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[][];
 };
 Matrix mul(Matrix a,Matrix b)
 {
     Matrix c;
     for(int i=;i<;i++)
     {
         for(int j=;j<;j++)
         {
             c.mat[i][j]=;
             for(int k=;k<;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,,sizeof(res.mat));
     for(int i=;i<;i++)
         res.mat[i][i]=;
     while(n)
     {
         if(n&)
             res=mul(res,x);
         x=mul(x,x);
         n>>=;
     }
     return res;
 }
 int main()
 {
     Matrix p;
     p.mat[][]=,p.mat[][]=,p.mat[][]=;
     p.mat[][]=p.mat[][]=p.mat[][]=;
     p.mat[][]=,p.mat[][]=,p.mat[][]=;
     cin>>T;
     while(T--)
     {
         cin>>n;
         Matrix ans=mod_pow(p,n);
         cout<<ans.mat[][]<<endl;
     }
 }