hdu1757-- A Simple Math Problem（矩阵快速幂优化）

A Simple Math Problem

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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 <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define LL __int64
struct node{
    LL a[12][12] ;
    int n ;
};
node mul(node p,node q,LL m)
{
    int i , j , k ;
    node s ;
    s.n = p.n ;
    for(i = 0 ; i < p.n ; i++)
        for(j = 0 ; j < p.n ; j++)
        {
            s.a[i][j] = 0 ;
            for(k = 0 ; k < p.n ; k++)
                s.a[i][j] = (s.a[i][j] + p.a[i][k]*q.a[k][j]) % m ;
        }
    return s ;
}
node pow(node p,LL k,LL m)
{
    if( k == 1 )
        return p ;
    node s = pow(p,k/2,m) ;
    s = mul(s,s,m) ;
    if( k%2 )
    {
        s = mul(s,p,m) ;
    }
    return s ;
}
int main()
{
    LL k , m , ans ;
    int i , j ;
    node p , s ;
    while( scanf("%I64d %I64d", &k, &m) != EOF )
    {
        if(k < 10)
        {
            printf("%I64d\n", k) ;
            continue ;
        }
        ans = 0 ;
        p.n = 10 ;
        for(i = 0 ; i < 10 ; i++)
        {
            for(j = 0 ; j < 10 ; j++)
                p.a[i][j] = 0 ;
            p.a[i][i+1] = 1 ;
        }
        for(i = 0 ; i < 10 ; i++)
            scanf("%I64d", &p.a[i][0]) ;
        s = pow(p,k-9,m) ;
        for(i = 0 ; i < 10 ; i++)
            ans = ( ans + (9-i)*s.a[i][0] ) % m ;
        printf("%I64d\n", ans) ;
    }
    return 0 ;
}