hdu 4291(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): 2487    Accepted Submission(s): 876

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

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define Mod1 1000000007
#define Mod2 222222224
#define Mod3 183120
ll Mod;
/*void find()
{
ll a,b,c,i,k,j;
a=0,b=1;
for(i=2;;i++)
{
c=(3*b+a)%1000000007;
if(c==1&&b==0)
break;
a=b;
b=c;
}
printf("%lld\n",i-1);
a=0,b=1;
for(k=2;;k++)
{
c=(3*b+a)%(i-1);
if(c==1&&b==0)
break;
a=b;
b=c;
}
printf("%lld\n",k-1);

a=0,b=1;
for(j=2;;j++)
{
c=(3*b+a)%(k-1);
if(c==1&&b==0)
break;
a=b;
b=c;
}
printf("%lld\n",j-1);
}
*/
struct Matrix
{
ll ma[2][2];
Matrix()
{
memset(ma,0,sizeof(ma));
}

void init(int a,int b,int c,int d)
{
ma[0][0]=a;
ma[0][1]=b;
ma[1][0]=c;
ma[1][1]=d;
}
Matrix operator * (const Matrix &m)
{
Matrix res;
for(int i=0; i<2; i++)
for(int j=0; j<2; j++)
for(int k=0; k<2; k++)
res.ma[i][j]=(res.ma[i][j]+m.ma[i][k]*ma[k][j])%Mod;
return res;
}
};

Matrix p;

Matrix quick_powm(Matrix m,ll n)
{
if(n==0)
{
p.init(1,0,0,1);
return p;
}
if(n==1)return m;
Matrix res=quick_powm(m,n/2);
Matrix ans=res*res;
if(n%2==1)
ans=ans*m;
return ans;
}

ll val;

int main()
{

while(~scanf("%lld",&val))
{
Matrix m,res;
m.init(3,1,1,0);
Mod=Mod3;
res=quick_powm(m,val);
Mod=Mod2;
res=quick_powm(m,res.ma[1][0]);
Mod=Mod1;
res=quick_powm(m,res.ma[1][0]);
printf("%lld\n",res.ma[1][0]);
}
return 0;
}