Blocks(矩阵快速幂自用模板)

最新推荐文章于 2021-03-28 10:34:18 发布

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本文介绍了一道关于涂色组合的数学问题：给定数量的方块，每个方块可以被涂成四种颜色中的一种，但红色和绿色方块的数量必须为偶数。文章提供了使用矩阵快速幂算法求解该问题的C++代码实现。

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Blocks

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7801		Accepted: 3797

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

过程就不写了。。。。=v=

#include <iostream>
#include <cstdio>

using namespace std;
const int mod = 10007;
typedef long long ll;
typedef struct mat
{
    int a[3][3];
}mat;
mat multi(mat a,mat b)
{
    mat c;
     for(int i=0;i<3;i++)for(int j=0;j<3;j++) c.a[i][j]=0;
      for(int i=0;i<3;i++)for(int j=0;j<3;j++)for(int k=0;k<3;k++)
        c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j])%mod;
      return c;
}
mat pow(mat a,int n)
{
    mat ans;//单位矩阵
    for(int i=0;i<3;i++)for(int j=0;j<3;j++) ans.a[i][j]=0;
    for(int i=0;i<3;i++) ans.a[i][i]=1;
    while(n)
    {
        if(n&1) ans=multi(ans,a);
        a=multi(a,a);
        n>>=1;
    }
    return ans;
}
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        mat ans,init;
        init.a[0][0]=2,init.a[0][1]=1,init.a[0][2]=0;
        init.a[1][0]=2,init.a[1][1]=2,init.a[1][2]=2;
        init.a[2][0]=0,init.a[2][1]=1,init.a[2][2]=2;
        ans=pow(init,n);
        printf("%d\n",ans.a[0][0]);
    }
    return 0;
}