A Short problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2711    Accepted Submission(s): 951

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  

 

Source 2012 ACM/ICPC Asia Regional Chengdu Online  

 

/*
* Author: lyucheng
* Created Time:  2017年05月20日 星期六 16时40分42秒
* File Name: HDU-4291-A_Short_problem.cpp
*/
/*
* 题意:让你求g(g(g(n)))mod 1e9+7,其中g(n)=3*g(n-1)+g(n-2)
*
*
* 思路:g(n)可以通过矩阵快速幂求出来,但是干后分别求出各自的循环节,能得到第一个循环节是222222224,
*      第二个循环节是183120
* */
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <time.h>
#define LL long long
#define maxn 3
using namespace std;
LL n;
LL mod;
/********************************矩阵快速幂**********************************/
class Matrix {
public:
LL a[maxn][maxn];

void init() {
memset(a,0,sizeof(a));
}

Matrix operator +(Matrix b) {
Matrix c;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
c.a[i][j] = (a[i][j] + b.a[i][j]) % mod;
return c;
}

Matrix operator +(LL x) {
Matrix c = *this;
for (int i = 0; i < 2; i++)
c.a[i][i] += x;
return c;
}

Matrix operator *(Matrix b)
{
Matrix p;
p.init();
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++)
p.a[i][j] = (p.a[i][j] + (a[i][k]*b.a[k][j])%mod) % mod;
return p;
}

Matrix power(LL t) {
Matrix ans,p = *this;
ans.init();
ans.a[0][0]=1;
ans.a[1][1]=1;
while (t) {
if (t & 1)
ans=ans*p;
p = p*p;
t >>= 1;
}
return ans;
}
}unit,init;
/********************************矩阵快速幂**********************************/
int main(int argc, char* argv[])
{
// freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout;
while(scanf("%lld",&n)!=EOF){
if(n<2){
printf("%lld\n",n);
continue;
}
//首先求最里面的g(n)
mod=183120;
unit.init();
unit.a[0][0]=1,unit.a[0][1]=0,unit.a[1][0]=0,unit.a[1][1]=0;
init.a[0][0]=3,init.a[0][1]=1,init.a[1][0]=1,init.a[1][1]=0;
init=init.power(n-1);
unit=unit*init;
n=unit.a[0][0];

if(n>=2){//很重要,要不然当n<2的时候矩阵乘法就会死循环
mod=222222224;
unit.init();
unit.a[0][0]=1,unit.a[0][1]=0,unit.a[1][0]=0,unit.a[1][1]=0;
init.a[0][0]=3,init.a[0][1]=1,init.a[1][0]=1,init.a[1][1]=0;
init=init.power(n-1);
unit=unit*init;
n=unit.a[0][0];
}

if(n>=2){//很重要,要不然当n<2的时候矩阵乘法就会死循环
mod=1000000007;
unit.init();
unit.a[0][0]=1,unit.a[0][1]=0,unit.a[1][0]=0,unit.a[1][1]=0;
init.a[0][0]=3,init.a[0][1]=1,init.a[1][0]=1,init.a[1][1]=0;
init=init.power(n-1);
unit=unit*init;
n=unit.a[0][0];
}
printf("%lld\n",n);
}
return 0;
}