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.

解题思路：

只要遇到“把一个向量V变成另一个向量V'，并且V'的每一个分量都是V各个分量的线性组合”的情况，就可以考虑用矩阵乘法来描述这个关系。

线性方程组：Ax = b，关键是构造矩阵A。

此题线性方程组如下：

< http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcD4KPHA+PGltZyBzcmM9"https://www.cppentry.com/upload_files/article/49/1_9wgfv__.png" alt="\">

接下来就是计算了A^n了

矩阵的幂：

对于矩阵的幂可以采用二分快速幂

My_Code:

#include 
  
   
#include 
   
     #include 
    
      using namespace std; typedef long long LL; const int N = 10; LL k, m; struct mtx { LL x[N+1][N+1]; mtx() { memset(x, 0, sizeof x ); } }; mtx operator *(const mtx &a, const mtx &b) { mtx c; for(int i=0; i
     
      >= 1) { if(n&1) ret = ret * a; a = a * a; } return ret; } int main() { int i; LL sum; mtx tmp; for(i=1; i
      
       >k>>m) { for(i=0; i
       
        >tmp.x[0][i]; mtx ans = tmp ^ (k-9); sum = 0; for(i=0; i