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hdu 2604 递推 矩阵快速幂

HDU 2604 Queuing (递推+矩阵快速幂)

这位作者讲的不错,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 递推 矩阵快速幂