hdu 1757 A Simple Math Problem 矩阵快速幂

矩阵快速幂模板

#include<bits/stdc++.h>
using namespace std;

const int N = 12;
int n,Mod;

struct Mat
{
int mat

;
};

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

int main()
{
//    freopen("Input.txt","r",stdin);
int k;
int aa[12];
n=10;
while(~scanf("%d%d",&k,&Mod))
{
for(int i=0;i<10;i++)
scanf("%d",aa+i);
if(k<10)
{
printf("%d\n",k);
continue;
}
Mat a;
memset(a.mat,0,sizeof(a.mat));
for(int i=0;i<10;i++)
a.mat[i][0]=aa[i];
for(int i=1;i<10;i++)
a.mat[i-1][i]=1;
a=a^(k-9);
int ans=0;
for(int i=0;i<10;i++)
ans=(ans+a.mat[i][0]*(9-i))%Mod;
printf("%d\n",ans);
}
return 0;
}