hdu 4291 矩阵幂 循环节

http://acm.hdu.edu.cn/showproblem.php?pid=4291

凡是取模的都有循环节-----常数有，矩阵也有，并且矩阵的更奇妙：
g(g(g(n))) mod 10⁹ + 7  最外层MOD=1e9+7  能够算出g(g(n))的循环节222222224。进而算出g(n)的循环节183120LL。然后由内而外计算就可以

凝视掉的是求循环节的代码

//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;

#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double EPS = 1e-8;
const double pi = acos(-1.0);
const int INF = 100000000;
const ll MOD[3] = {183120LL,222222224LL,1000000007LL};
const int N = 2;

struct Matrix{
    ll m[N][N];
    //int sz;//矩阵的大小
};

Matrix I= {3LL,1LL,//要幂乘的矩阵
           1LL,0LL,
          };
Matrix unin={1LL,0LL,//单位矩阵
             0LL,1LL,
            };
Matrix matrixmul(Matrix a,Matrix b,long long  mod)//矩阵a乘矩阵b
{
    Matrix c;
    for(int i=0; i<N; i++)
        for(int j=0; j<N; j++)
        {
            c.m[i][j]=0LL;
            for(int k=0; k<N; k++)
                c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod;
            c.m[i][j]%=mod;
        }
    return c;
}
Matrix quickpow(long long n,long long  mod)
{
    Matrix m=I,b=unin;//求矩阵I的n阶矩阵
    while(n>=1)
    {
        if(n&1)
            b=matrixmul(b,m,mod);
        n=n>>1;
        m=matrixmul(m,m,mod);
    }
    return b;
}

ll solve(ll n)
{
    ll ans;
    Matrix ret;
    ret.m[0][0]=n;
    for(int i=0;i<3;i++)
    {
        if(ret.m[0][0]!=0 && ret.m[0][0]!=1)ret=quickpow(ret.m[0][0]-1,MOD[i]);
    }

    return ret.m[0][0];
}

int main()
{
    //precal();
    ll n;
    while(~scanf("%I64d",&n))
    {
        if(n==0){puts("0");continue;}
        if(n==1){puts("1");continue;}
        //printf("%I64d
",solve(n));
        cout << solve(n)%1000000007LL << endl;
    }
    return 0;
}

hdu 4291 矩阵幂  循环节