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:
1 #include<set> 2 #include<queue> 3 #include<cstdio> 4 #include<cstring> 5 #include<cstdlib> 6 #include<iostream> 7 #include<algorithm> 8 using namespace std; 9 #define For(i,n) for(int i=1;i<=n;i++) 10 #define Rep(i,l,r) for(int i=l;i<=r;i++) 11 12 struct Matrix{ 13 int A[17][17]; 14 Matrix(){ 15 memset(A,0,sizeof(A)); 16 } 17 }Unit,Ans,TAns; 18 19 int opt[20],n,Mod; 20 21 Matrix operator * (Matrix A,Matrix B){ 22 Matrix C; 23 Rep(i,0,15) 24 Rep(j,0,15) 25 Rep(k,0,15) 26 C.A[i][j] = ( C.A[i][j] + (A.A[i][k] * B.A[k][j]) % Mod ) % Mod; 27 return C; 28 } 29 30 bool Check(int s1,int s2){ 31 if((s1|s2)!=15) return false; 32 for(int i=0;i<=15;){ 33 if( (s1&(1<<i)) != (s2&(1<<i)) ) i++; 34 else{ 35 if(i==15) return false; else 36 if( ( s1 & (1<<(i+1)) ) != ( s2 & (1<<(i+1)) ) ) return false; 37 i+=2; 38 } 39 } 40 return true; 41 } 42 43 int main(){ 44 Rep(i,0,15){ 45 Rep(j,0,15) 46 if(Check(i,j)) Ans.A[i][j] = TAns.A[i][j] = 1; 47 if(Check(i,15)) 48 opt[i] = 1; 49 } 50 while(scanf("%d%d",&n,&Mod),Mod+n){ 51 Rep(i,0,15){ 52 Rep(j,0,15){ 53 Ans.A[i][j] = TAns.A[i][j]; 54 Unit.A[i][j] = 0; 55 } 56 Unit.A[i][i] = 1; 57 } 58 while(n){ 59 if(n&1) Unit = Unit * Ans; 60 Ans = Ans * Ans; 61 n = n >> 1; 62 } 63 int ans = 0; 64 Rep(i,0,15) 65 ans = (ans % Mod + (opt[i] * Unit.A[0][i] % Mod) % Mod) % Mod; 66 printf("%d ",ans); 67 } 68 return 0; 69 }