HDU1757(矩阵快速幂+简单的矩阵构造)

很简单的矩阵构造看似很麻烦，其实直接写就ok了。

⎡⎣⎢⎢⎢⎢f(n)f(n−1)....f(n−9)⎤⎦⎥⎥⎥⎥=⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢a100....0b0100c0010d0000e0000f0000g0000h0001i0000⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎡⎣⎢⎢⎢⎢f(n−1)f(n−2)....f(n−10)⎤⎦⎥⎥⎥⎥

code：

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>#include<iostream>#include<string>#include <set>#include<time.h>//a&3==a%4using namespace std ;#define ll int#define mem(a) memset(a,0,sizeof(a))const double eps = 1e-8;const int maxn = 110010;const int inf = 0x3f3f3f3f;ll mod;struct matrix{    ll mat[30][30];    matrix()    {        memset(mat,0,sizeof(mat));    }};matrix multiply(matrix a,matrix b,int n){    matrix res;    for(int i=0;i<n;i++)    {        for(int j=0;j<n;j++)        {            for(int k=0;k<n;k++)            {                res.mat[i][j]+=a.mat[i][k]*b.mat[k][j]%mod;                res.mat[i][j]%=mod;            }        }    }    return res;}matrix quickmi(matrix a,ll b,int n){    matrix E;    E.mat[0][0]=1;    matrix res;    if(b==0)    {        return E;    }    for(int i=0;i<n;i++)    {        E.mat[i][i]=1;        res.mat[i][i]=1;    }    E=multiply(a,E,n);    while(b>0)    {        if(b%2==1)        {            res=multiply(res,E,n);        }        b=b/2;        E=multiply(E,E,n);    }    return res;}void out(matrix res,int n){    for(int i=0;i<n;i++)    {        cout<<res.mat[i][0];        for(int j=1;j<n;j++)        {            cout<<" "<<res.mat[i][j];        }        cout<<endl;    }}int main(){    matrix res;    matrix one;    for(int i=0;i<10;i++)    {        one.mat[i][0]=i;    }    matrix sum;    ll k,m;    while(cin>>k>>m)    {        if(k>=10)        {            for(int i=0;i<10;i++)            {                scanf("%d",&res.mat[0][i]);                if(i>=1)                    res.mat[i][i-1]=1;            }            mod=m;            sum=quickmi(res,k-9,10);            //out(sum,10);            int hehe=0;            for(int i=9;i>=0;i--)            {                hehe+=(i*sum.mat[0][9-i])%mod;                hehe=hehe%mod;            }            cout<<hehe<<endl;        }        else        {           for(int i=0;i<10;i++)            {                scanf("%d",&res.mat[0][i]);            }            cout<<k<<endl;        }    }    return 0;}