HDU 4291 A Short problem （2012成都网络赛，矩阵快速幂+循环节）

链接： click here~~

题意：

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 10 ⁹ + 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
   
   
   
   

【解题思路】具体参考前一篇博客，基本上一样的思路： 点击这里

 代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
using namespace std;
#define LL long long
const long long mod1=1e9+7;
const long long mod2=222222224;
const long long  mod3=183120;
struct Matrix
{
    long long mapp[2][2];
};
Matrix p= {0,1,1,0};
Matrix p1= {0,1,1,3};
Matrix unin= {1,0,0,1};
Matrix powmul(Matrix a,Matrix b,long long mod)
{
    Matrix c;
    for(int i=0; i<2; i++)
        for(int j=0; j<2; j++)
        {
            c.mapp[i][j]=0;
            for(int k=0; k<2; k++)
                c.mapp[i][j]+=(a.mapp[i][k]*b.mapp[k][j])%mod;
            c.mapp[i][j]%=mod;
        }
    return c;
}
Matrix powexp(long long n,long long mod)
{
    Matrix m=p1,b=unin;
    while(n>=1)
    {
        if(n&1) b=powmul(b,m,mod);
        n>>=1;
        m=powmul(m,m,mod);
    }
    return powmul(p,b,mod);
}
long long T;
int main()
{
    while(~scanf("%lld",&T))
    {
        Matrix ans;
        ans=powexp(T,mod3);
        ans=powexp(ans.mapp[0][0],mod2);
        ans=powexp(ans.mapp[0][0],mod1);
        cout<<ans.mapp[0][0]<<endl;
        // printf("%lld\n",ans.mapp[0][0]);
    }
    return 0;
}