HDU2604 Queuing

http://acm.hdu.edu.cn/showproblem.php?pid=2604

矩阵快速幂优化DP

\[dp[i][0]表示以00结尾\\ dp[i][1]表示以01结尾\\ dp[i][2]表示以10结尾\\ dp[i][3]表示以11结尾\\ dp[i][0]=dp[i-1][0]+dp[i-1][2]\\ dp[i][1]=dp[i-1][0]\\ dp[i][2]=dp[i-1][1]+dp[i-1][3]\\ dp[i][3]=dp[i-1][1]\\ 初始矩阵： \begin{equation} \left[ \begin{array}{c} dp_{i-1,0}\\ dp_{i-1,1}\\ dp_{i-1,2}\\ dp_{i-1,3}\\ \end{array} \right] \end{equation} \\ \]

\[转移矩阵： \begin{equation} \left[ \begin{array}{c} 1 & 0 & 1 & 0\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 1 & 0 & 0\\ \end{array} \right] \end{equation} \]

\(C++ Code:\)

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#define N 1000005
#define n 4
using namespace std;
int l,m;
struct mat
{
    int a[5][5];
    mat operator * (mat b)
    {
        mat c;
        for (int i=0;i<=n;i++)
            for (int j=0;j<=n;j++)
                c.a[i][j]=0;
        for (int k=1;k<=n;k++)
            for (int i=1;i<=n;i++)
                for (int j=1;j<=n;j++) 
                    c.a[i][j]=(c.a[i][j]+a[i][k]*b.a[k][j])%m;
        return c;
    }
}zero,zy,cs;
mat ksm(mat x,int y)
{
    mat ans=zero;
    while (y)
    {
        if (y&1)
            ans=ans*x;
        x=x*x;
        y >>=1;
    }
    return ans;
}
int main()
{
    zy.a[1][1]=1,zy.a[1][3]=1,zy.a[2][1]=1;
    zy.a[3][2]=1,zy.a[3][4]=1,zy.a[4][2]=1;
    for (int i=1;i<=n;i++)
        zero.a[i][i]=1;
    cs.a[1][1]=2,cs.a[2][1]=1,cs.a[3][1]=2,cs.a[4][1]=1;
    while (scanf("%d%d",&l,&m)!=EOF)
    {
        if (l<=2)
        {
            switch (l)
            {
                case 0:
                    cout << 0 << endl;
                    break;
                case 1:
                    cout << 2 << endl;
                    break;
                case 2:
                    cout << 4 << endl;
                    break;
            }
            continue;
        }
        mat ans=ksm(zy,l-3)*cs;
        int g=(ans.a[1][1]+ans.a[2][1]+ans.a[3][1]+ans.a[4][1])%m;
        cout << g << endl;
    }
}