[动态规划+矩阵快速幂优化]Pku3420--Quad Tiling

Quad Tiling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1801		Accepted: 698

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

感慨：这道题我郁闷了很久..因为网上没有很好的解题报告...而且标程写得..我看不懂= =所以一直没有理解..现在静静地想一想，其实一切都顺理成章嘛..一次AC，庆祝下，虽然很爽，但郁闷的是，刚blog出了bug，写好的解题报告没了，郁闷= =以后一定记得发之前保存个..

题目大意：给出一个4*n的矩阵，问用1*2的骨牌将它完全覆盖有多少种方案。输出方案数mod m。

分析：传统方法因为n最大有10^9之多，所以会超时。若定义f[i,k]表示前i-1行已经完全覆盖，第i行状态为k的方案数的话，注意到f[i]只和f[i-1]有关，而且关系是线性的，且f的i域范围很大，但k域范围很小（0~15），再者每相邻两行的状态转移方程完全相同，所以可以用矩阵来表示这种转移方法。

构造16*16的矩阵G，使得f[i]*G=f[i+1]。这里，将f[i]看成是一个1*16的矩阵。

只要对矩阵乘法理解透彻，就不难用dfs来构造出矩阵G。然后，只要将第一行可能的状态org乘上Gn次，就能得到f[n+1]，f[n+1,0]即答案。

由于矩阵乘法满足结合律，所以可以用快速幂求出G^N，再乘以org。至此，问题解决。

codes:

type arr=array[0..16,0..16] of longint; const org:array[0..15] of longint=(1,0,0,1,0,0,1,0,0,0,0,0,1,0,0,1);//第一行可能放置出来的状态 var g,ans,temp:array[0..16,0..16] of longint; n,m:longint; procedure dfs(x,old,new:longint); begin if x=4 then inc(g[new,old]); if x>3 then exit; if old and (1 shl x)=0 then dfs(x+1,old,new shl 1 or 1); if old and (1 shl x)>0 then begin if old and (1 shl (x+1))>0 then dfs(x+2,old,new shl 2 or 3); dfs(x+1,old,new shl 1); end; end; procedure mul(a,b:arr;var c:arr); var i,j,k:longint; t:arr; begin fillchar(t,sizeof(t),0); for i:=0 to 15 do for j:=0 to 15 do begin for k:=0 to 15 do t[i,j]:=t[i,j]+a[i,k]*b[k,j]; t[i,j]:=t[i,j] mod m; end; c:=t; end; procedure cal(x:longint); begin if x=1 then begin ans:=g; exit; end; cal(x div 2); mul(ans,ans,temp); ans:=temp; if x and 1>0 then begin mul(ans,g,temp); ans:=temp; end; end; procedure print; var i:longint; an:int64; begin an:=0; for i:=0 to 15 do an:=(an+ans[0,i]*org[i]) mod m; writeln(an); end; procedure main; var i:longint; begin for i:=0 to 15 do dfs(0,i,0); readln(n,m); while n+m<>0 do begin fillchar(ans,sizeof(ans),0); cal(n); print; readln(n,m); end; end; begin main; end.