hdu 2604 Queuing 递推+矩阵快速幂

[code]#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=10;
int n=4,m;

struct Mat
{
    int mat

;
};

Mat operator * (Mat a, Mat b)
{
    Mat c;
    memset(c.mat, 0, sizeof(c.mat));
    int i, j, k;
    for(k = 0; k < n; ++k)
    {
        for(i = 0; i < n; ++i)
        {
            for(j = 0; j < n; ++j)
            {
                c.mat[i][j] += a.mat[i][k] * b.mat[k][j];
                c.mat[i][j]=c.mat[i][j]%m;
            }
        }
    }
    return c;
}

Mat operator ^ (Mat a, int k)
{
    Mat c;
    int i, j;
    for(i = 0; i < n; ++i)
        for(j = 0; j < n; ++j)
            c.mat[i][j] = (i == j);    //初始化为单位矩阵

    for(; k; k >>= 1)
    {
        if(k&1) c = c*a;
        a = a*a;
    }
    return c;
}

int main()
{
    int k,i,j;
    while(~scanf("%d%d",&k,&m))
    {
        if(k==0) printf("%d\n",k);
        else if(k==1) printf("%d\n",2%m);
        else if(k==2) printf("%d\n",4%m);
        else if(k==3) printf("%d\n",6%m);
        else
        {
            Mat a,b;
            memset(a.mat,0,sizeof(a.mat));
            a.mat[0][0]=1;
            a.mat[0][2]=1;
            a.mat[0][3]=1;
            a.mat[1][0]=1;
            a.mat[2][1]=1;
            a.mat[3][2]=1;
            a=a^(k-4);
            /*for(i=0; i<4; i++)
            {
                for(j=0; j<4; j++)
                    printf("%d ",a.mat[i][j]);
                printf("\n");
            }*/
            memset(b.mat,0,sizeof(b.mat));
            b.mat[0][0]=9;
            b.mat[1][0]=6;
            b.mat[2][0]=4;
            b.mat[3][0]=2;
            int ans=0;
            for(i=0; i<4; i++)
            {

                ans=(ans+b.mat[i][0]*a.mat[0][i])%m;
                //printf("%d %d %d\n",b.mat[i][0],a.mat[0][i],ans);
            }
            printf("%d\n",ans%m);
        }
    }
    return 0;
}