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 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

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 10
ll n, m;
struct mtx{
int x

;
};
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"
4000
, "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