HDU 2604 Queuing (遞推+矩陣快速冪)

這位作者講的不錯，可以看看他的

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

const int N = 5;
int msize, Mod;

struct Mat
{
    int mat[N][N];
};

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

Mat operator ^(Mat a, int k)
{
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    for(int i = 0; i < msize; ++i)
        c.mat[i][i]=1;
    for(; k; k >>= 1)
    {
        if(k&1) c = c*a;
        a = a*a;
    }
    return c;
}

int main()
{
//    freopen("in.txt","r",stdin);
 

    int n;
    msize = 4;
    int f[] = {9, 6, 4, 2};
    while(~scanf("%d%d", &n, &Mod))
    {
        if(n <= 4)
        {
            printf("%d\n", f[4-n] % Mod);
            continue;
        }
        Mat A;
        memset(A.mat,0,sizeof(A.mat));
        for(int i = 0; i < msize; i++)
            A.mat[0][i] = 1;
        A.mat[0][1] = 0;
        for(int i = 1; i < msize; i++)
            A.mat[i][i-1] = 1;
        A = A^(n - msize);
        int ans = 0;
        for(int i=0; i < msize; i++)
            ans = (ans + A.mat[0][i]*f[i]) % Mod;
        printf("%d\n", ans);
    }
    return 0;
}

hdu 2604 遞推 矩陣快速冪