hdu 1757 矩阵连乘

http://acm.hdu.edu.cn/showproblem.php?pid=1757

Problem Description

Lele now is thinking about a simple function f(x).

If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .

Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.

 

Input

The problem contains mutiple test cases.Please process to the end of file.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.

 

Output

For each case, output f(k) % m in one line.

 

Sample Input

   
   
   
   

    
    
    
    10 9999
1 1 1 1 1 1 1 1 1 1
20 500
1 0 1 0 1 0 1 0 1 0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    45
104
   
   
   
   

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <math.h>
using namespace std;
typedef long long LL;
const int N=10;
LL MOD,n;
int a[10];
struct Matrix
{
    LL m[N][N];
};
Matrix I=
{
    1,0,0,0,0,0,0,0,0,0,
    0,1,0,0,0,0,0,0,0,0,
    0,0,1,0,0,0,0,0,0,0,
    0,0,0,1,0,0,0,0,0,0,
    0,0,0,0,1,0,0,0,0,0,
    0,0,0,0,0,1,0,0,0,0,
    0,0,0,0,0,0,1,0,0,0,
    0,0,0,0,0,0,0,1,0,0,
    0,0,0,0,0,0,0,0,1,0,
    0,0,0,0,0,0,0,0,0,1
};
Matrix multi(Matrix a,Matrix b)
{
    Matrix c;
    for(int i=0; i<N; i++)
        for(int j=0; j<N; j++)
        {
            c.m[i][j]=0;
            for(int k=0; k<N; k++)
            {
                c.m[i][j]+=a.m[i][k]*b.m[k][j]%MOD;
            }
            c.m[i][j]=c.m[i][j]%MOD;
        }
    return c;
}
Matrix quick_mod(Matrix a,LL k)
{
    Matrix ans=I;
    while(k!=0)
    {
        if(k&1)
        {
            ans=multi(ans,a);
        }
        k>>=1;
        a=multi(a,a);
    }
    return ans;
}
int main()
{
    while(~scanf("%I64d%I64d",&n,&MOD))
    {
        for(int i=0;i<10;i++)
            scanf("%d",&a[i]);
        if(n<=9)
        {
            printf("%I64d\n",n);
            continue;
        }
        Matrix A={a[0],a[1],a[2],a[3],a[4],a[5],a[6],a[7],a[8],a[9],
                    1,0,0,0,0,0,0,0,0,0,
                    0,1,0,0,0,0,0,0,0,0,
                    0,0,1,0,0,0,0,0,0,0,
                    0,0,0,1,0,0,0,0,0,0,
                    0,0,0,0,1,0,0,0,0,0,
                    0,0,0,0,0,1,0,0,0,0,
                    0,0,0,0,0,0,1,0,0,0,
                    0,0,0,0,0,0,0,1,0,0,
                    0,0,0,0,0,0,0,0,1,0
                    };
        Matrix P=quick_mod(A,n-9);
        LL x=(P.m[0][0]*9%MOD+P.m[0][1]*8%MOD+P.m[0][2]*7%MOD+P.m[0][3]*6%MOD+P.m[0][4]*5%MOD+P.m[0][5]*4%MOD+P.m[0][6]*3%MOD+P.m[0][7]*2%MOD+P.m[0][8])%MOD;
        if(x<0)
            x+=MOD;
        printf("%I64d\n",x);
    }
    return 0;
}