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 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

012

 

Sample Output

0142837

【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 longconst 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;}