hdu 1757 A Simple Math Problem （矩阵快速幂）

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

这就算我写的第一道矩阵快速幂的题了！
矩阵快速幂的给我的感觉就是先找规律，YY出一个n*n的矩阵，矩阵左边写一列数，矩阵右边写一列数。让矩阵乘上右边那一列数等于左边的那列数。盗张图！！

然后......然后就没有然后了QWQ，输出答案就行了。
代码如下：

 #include <bits/stdc++.h>

 using namespace std;
 const int N=;
 int k,m;
 struct Matrix//用结构体来存矩阵
 {
     long long int mat[N][N];//直接上long long int 别找事
 }matrix;
 void init()
 {
     for (int i=;i<N;++i)
     scanf("%lld",&matrix.mat[][i]);
     for (int i=;i<N;++i)
     {
         for (int j=;j<N;++j)
         {
             if (i==j+)
             matrix.mat[i][j]=;
             else
             matrix.mat[i][j]=;
         }
     }
 }
 Matrix operator * (Matrix a,Matrix b)//定义乘号
 {
     Matrix c;
     for (int i=;i<N;++i)
     {
         for (int j=;j<N;++j)
         {
             c.mat[i][j]=;//初始化值为0
             for (int k=;k<N;++k)
             c.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
             c.mat[i][j]%=m;//记得取模，否则就炸了
         }
     }
     return c;
 }
 Matrix Pow (int n)//定义快速幂函数
 {
     Matrix t;
     if (n==)
     return matrix;
     if (n&)
     return matrix*Pow(n-);
     else
     {
         Matrix temp=Pow(n>>);
         return temp*temp;
     }
 }
 int main()
 {
     //freopen("de.txt","r",stdin);
     while (~scanf("%d%d",&k,&m))
     {
         init();
         if (k<)
         {
             printf("%lld\n",matrix.mat[][k]%m);
             continue;
         }
         Matrix temp=Pow(k-);
         long long int ans=;
         for (int i=;i<N;++i)
         {
             ans += temp.mat[][i]*(N-i-);
             ans%=m;
         }
         printf("%lld\n",ans);
     }
     return ;
 }