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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：
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;
}
