Quad Tiling
| Time Limit: 1000MS | Memory Limit: 65536K |
| Total Submissions: 4107 | Accepted: 1878 |
Description
Tired of(厌烦) the Tri Tiling gamefinally, Michael turns to(转向) a morechallengeable game, Quad Tiling:
In how many ways can you tile(铺瓷砖) a 4 × N (1 ≤ N ≤109) rectangle(矩形) with 2 × 1dominoes(多米诺骨牌)? For the answerwould be very big, output the answer modulo M (0 < M ≤105).
Input
Input consists of several test casesfollowed by a line containing double 0. Each test case consists of twointegers, N and M, respectively(分别的).
Output
For each test case, output the answermodules M.
Sample Input
1 10000
3 10000
5 10000
0 0
Sample Output
1
11
95
Source
POJMonthly--2007.10.06, Dagger
题意：在一个4*N的矩阵中用2*1的多米诺骨牌铺满，问有多少种铺法?
模板：
递推式：a[i]=a[i-1]+5*a[i-2]+a[i-3]-a[i-4];
由于N高达10^9，所以要用矩阵进行优化。
|0 1 0 0|
|0 0 1 0|
|0 0 0 1|
|-1 1 5 1|
与
|a[i-3]|
|a[i-2]|
|a[i-1]|
|a[i]|
相乘
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <cstdlib>
#include <algorithm>
#define LL long longusing namespace std;
const int Max = 10;int Mod;
struct Matrix
{int n,m;int a[Max][Max];void clear()//清空矩阵{n=0;m=0;memset(a,0,sizeof(a));}Matrix operator * (const Matrix &b)const//矩阵相乘{Matrix tmp;tmp.clear();tmp.n=n;tmp.m=b.m;for(int i=0;i<n;i++){for(int j=0;j<b.m;j++){for(int k=0;k<m;k++){tmp.a[i][j]=(tmp.a[i][j]+(a[i][k]%Mod)*(b.a[k][j]%Mod))%Mod;}}}return tmp;}
};void Pow(int m)
{Matrix s;s.clear();s.n=4;s.m=4;s.a[3][3]=1;s.a[3][2]=5;s.a[3][1]=1;s.a[3][0]=-1;s.a[1][2]=1;s.a[2][3]=1;s.a[0][1]=1;Matrix ans;ans.clear();ans.n=4;ans.m=1;ans.a[0][0]=1;ans.a[1][0]=5;ans.a[2][0]=11;ans.a[3][0]=36;while(m)//快速幂{if(m&1){ans=s*ans;}s=s*s;m>>=1;}printf("%d\n",ans.a[3][0]);
}int main()
{ int n;while(scanf("%d %d",&n,&Mod),n){if(n<4){switch(n){case 1:printf("%d\n",1%Mod);break;case 2:printf("%d\n",5%Mod);break;case 3:printf("%d\n",11%Mod);break;}continue;}Pow(n-4);}return 0;
}
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