poj 3734 Blocks

http://poj.org/problem?id=3734

Blocks

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4626		Accepted: 2102

Description

Panda has received an assignment of painting a line of blocks. Since Panda is such an intelligent boy, he starts to think of a math problem of painting. Suppose there are N blocks in a line and each block can be paint red, blue, green or yellow. For some myterious reasons, Panda want both the number of red blocks and green blocks to be even numbers. Under such conditions, Panda wants to know the number of different ways to paint these blocks.

Input

The first line of the input contains an integer T(1≤T≤100), the number of test cases. Each of the next T lines contains an integer N(1≤N≤10^9) indicating the number of blocks.

Output

For each test cases, output the number of ways to paint the blocks in a single line. Since the answer may be quite large, you have to module it by 10007.

Sample Input

2
1
2

Sample Output

2
6
红绿都是偶数的方案为ai，红绿恰有一个偶数的方案为bi，红绿都是奇数的方案为ci
ai+1=2*ai+bi；
bi+1=2*ai+2*bi+2*ci；
ci+1=bi+2*ci;

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
typedef vector<int> vec;
typedef vector<vec> mat;
typedef long long ll;
const int M=10007;
int N;
mat mul( mat &A,mat &B)
{

    mat C(A.size(),vec(B[0].size()));
     for(int i=0;i<A.size();i++)
     {
      for(int k=0;k<B.size();k++)
      {
        for(int j=0;j<B[0].size();j++)
        {
            C[i][j]=(C[i][j]+A[i][k]*B[k][j])%M;
        }
      }

     }
     return C;
}
mat pow(mat A,ll n)
{
    mat B(A.size(),vec(A.size()));
     for(int i=0;i<A.size();i++)
    {
        B[i][i]=1;
    }
    while(n>0)
    {
        if(n&1)
        B=mul(B,A);
        A=mul(A,A);
        n>>=1;
    }
    return B;
}
void solve()
{
    mat A(3,vec(3));
    A[0][0]=2;A[0][1]=1;A[0][2]=0;
    A[1][0]=2;A[1][1]=2;A[1][2]=2;
    A[2][0]=0;A[2][1]=1;A[2][2]=2;
    A=pow(A,N);
    printf("%d
",A[0][0]);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
    scanf("%d",&N);
    solve();
    }
    return 0;
}