文章标题 POJ 3734 : Blocks （矩阵快速幂）

Blocks

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

/*

假设从左边开始染色。
设染到第i个时，红绿都是偶数的方案书为a(i)，
红绿恰有一个是偶数的方案数为b(i)，
红绿都为奇数的方案数为c(i)，
那么染到第i+1个方块时，红绿都为偶数的方案只有两种可能
1：染到第i 个后，红绿的数量都为偶数，第i+1 个方块染成黄色或者蓝色；
2：染到第i 个后，红绿数量恰有一个偶数，第i+1 个方块染成红绿中数量是奇数的那种颜色
因此得到递推公式：a(i+1)=2*a(i)+b(i),
同样的递推，能得到：
b(i+1)=2*a(i)+2*b(i)+2*c(i)
c(i+1)=b(i)+2*c(i)

使用矩阵来表示此递推式可得：
| a(i+1) |     | 2 1 0 |      | a(i) | 
| b(i+1) |  = | 2 2 2 |  *  | b(i) | 
| c(i+1) |     | 0 1 2 |      | c(i) | 

*/ 
#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<map>
#include<queue> 
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
typedef pair<int,int> pii;
typedef long long ll;


int mod=10007;
int n,N,T;

struct Matrix {
    ll mat[3][3];
    Matrix (){}
    Matrix(int tmp[3][3]){
        for (int i=0;i<3;i++){
            for (int j=0;j<3;j++){
                mat[i][j]=tmp[i][j];
            }
        }
    }
    Matrix operator * (const Matrix m)const {
        Matrix tmp;
        for (int i=0;i<3;i++){
            for (int j=0;j<3;j++){
                tmp.mat[i][j]=0;
                for (int k=0;k<3;k++){
                    tmp.mat[i][j]+=mat[i][k]*m.mat[k][j]%mod;
                    tmp.mat[i][j]%=mod;
                }
            }
        }
        return tmp;
    }
};

Matrix Pow(Matrix &m,int k){
    Matrix ans;
    memset (ans.mat,0,sizeof (ans.mat));
    for (int i=0;i<3;i++)ans.mat[i][i]=1;
    while (k){
        if (k&1)ans=ans*m;
        k>>=1;
        m=m*m;
    }
    return ans;
}

int main()
{
    scanf ("%d",&T);
    while (T--){
        scanf ("%d",&N);
        int tmp[3][3]={2,1,0,2,2,2,0,1,2};
        Matrix ans=Matrix(tmp);
        ans=Pow(ans,N);
        int res=ans.mat[0][0];
        printf ("%d\n",res);
    }
    return 0; 
}