Blocks（POJ 3734 矩阵快速幂）

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　　//T
1　　//N
2

Sample Output

2
6
给定n方块染色，颜色有红黄绿蓝，问红绿都是偶数的情况有多少种。先要写出递推公式，见：


最开始的情况是2，2，0，乘以该矩阵，当然直接求n次幂答案也是对的

#include <cstring>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
using namespace std;
#define MOD 10007
typedef long long LL;
int T,n;
struct Matrix
{
    LL mat[3][3];
};
Matrix mul(Matrix a,Matrix b)
{
    Matrix c;
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<3;j++)
        {
            c.mat[i][j]=0;
            for(int k=0;k<3;k++)
                c.mat[i][j]=(c.mat[i][j]+a.mat[i][k]*b.mat[k][j])%MOD;
        }
    }
    return c;
}
Matrix mod_pow(Matrix x,LL n)
{
    Matrix res;
    memset(res.mat,0,sizeof(res.mat));
    for(int i=0;i<3;i++)
        res.mat[i][i]=1;
    while(n)
    {
        if(n&1)
            res=mul(res,x);
        x=mul(x,x);
        n>>=1;
    }
    return res;
}
int main()
{
    Matrix p;
    p.mat[0][0]=2,p.mat[0][1]=1,p.mat[0][2]=0;
    p.mat[1][0]=p.mat[1][1]=p.mat[1][2]=2;
    p.mat[2][0]=0,p.mat[2][1]=1,p.mat[2][2]=2;
    cin>>T;
    while(T--)
    {
        cin>>n;
        Matrix ans=mod_pow(p,n);
        cout<<ans.mat[0][0]<<endl;
    }
}