hdu 6185
题意：有4*n的空地，问用1*2的瓷砖铺满不重叠，方案数
思路：a[n]=a[n-1]+5*a[n-2]+a[n-3]-a[n-4]，矩阵快速幂搞搞
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const LL MOD=1e9+7; 5 struct Matrix 6 { 7 LL a[4][4]; 8 Matrix() 9 { 10 memset(a,0,sizeof(a)); 11 } 12 void init() 13 { 14 for(int i=0;i<4;i++) 15 for(int j=0;j<4;j++) 16 a[i][j]=(i==j); 17 } 18 Matrix operator + (const Matrix &B)const 19 { 20 Matrix C; 21 for(int i=0;i<4;i++) 22 for(int j=0;j<4;j++) 23 C.a[i][j]=(a[i][j]+B.a[i][j])%MOD; 24 return C; 25 } 26 Matrix operator * (const Matrix &B)const 27 { 28 Matrix C; 29 for(int i=0;i<4;i++) 30 for(int k=0;k<4;k++) 31 for(int j=0;j<4;j++) 32 C.a[i][j]=(C.a[i][j]+1LL*a[i][k]*B.a[k][j])%MOD; 33 return C; 34 } 35 Matrix operator ^ (const LL &t)const 36 { 37 Matrix A=(*this),res; 38 res.init(); 39 LL p=t; 40 while(p) 41 { 42 if(p&1)res=res*A; 43 A=A*A; 44 p>>=1; 45 } 46 return res; 47 } 48 }; 49 Matrix base; 50 void init() 51 { 52 53 base.a[0][0]=0;base.a[0][1]=0;base.a[0][2]=0;base.a[0][3]=-1; 54 base.a[1][0]=1;base.a[1][1]=0;base.a[1][2]=0;base.a[1][3]=1; 55 base.a[2][0]=0;base.a[2][1]=1;base.a[2][2]=0;base.a[2][3]=5; 56 base.a[3][0]=0;base.a[3][1]=0;base.a[3][2]=1;base.a[3][3]=1; 57 } 58 LL query(LL x) 59 { 60 if(x==1)return 1; 61 if(x==2)return 5; 62 if(x==3)return 11; 63 if(x==4)return 36; 64 Matrix ans=base^(x-4); 65 return ((36*ans.a[3][3]+11*ans.a[2][3]+5*ans.a[1][3]+ans.a[0][3])%MOD+MOD)%MOD; 66 } 67 int main() 68 { 69 init(); 70 LL n; 71 while(~scanf("%lld",&n)) 72 { 73 printf("%lld\n",query(n)); 74 } 75 return 0; 76 }
