hdu 3519 Lucky Coins Sequence 矩阵快速幂

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本文探讨了如何通过矩阵快速幂算法高效计算特定条件下硬币序列的数量。在N枚硬币排成一行的情况下，研究了至少连续三枚硬币状态相同的序列数量，即幸运硬币序列的数量，并给出了具体实现。

Lucky Coins Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 921 Accepted Submission(s): 488

Problem Description

As we all know,every coin has two sides,with one side facing up and another side facing down.Now,We consider two coins's state is same if they both facing up or down.If we have N coins and put them in a line,all of us know that it will be 2^N different ways.We call a "N coins sequence" as a Lucky Coins Sequence only if there exists more than two continuous coins's state are same.How many different Lucky Coins Sequences exist?

Input

There will be sevaral test cases.For each test case,the first line is only a positive integer n,which means n coins put in a line.Also,n not exceed 10^9.

Output

You should output the ways of lucky coins sequences exist with n coins ,but the answer will be very large,so you just output the answer module 10007.

Sample Input

Sample Output

硬币所有可能排列情况为2^n,朝向相同硬币连续三个以下的排列情况为a(1)=2,a(2)=4,a(3)=6,a(4)=10,观察或者推算得a(n)=a(n-1)+a(n-2);

那么连续三个或三个以上硬币的所有组合情况为b(n)=2^n-a(n);

化为只含b(n)的递推式即 b(n)=b(n-1)+b(n-2)+2^(n-2);b(1)=0;b(2)=0;b(3)=2;b(4)=6;用矩阵快速幂可快速算得b(n);

在这里讲一下矩阵快速幂求递推式的原理:

b(n) 1 1 1 b(n-1)

b(n-1) =( 1 0 0 ) * ( b(n-2) )

2^(n-1) 0 0 2 2^(n-2)

显然你要求b(n)需调用n次上式,即

b(n) 1 1 1 b(2)

b(n-1) =( 1 0 0 )^n *( b(1) )

2^(n-1) 0 0 2 2

1 1 1

我们有矩阵快速幂,计算 ( 1 0 0 )^n时间复杂度只用log n,和快速幂的时间复杂度一样,所以此法可行.

0 0 2

1 1 1

那么b(n)就是 ( 1 0 0 )^n 的第一行第三列与2的乘积(因为b(1)和b(2)都是0);

0 0 2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

const int mod=10007;
struct node
{
    int a[4][4];
    node()
    {
        a[1][1]=a[1][2]=a[1][3]=a[2][1]=1;
        a[2][2]=a[2][3]=a[3][1]=a[3][2]=0;
        a[3][3]=2;
    }
    void init()           //将其初始化为单位矩阵
    {
        memset(a,0,sizeof(a));
        for(int i=1; i<4; i++)
            a[i][i]=1;
    }
};

node mul(node a,node b)  //(a*b)%mod  矩阵乘法
{
    node ans;
    for(int i=1; i<4; i++)
        for(int j=1; j<4; j++)
        {
            ans.a[i][j]=0;
            for(int k=1; k<4; k++)
                ans.a[i][j]+=a.a[i][k]*b.a[k][j];
            ans.a[i][j]%=mod;
        }
    return ans;
}
node power(node a,int n)    //(a^n)%mod  //矩阵快速幂
{
    node ans;
    ans.init();
    while(n)
    {
        if(n%2)//n&1
            ans=mul(ans,a);
        n/=2;
        a=mul(a,a);
    }
    return ans;
}
int main()
{
    int i,j,n;
    while(~scanf("%d",&n))
    {
        if(n<3){printf("0\n");continue;}
        node ans;
        ans=power(ans,n-2);
        /*     for(i=1;i<4;i++)
             {for(j=1;j<4;j++)
              printf("%d",ans.a[i][j]);
             printf("\n");}*/
        printf("%d\n",(ans.a[1][3]*2)%mod);
    }
    return 0;
}