矩阵快速幂 poj 3734

最新推荐文章于 2021-07-08 09:30:58 发布

sky_zdk

最新推荐文章于 2021-07-08 09:30:58 发布

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分类专栏： ACM poj 文章标签：矩阵快速幂

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本文介绍了一个有趣的数学问题——熊猫希望对一排方块进行涂色，要求红色和绿色方块的数量均为偶数，求解在颜色限制条件下不同的涂色方案数量。文章通过递推公式和矩阵快速幂的方法解决了这一问题，并提供了完整的代码实现。

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Blocks

Time Limit: 1000MS		Memory Limit: 65536K
Total 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 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

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]);
     }


}