HDU 1757 A Simple Math Problem

http://acm.hdu.edu.cn/showproblem.php?pid=1757

矩阵快速幂

/*
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);

|f(10) |      |a0 a1 a2 ...a8 a9|   |f(9)|
| f(9) |      | 1  0  0 ... 0  0|   |f(8)|
| .....|  =   |.. ... ... ... ..|   | .. |
| f(2) |      | 0  0  0 ... 0  0|   |f(1)|
| f(1) |      | 0  0  0 ... 1  0|   |f(0)|

另A举证为10*10的举证，如上图。
可以推出：
(f(n),f(n-1),...,f(n-9))^(-1) = A^(n-9)*(f(9),f(8),...,f(0))^(-1)
*/

#include<iostream>
using namespace std;
__int64 k,m;
struct mat
{
    int mar[10][10];
}a,b,tmp;

mat matrixmul(mat a,mat b)
{
    int i,j,k;
    for(i = 0;i < 10;i++)
        for(j = 0;j < 10;j++)
        {
            tmp.mar[i][j] = 0;
            for(k = 0;k < 10;k++)
                tmp.mar[i][j] += (a.mar[i][k] * b.mar[k][j]) % m;
            tmp.mar[i][j] %= m;
        }
    return tmp;
}

void matrix_binary()
{
    while(k)
    {
        if(k & 1)
            b = matrixmul(b,a);
        a = matrixmul(a,a);
        k = k >> 1;
    }
}
int main()
{
    int i;
    while (scanf("%I64d%I64d",&k,&m) != EOF)
    {
        memset(a.mar,0,sizeof(a.mar));
        for(i = 1;i < 10;i++)
            a.mar[i][i-1] = 1;
        memset(b.mar,0,sizeof(b.mar));
        for(i = 0;i < 10;i++)
            b.mar[i][i] = 1;
        for(i = 0;i < 10;i++)
            scanf("%d",&a.mar[0][i]);
        if(k < 10)
        {
            printf("%d\n", k % m);
            continue;
        }
        k -= 9;
        matrix_binary();
        int res = 0;
        for(i = 0;i < 10;i++)
            res += (b.mar[0][i] * (9-i)) % m;
        printf("%d\n",res%m);
    }
    return 0;
}