Quad Tiling

Time Limit:?1000MS	Memory Limit:?65536K
Total Submissions:?4107	Accepted:?1878

Description

Tired of(厭煩) the?Tri Tiling?gamefinally, Michael turns to(轉向) a morechallengeable game, Quad Tiling:

In how many ways can you tile(鋪瓷磚) a 4 ×?N?(1 ≤?N?≤10⁹) rectangle(矩形) with 2 × 1dominoes(多米諾骨牌)? 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(分別的).

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

題意：在一個4*N的矩陣中用2*1的多米諾骨牌鋪滿，問有多少種鋪法?

模板：

遞推式：a[i]=a[i-1]+5*a[i-2]+a[i-3]-a[i-4];

由于N高達10^9，所以要用矩陣進行優化。?
|0 1 0 0|?
|0 0 1 0|?
|0 0 0 1|?
|-1 1 5 1|?
與?
|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()//清空矩陣{n=0;m=0;memset(a,0,sizeof(a));}Matrix operator * (const Matrix &b)const//矩陣相乘{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)//快速冪{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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