hdu 1757 A Simple Math Problem（矩陣快速冪）

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2512 Accepted Submission(s): 1461


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.
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

Author linle

題解及代碼：

#include 
#include 
#include 
using namespace std;
const int mod=1e9;
struct mat
{
    __int64 t[10][10];
    void set()
    {
        memset(t,0,sizeof(t));
    }
} a,b,c;

mat multiple(mat a,mat b,int n,int p)
{
    int i,j,k;
    mat temp;
    temp.set();
    for(i=0; i>=1;
        b=multiple(b,b,n,p);
    }
    return t;
}

void init(int a[])
{
   b.set();
   for(int i=0;i<10;i++)
   {
       for(int j=0;j<10;j++)
       if(j==9)
       b.t[i][j]=a[i];
       else if(i==j+1) b.t[i][j]=1;
   }
   /*for(int i=0;i<=9;i++)
   {
       for(int j=0;j<=9;j++)
        cout<>k>>M)
    {
        for(int i=9;i>=0;i--)
            cin>>s[i];
        if(k<10)
        {
            cout<