hdu 1757 A Simple Math Problem

A Simple Math Problem
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 99991 1 1 1 1 1 1 1 1 120 5001 0 1 0 1 0 1 0 1 0

Sample Output

45104

题目意思还是很清晰，刚开始觉得会超时，但是没有。
看着写的

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define ll long long#define N 18#define M 10ll n, m;struct mtx{    int x[N][N];};mtx uni, ini;void init(){    memset( ini.x, 0, sizeof( ini.x ));    for( int i = 1 ; i < M ; i ++ ){        ini.x[i][i-1] = 1;    }    memset( uni.x, 0, sizeof( uni.x ));///    for( int i = 0 ; i < M ; i ++ ){///        uni.x[i][i] = 1;    }}mtx g_m( mtx a, mtx b ){    mtx c;    for( int i = 0 ; i < M ; i ++ ){        for( int j = 0 ; j < M ; j ++ ){            c.x[i][j] = 0;            for( int k = 0 ; k < M ; k ++ ){                c.x[i][j] += (a.x[i][k] * b.x[k][j]) % m;            }            c.x[i][j] %= m;        }    }    return c;}mtx s_m( mtx a , mtx b , int t ){    t %= m;    while( t ){        if( t & 1 ) b = g_m( a , b );        a = g_m( a, a );        t >>= 1;    }    return b;}int main(){    freopen( "in.txt", "r", stdin );    while( scanf( "%d%d", &n, &m ) != EOF ){//cout<<n<<" "<<m<<endl;        init();        for( int i = 0 ; i < M ; i ++ ){            scanf( "%d", &ini.x[0][i] );        }        if( n < M ){            printf( "%d\n", n%m );        }        else{            mtx ant = s_m( ini, uni, n-9 );///            int sum = 0;            for( int i = 0 ; i < M ; i ++ ){                sum += ( ant.x[0][i] * (9-i) ) % m;///            }            printf( "%d\n", sum%m );        }    }    return 0;}

模版