矩阵快速幂 poj 3734

Blocks

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6461 Accepted: 3111

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 nextT 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

212

Sample Output

26

Source

我们可以从左边开始染色，设染到第i个方块为止，红绿都是偶数的方案数为ai,红绿恰有一个是偶数的方案数是bi,红绿都是奇数的方案数是ci。这样染到第i+1为止，可以有如下的递推式：

a(i+1)=2*ai+bi;

b(i+1)=2*ai+2*bi+2*ci;

c(i+1)=bi+2*ci;

这样我们可以建立一个矩阵

a(i+1)     |2   1   0|  ai

b(i+1)  = |2   2   2|  bi

c(i+1)     |0   1   2|  ci

这样就和斐波那契数列一样了

代码如下：

#include<stdio.h>
#define M 10007
struct node
{
     int c[3][3];
}FAB;
struct node mul(struct node *a,struct node *b)
{
     struct node t;
     int i,j,k;
     for(i=0;i<3;i++)
          for(j=0;j<3;j++)
               t.c[i][j]=0;
     for(i=0;i<3;i++)
         for(j=0;j<3;j++)
             for(k=0;k<3;k++)
             {
                  t.c[i][j]=(t.c[i][j]+a->c[i][k]*b->c[k][j])%M;
             }
    
      return t;
}
struct node power(struct node *a,int n)
{
        struct node temp;
        temp.c[0][0]=1;temp.c[0][1]=0;temp.c[0][2]=0;
        temp.c[1][0]=0;temp.c[1][1]=1;temp.c[1][2]=0;
        temp.c[2][0]=0;temp.c[2][1]=0;temp.c[2][2]=1;
        while(n>0)
        {
              if(n&1)
              temp=mul(&temp,a);
              *a=mul(a,a);
               n>>=1;
        }
        return temp;
}
int main(void)
{
 
     int n,t;
     scanf("%d",&t);
     while(t--)
     {
        scanf("%d",&n);
        FAB.c[0][0]=2;FAB.c[0][1]=1;FAB.c[0][2]=0;
        FAB.c[1][0]=2;FAB.c[1][1]=2;FAB.c[1][2]=2;
        FAB.c[2][0]=0;FAB.c[2][1]=1;FAB.c[2][2]=2;
        FAB=power(&FAB,n);
        printf("%d\n",FAB.c[0][0]);
     }
 
   
}