POJ_3420_Quad Tiling

Quad Tiling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4410		Accepted: 2013

Description

Tired of the Tri Tiling game finally, Michael turns to a more challengeable game, Quad Tiling:

In how many ways can you tile a 4 × N (1 ≤ N ≤ 109) rectangle with 2 × 1 dominoes? For the answer would be very big, output the answer modulo M (0 < M ≤ 105).

Input

Input consists of several test cases followed by a line containing double 0. Each test case consists of two integers, N and M, respectively.

Output

For each test case, output the answer modules M.

Sample Input

1 10000
3 10000
5 10000
0 0

Sample Output

1
11
95

Source
POJ Monthly--2007.10.06, Dagger

递推公式及矩阵，快速幂加速

f
=f[n-1]+5*f[n-2]+f[n-3]-f[n-4];
LL a[4][4]={
{11, 5, 1, 1},
{ 0, 0, 0, 0},
{ 0, 0, 0, 0},
{ 0, 0, 0, 0}
},
c[4][4]={
{ 1, 1, 0, 0},
{ 5, 0, 1, 0},
{ 1, 0, 0, 1},
{-1, 0, 0, 0}
};

#include <iostream>
#include <string>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
typedef long long           LL ;
typedef unsigned long long ULL ;
const int    maxn = 1000 + 10  ;
const int    inf  = 0x3f3f3f3f ;
const int    npos = -1         ;
const int    mod  = 1e9 + 7    ;
const int    mxx  = 100 + 5    ;
const double eps  = 1e-6       ;

LL m;
struct Matrix{
LL a[4][4];
Matrix operator * (const Matrix B){
Matrix C;
memset(&C,0,sizeof(C));
for(int k=0;k<4;k++)
for(int i=0;i<4;i++)
if(a[i][k])
for(int j=0;j<4;j++)
if(B.a[k][j])
C.a[i][j]=((C.a[i][j] + (a[i][k]*B.a[k][j])%m)%m + m*2)%m;
return C;
}
};
LL a[4][4]={
{11, 5, 1, 1},
{ 0, 0, 0, 0},
{ 0, 0, 0, 0},
{ 0, 0, 0, 0}
},
c[4][4]={
{ 1, 1, 0, 0},
{ 5, 0, 1, 0},
{ 1, 0, 0, 1},
{-1, 0, 0, 0}
};
LL cal(int k){
if(k>0){
Matrix e, f;
for(int i=0;i<4;i++)
for(int j=0;j<4;j++){
e.a[i][j]=a[i][j];
f.a[i][j]=c[i][j];
}
while(k){
if(1&k){
e=e*f;
}
f=f*f;
k>>=1;
}
return e.a[0][0]%m;
}else{
return a[0][-k]%m;
}
}
int n;
int main(){
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
while((~scanf("%d %lld",&n,&m)) && n>0){
printf("%lld\n",cal(n-3));
}
return 0;
}