hdu 4291 A Short problem（矩阵+取模循环节）

A Short problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1785 Accepted Submission(s): 651



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

矩阵非常easy构造

矩阵A
1 0
0 0
递推矩阵B
3 1
1 0
g(n)=A*B^(n-1)的第1行第1列。

如今是一个多重函数在最外层取模，n为10^18,当到最外层取模肯定不能够，这时候就要寻找循环节了。首先是最
外层的模。通过暴力判循环节

long long x3, x1=0,x2=1;
long long temp=1000000007;
for(long long i=2;;i++)
{
x3=(3*x2+x1)%temp;
if(x3==1&&x2==0)
{
printf("%I64d\n",i-1);
break;
}
x1=x2;
x2=x3;
}

求出第2层队222222224 取模。同理第3层对183120取模。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
struct matrix
{
long long ma[3][3];
};
long long mod[3];
matrix multi(matrix x,matrix y,long long m)
{
matrix ans;
memset(ans.ma,0,sizeof(ans.ma));
for(int i=1; i<=2; i++)
{
for(int j=1; j<=2; j++)
{
if(x.ma[i][j])
{
for(int k=1; k<=2; k++)
{
ans.ma[i][k]=(ans.ma[i][k]+(x.ma[i][j]*y.ma[j][k])%m)%m;
}
}
}
}
return ans;
}
int main()
{
long long n;
while(~scanf("%I64d",&n))
{
mod[0]=183120;
mod[1]=222222224;
mod[2]=1000000007;
for(int l=0; l<3; l++)
{
if(n==0)
continue;
n=n-1;
matrix a;
a.ma[1][1]=1;
a.ma[1][2]=a.ma[2][1]=a.ma[2][2]=0;
matrix b;
b.ma[1][1]=3;
b.ma[1][2]=1;
b.ma[2][1]=1;
b.ma[2][2]=0;

matrix ans;
for(int i=1; i<=2; i++)
{
for(int j=1; j<=2; j++)
{
if(i==j)
ans.ma[i][j]=1;
else
ans.ma[i][j]=0;
}
}
while(n)
{
if(n&1)
ans=multi(ans,b,mod[l]);
b=multi(b,b,mod[l]);
n=n>>1;
}
n=ans.ma[1][1];
}
printf("%I64d\n",n);
}
return 0;
}