HDU——4291A Short problem（矩阵快速幂+循环节）

A Short problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2461 Accepted Submission(s): 864


[align=left]Problem Description[/align]
　　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

[align=left]Input[/align]
　　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).

[align=left]Output[/align]
　　For each test case, please print a single line with a integer, the corresponding answer to this case.

[align=left]Sample Input[/align]

0
1
2

[align=left]Sample Output[/align]

0
1
42837

三层复合函数。写完交上去TLE，不解。查了题解发现有循环节。循环节的话用set::find函数来查找是否存在过，若存在则输出循环节，不存在则插入当前数值继续循环。
然后就是矩阵快速幂了。递推式：

（当某一层n为0的时候加上mod否则可能出错）
代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long LL;
#define INF 0x3f3f3f3f
const long long mod=1000000007;
struct mat
{
LL pos[2][2];
mat(){memset(pos,0,sizeof(pos));}
};
inline mat mul(const mat &a,const mat &b,const LL &mod)
{
mat c;
for (int i=0; i<2; i++)
{
for (int j=0; j<2; j++)
{
for (int k=0; k<2; k++)
{
c.pos[i][j]+=((a.pos[i][k])%mod*(b.pos[k][j])%mod)%(mod);
}
}
}
return c;
}
inline mat matpow(mat a,LL b,const LL &mod)
{
mat r;
r.pos[1][1]=r.pos[0][0]=1;
mat bas=a;
while (b!=0)
{
if(b&1)
r=mul(r,bas,mod);
bas=mul(bas,bas,mod);
b>>=1;
}
return r;
}
int main(void)
{
mat one,t;
LL n,ans;
one.pos[0][0]=1;
one.pos[0][1]=0;
t.pos[0][0]=3;
t.pos[0][1]=t.pos[1][0]=1;
mat poww,initt;
while (cin>>n)
{
if(n==0)
{
cout<<"0"<<endl;
continue;
}
poww=t,initt=one;
poww=matpow(poww,n-1,183120);
initt=mul(initt,poww,183120);
ans=initt.pos[0][0];
ans+=183120;

poww=t,initt=one;
poww=matpow(poww,ans-1,222222224);
initt=mul(initt,poww,222222224);
ans=initt.pos[0][0];
ans+=222222224;

poww=t,initt=one;
poww=matpow(poww,ans-1,1000000007);
initt=mul(initt,poww,1000000007);
cout<<initt.pos[0][0]%1000000007<<endl;
}
return 0;
}