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思路：矩阵快速幂

分析:

1 裸题

代码：

/************************************************
 * By: chenguolin                               * 
 * Date: 2013-08-23                             *
 * Address: http://blog.csdn.net/chenguolinblog *
 ***********************************************/
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

typedef __int64 int64;
const int N = 10;

int k , MOD;
int arr[N] , f[N];
struct Matrix{
    int64 mat[N][N];
    Matrix operator*(const Matrix& m)const{
        Matrix tmp;
        for(int i = 0 ; i < N ; i++){
            for(int j = 0 ; j < N ; j++){
                tmp.mat[i][j] = 0;
                for(int k = 0 ; k < N ; k++){
                    tmp.mat[i][j] += mat[i][k]*m.mat[k][j]%MOD;
                    tmp.mat[i][j] %= MOD;
                }
            } 
        }
        return tmp;
    }
};

int64 Pow(Matrix &m , int k){
    Matrix ans;
    memset(ans.mat , 0 , sizeof(ans.mat));
    for(int i = 0 ; i < N ; i++)
        ans.mat[i][i] = 1;
    k -= 9;
    while(k){
        if(k&1)
           ans = ans*m;
        k >>= 1;
        m = m*m;
    } 
    int64 sum = 0;
    for(int i = 0 ; i < N ; i++){
        sum += ans.mat[0][i]*f[N-i-1]%MOD;
        sum %= MOD;
    }
    return sum;
}

void init(Matrix &m){
    memset(m.mat , 0 , sizeof(m.mat));
    for(int i = 0 ; i < N ; i++)
        m.mat[0][i] = arr[i];
    for(int i = 0 ; i < N-1 ; i++)
        m.mat[i+1][i] = 1;
    for(int i = 0 ; i < N ; i++)
        f[i] = i;
}

int main(){
    Matrix m;
    while(scanf("%d%d" , &k , &MOD) != EOF){
         for(int i = 0 ; i < N ; i++)
             scanf("%d" , &arr[i]);
         init(m);
         if(k < 10)
             printf("%d\n" , k%MOD);
         else
             printf("%I64d\n" , Pow(m , k));
    }
    return 0;
}