POJ3420 Quad Tiling (矩阵加速状压dp）

传送门：http://poj.org/problem?id=3420

Quad Tiling

Time Limit: 1000MS		Memory Limit: 65536K

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

 

这题是POJ2411的放大版。。记得JSOI2013第一轮考过一题5*N的。。当时cxt还和我炫耀。。现在看看还是很简单的

N=1e9 M=1e5的时候。。需要longlong，不过我侥幸没加longlong过了。。

做法就是开个16*16的矩阵。。不过根据矩阵似乎就可以直接推出来递推式了。。好神orz...

Codes:

#include<set>
#include<queue>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,l,r) for(int i=l;i<=r;i++)

struct Matrix{
    int A[17][17];
    Matrix(){
        memset(A,0,sizeof(A));
    }
}Unit,Ans,TAns;

int opt[20],n,Mod;

Matrix operator * (Matrix A,Matrix B){
   Matrix C;
    Rep(i,0,15)
      Rep(j,0,15)
        Rep(k,0,15)
           C.A[i][j] = ( C.A[i][j] + (A.A[i][k] * B.A[k][j]) % Mod ) % Mod;
    return C;
}

bool Check(int s1,int s2){
    if((s1|s2)!=15) return false;
    for(int i=0;i<=15;){
        if( (s1&(1<<i)) != (s2&(1<<i)) ) i++;
        else{
            if(i==15) return false; else
            if(  ( s1 & (1<<(i+1)) ) != ( s2 & (1<<(i+1)) ) ) return false;
            i+=2;
        }
    }
    return true;
}

int main(){
    Rep(i,0,15){
        Rep(j,0,15)
            if(Check(i,j)) Ans.A[i][j] = TAns.A[i][j] = 1;
        if(Check(i,15)) 
            opt[i] = 1;
    }
    while(scanf("%d%d",&n,&Mod),Mod+n){        
        Rep(i,0,15){
            Rep(j,0,15){
                Ans.A[i][j] = TAns.A[i][j];
                Unit.A[i][j] = 0;
            }
            Unit.A[i][i] = 1;
        } 
        while(n){
            if(n&1) Unit = Unit * Ans;
            Ans = Ans * Ans;
            n = n >> 1;
        }
        int ans = 0;
        Rep(i,0,15) 
            ans = (ans % Mod + (opt[i] * Unit.A[0][i] % Mod) % Mod) % Mod;
        printf("%d
",ans);
    }
    return 0;
}