A Simple Math Problem(矩陣快速冪)

A Simple Math Problem

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

完整代碼：

#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn=10;
ll n,mod;
struct mat
{
    ll m[maxn][maxn];
};
mat mat_mul(mat a,mat b)        /*矩陣乘法*/
{
    mat res;
    memset(res.m,0,sizeof(res.m));
    for(ll k=0;k<maxn;k++)
    {
        for(ll i=0;i<maxn;i++)
        {
            if(!a.m[i][k])
            {
                continue;
            }
            for(ll j=0;j<maxn;j++)
            {
                if(!b.m[k][j])
                {
                    continue;
                }
                res.m[i][j]=(res.m[i][j]+a.m[i][k]*b.m[k][j])%mod;
            }
        }
    }
    return res;
}
mat mat_pow(mat a,ll b)         /**矩陣快速冪**/
{
    mat res;
    memset(res.m,0,sizeof(res.m));
    for(ll i=0;i<maxn;i++)      //單位矩陣
    {
        res.m[i][i]=1;
    }
    while(b)
    {
        if(b&1)
        {
            res=mat_mul(res,a);
        }
        a=mat_mul(a,a);
        b>>=1;
    }
    return res;
}

int main()
{
    while(cin>>n>>mod)
    {
        mat a;
        memset(a.m,0,sizeof(a.m));
        for(ll i=0;i<maxn;i++)
        {
            cin>>a.m[i][0];
        }
        //按題目說的小於10，和大於10兩個區間
        if(n<10)
        {
            cout<<n<<endl;
        }
        else
        {
            for(ll i=1;i<maxn;i++)
            {
                a.m[i-1][i]=1;
            }
            a=mat_pow(a,n-9);
            ll res=0;
            for(int i=0;i<maxn;i++)
            {
                res=(res+(9-i)*a.m[i][0])%mod;
            }
            cout<<res<<endl;

        }
    }
}