Poj 3734 rapid matrix power + YY

Original Intention: N blocks are arranged in a column, each of which can be marked as red, blue, green, or yellow. Ask the number of solutions where both the red and green squares are even numbers.

 

SOL: Finding regular column recursion + rapid matrix power

Set the number of I + 1 blocks that have been dyed.

In a [I] = 1-I blocks, the number of red and green blocks is an even number of solutions.

In B [I] = 1-I blocks, the number of red and green blocks is an even number and the number of solutions is an odd number (Red even green odd or red odd Green even)

C [I] = in 1-I blocks, the number of red and green blocks is an odd number of solutions.

 

The recursive formula can be obtained:

A [I + 1] = 2 * A [I] + B [I]

B [I + 1] = 2 * A [I] + 2 * B [I] + 2 * C [I]

C [I + 1] = B [I] + 2 * C [I]

How to combine with a matrix? Write the formula like this,

A [I + 1] = 2 * A [I] + 1 * B [I] + 0 * C [I]

B [I + 1] = 2 * A [I] + 2 * B [I] + 2 * C [I]

C [I + 1] = 0 * A [I] + 1 * B [I] + 2 * C [I]

 

Then, YY generates a matrix:

Therefore,

In this way, you can use the Matrix to quickly calculate the power.

 

 1 #include "iostream" 2 #include "vector" 3 #include "cstring" 4 using namespace std; 5  6 typedef unsigned long int ULL; 7 typedef vector<ULL> vec; 8 typedef vector<vec> mat; 9 const ULL P=10007;10 int n,m;11 12 mat mul(mat &A,mat &B)      //return A*B13 {14     mat C(A.size(),vec(B[0].size()));15     for (int i=0;i<(int)A.size();i++)16     {17         for (int k=0;k<(int)B.size();k++)18         {19             for (int j=0;j<(int)B[0].size();j++)20             {21                 C[i][j]=(C[i][j]+A[i][k]*B[k][j])%P;22             }23         }24     }25     return C;26 }27 28 mat m_pow(mat A,int m)      //return A^m29 {30     mat B(A.size(),vec(A.size()));31     for (int i=0;i<(int)A.size();i++)32         B[i][i]=1;33     while (m>0)34     {35         if (m&1)    B=mul(B,A);36         A=mul(A,A);37         m>>=1;38     }39     return B;40 }41 42 int main()43 {44     int T,N;45     cin>>T;46     while (T--)47     {48         cin>>N;49         mat A(3,vec(3));50         A[0][0]=2;  A[0][1]=1;  A[0][2]=0;51         A[1][0]=2;  A[1][1]=2;  A[1][2]=2;52         A[2][0]=0;  A[2][1]=1;  A[2][2]=2;53 54         A=m_pow(A,N);55 56         cout<<A[0][0]<<endl;57     }58     return 0;59 }60 61 /*62 int main()63 {64     int T;65     cin>>T;66     while (T--)67     {68         cin>>n>>m;69         mat A(n,vec(n));70 71         for (int i=0;i<n;i++)72             for (int j=0;j<n;j++)73                 cin>>A[i][j];74 75         A=m_pow(A,m);76 77         ULL ans=0;78         for (int i=0;i<n;i++)79         {80             ans+=A[i][i];81             ans=ans%P;82         }83         cout<<ans<<endl;84     }85     return 0;86 }87 */

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Poj 3734 rapid matrix power + YY