A Short problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2461    Accepted Submission(s): 864

Problem Description

　　According to a research, VIM users tend to have shorter fingers, compared with Emacs users.
　　Hence they prefer problems short, too. Here is a short one:
　　Given n (1 <= n <= 10¹⁸), You should solve for 
g(g(g(n))) mod 10⁹ + 7
　　where
g(n) = 3g(n - 1) + g(n - 2)
g(1) = 1
g(0) = 0

Input

　　There are several test cases. For each test case there is an integer n in a single line.
　　Please process until EOF (End Of File).

Output

　　For each test case, please print a single line with a integer, the corresponding answer to this case.

Sample Input

0
1
2

Sample Output

0
1
42837

三层复合函数。写完交上去TLE，不解。查了题解发现有循环节。循环节的话用set::find函数来查找是否存在过，若存在则输出循环节，不存在则插入当前数值继续循环。

然后就是矩阵快速幂了。递推式：（当某一层n为0的时候加上mod否则可能出错）

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long LL;
#define INF 0x3f3f3f3f
const long long mod=1000000007;
struct mat
{
    LL pos[2][2];
    mat(){memset(pos,0,sizeof(pos));}
};
inline mat mul(const mat &a,const mat &b,const LL &mod)
{
    mat c;
    for (int i=0; i<2; i++)
    {
        for (int j=0; j<2; j++)
        {
            for (int k=0; k<2; k++)
            {
                c.pos[i][j]+=((a.pos[i][k])%mod*(b.pos[k][j])%mod)%(mod);
            }
        }
    }
    return c;
}
inline mat matpow(mat a,LL b,const LL &mod)
{
    mat r;
    r.pos[1][1]=r.pos[0][0]=1;
    mat bas=a;
    while (b!=0)
    {
        if(b&1)
            r=mul(r,bas,mod);
        bas=mul(bas,bas,mod);
        b>>=1;
    }
    return r;
}
int main(void)
{
    mat one,t;
    LL n,ans;
    one.pos[0][0]=1;
    one.pos[0][1]=0;
    t.pos[0][0]=3;
    t.pos[0][1]=t.pos[1][0]=1;
    mat poww,initt;
    while (cin>>n)
    {
        if(n==0)
        {
            cout<<"0"<<endl;
            continue;
        }
        poww=t,initt=one;
        poww=matpow(poww,n-1,183120);
        initt=mul(initt,poww,183120);
        ans=initt.pos[0][0];
        ans+=183120;

        poww=t,initt=one;
        poww=matpow(poww,ans-1,222222224);
        initt=mul(initt,poww,222222224);
        ans=initt.pos[0][0];
        ans+=222222224;

        poww=t,initt=one;
        poww=matpow(poww,ans-1,1000000007);
        initt=mul(initt,poww,1000000007);
        cout<<initt.pos[0][0]%1000000007<<endl;
    }
    return 0;
}