        {
            if(i * i == n) fac[c++] = i;
            else
            {
                fac[c++] = i;
                fac[c++] = n / i;
            }
        }
    }
}

LL find_loop(LL n)
{
    Solve(n,pri,num);
    LL ans=1;
    for(int i=0; i<cnt; i++)
    {
        LL record=1;
        if(pri[i]==2)
            record=3;
        else if(pri[i]==3)
            record=8;
        else if(pri[i]==5)
            record=20;
        else
        {
            if(legendre(5,pri[i])==1)
                Work(pri[i]-1);
            else
                Work(2*(pri[i]+1));
            sort(fac,fac+c);
            for(int k=0; k<c; k++)
            {
                Matrix a = power(A,fac[k]-1,pri[i]);
                LL x = (a.m[0][0]%pri[i]+a.m[0][1]%pri[i])%pri[i];
                LL y = (a.m[1][0]%pri[i]+a.m[1][1]%pri[i])%pri[i];
                if(x==1 && y==0)
                {
                    record = fac[k];
                    break;
                }
            }
        }
        for(int k=1; k<num[i]; k++)
            record *= pri[i];
        ans = ans/gcd(ans,record)*record;
    }
    return ans;
}

void Init()
{
    A.m[0][0] = 1;
    A.m[0][1] = 1;
    A.m[1][0] = 1;
    A.m[1][1] = 0;
}

int main()
{
    LL n;
    Init();
    isprime();
    while(cin>>n)
        cout<<find_loop(n)<<endl;
    return 0;
}