题解思路：
构造矩阵，矩阵乘法计算还是t;
需要找循环节; (注意因为是复合函数，不可以在里面取mod)
暴力跑只有可以找到g(222222224)%1e9==g(0)%1e9;
所以 g(g(n)%222222224)%1e9==g(g(n));
之后还可以跑出2个循环节
从内到外
240 183120 222222224 1e9+7
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<stack>#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
#define int long longconst int maxn=65000+10;using namespace std;int prime[maxn],is_prime[maxn];struct Matrix
{int nmap[3][3];
};Matrix initA={0,0,0,0,1,0,0,0,1,
};Matrix T=
{0,0,0,0,0,0,0,0,0,
};void is_p(){for(int i=2;i<=maxn;i++){if(!is_prime[i])for(int j=i+i;j<maxn;j+=i){is_prime[j]=1;}}
}ll pow(ll a,ll b,ll mod)
{ll sum=1;while(b){if(b&1) sum=(sum*a)%mod;a=(a*a)%mod;b>>=1;}return sum%mod;
}void put(Matrix x)
{for(int i=1;i<=2;i++){for(int j=1;j<=2;j++){cout<<x.nmap[i][j]<<" ";}cout<<endl;}cout<<endl;
}Matrix Matrix_mul(Matrix a,Matrix b,int mod)
{Matrix A=T;for(int k=1;k<=2;k++){for(int i=1;i<=2;i++){if(a.nmap[i][k])for(int j=1;j<=2;j++){A.nmap[i][j]+=(a.nmap[i][k]*b.nmap[k][j])%mod;A.nmap[i][j]%=mod;}}}return A;
}Matrix Matrix_smul(Matrix a,int b,int mod)
{Matrix A=initA;if(b==-1) return a;while(b){if(b&1) A=Matrix_mul(A,a,mod);a=Matrix_mul(a,a,mod);b>>=1;}return A;
}#undef int
int main(){
#define int long longMatrix C={0,0,0,0,3,1,0,1,0,};Matrix D={0,0,0,0,0,0,0,1,0,};int n=1,mod=1e9+7;while(~scanf("%lld",&n)){n%=240;mod=183120;Matrix B=Matrix_smul(C,n,mod);B=Matrix_mul(B,D,mod);// put(B)mod=222222224;B=Matrix_smul(C,B.nmap[1][1],mod);B=Matrix_mul(B,D,mod);// put(B);mod=1e9+7;B=Matrix_smul(C,B.nmap[1][1],mod);B=Matrix_mul(B,D,mod);// put(B);printf("%lld\n",B.nmap[1][1]);}// }
/* int a=1,b=0,c,mod=240,ok=0;//跑循环节的代码...for(int i=1;;i++){c=(3*a+b)%mod;b=a%mod;a=c%mod;if(a==1&&b==0){cout<<i<<endl;ok++;if(ok>10) break;}}
*/return 0;
}
)