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hdu 2604 Queuing 遞推+矩陣快速冪

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

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

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); //初始化為單位矩陣 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; }