A Simple Math Problem
Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6592    Accepted Submission(s): 4047

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

程式碼：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
const ll maxn = 11;
ll m,mod;
struct matrix
{
    int n;
    ll d[maxn][maxn];
    void init(int n)
    {
       this -> n = n;
       mem(d,0);
    }
    matrix operator *(matrix & b)
    {
        matrix ans;
        ans.init(n);
        for (int i = 0;i < n;i ++)
            for (int j = 0;j < n;j ++)
            {
                for (int k = 0;k < n;k ++)
                    ans.d[i][j]=(ans.d[i][j]+d[i][k]*b.d[k][j])%mod;
            }
        return ans;
    }
};
matrix quick(matrix a,ll b)
{
    matrix res;
    res.init(a.n);
    for (int i = 0;i < res.n;i ++)
        res.d[i][i] = 1;
    while (b)
    {
        if (b & 1) res = res * a;
        a = a * a;
        b >>= 1;
    }
    return res;
}
int main()
{
    while (~scanf("%lld %lld",&m,&mod))
    {
        matrix x;
        x.init(10);
        for (int i = 0;i < 10;i ++)
            scanf("%lld",&x.d[9 - i][9]);
        if (m < 10)
        {
            printf("%lld\n",m % mod);
            continue;
        }
        for (int i = 0;i < 9;i ++)
            x.d[i + 1][i] = 1;
        x = quick(x,m - 9);
        matrix ans;
        ans.init(10);
        for (int i = 0;i < 10;i ++) ans.d[0][i] = i;
        ans = ans * x;
        printf("%lld\n",ans.d[0][9]);
    }
    return 0;
}