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

• 遞推公式及矩陣，快速冪加速

• f[n]=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;
}