poj 3734 Blocks（矩阵快速幂）

最新推荐文章于 2021-09-11 19:55:36 发布

原创最新推荐文章于 2021-09-11 19:55:36 发布 · 483 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#dp

DP 同时被 3 个专栏收录

16 篇文章

订阅专栏

模板

7 篇文章

订阅专栏

DP——矩阵快速幂

1 篇文章

订阅专栏

本文探讨了一个有趣的数学问题：如何计算在特定条件下（红色和绿色方块数量必须为偶数），用四种颜色对一排方块进行涂色的不同方式的数量。通过动态规划方法转化为矩阵运算，并利用快速幂求解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Blocks

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6472		Accepted: 3117

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

PKU Campus 2009 (POJ Monthly Contest – 2009.05.17), Simon

题解：这道题第一感觉是dp，然后仔细考虑考虑到第i个方块，设同时为偶数的方案数为ai，有一个为偶数的方案数为bi，全是奇数的方案数是ci

a（i+1）=2*ai+bi

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

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

然后可以得到矩阵，之后求快速幂就行了

代码

#include <vector>
#include <cstdio>
#include <iostream>
//#include <bits/stdc++.h>

using namespace std;
const int M=10007;  //修改余数
typedef vector<int>  vec;
typedef vector<vec>  mat;
typedef long long LL;
mat mul(mat &A,mat &B)
{
    mat C(A.size(),vec(B[0].size()));
    for(int i=0;i<A.size();i++)
    {
        for(int k=0;k<B.size();k++)
        {
            for(int j=0;j<B[0].size();j++)
                    C[i][j]=(C[i][j]+A[i][k]*B[k][j])%M;
        }
    }
    return C;
}

mat pow(mat A,LL n)
{
    mat B(A.size(),vec(A.size()));
    for(int i=0;i<A.size();i++)
        B[i][i]=1;
    while(n>0)
    {
        if(n&1) B=mul(B,A);
        A=mul(A,A);
        n>>=1;
    }
    return B;
}

LL n;
int main()
{
    int t;
    cin>>t;
    while(t--){
        cin>>n;
        mat A(3,vec(3));
        A[0][0]=A[1][1]=A[1][2]=A[1][0]=A[2][2]=2;
        A[0][1]=A[2][1]=1;
        A[2][0]=A[0][2]=0;
        //cout<<1<<endl;
        A=pow(A,n);
        cout<<A[0][0]<< endl;
    }
    return 0;
}