A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;

struct matrix {
    ll f[11][11];
};
int n,m;

matrix mul(matrix a,matrix b){
    matrix s;
    memset(s.f,0,sizeof s.f);
    int i,j,k;
    for(k=0;k<10;k++)
    for(i=0;i<10;i++){
        if(!a.f[i][k])  continue;
        for(j=0;j<10;j++){
            if(!b.f[k][j])  continue;
            s.f[i][j]+=a.f[i][k]*b.f[k][j];
            s.f[i][j]%=m;
        }
    }
    return s;
}

matrix pow_mod(matrix a,int k){
    matrix s;
    memset(s.f,0,sizeof s.f);
    for(int i=0;i<10;i++)
        s.f[i][i]=1;
    while(k){
        if(k&1)   s=mul(s,a);
        a=mul(a,a);
        k>>=1;
    }
    return s;
}

int main(){
    while(cin>>n>>m){
        int i;
        matrix e;
        memset(e.f,0,sizeof e.f);
        for(i=0;i<10;i++)
            cin>>e.f[i][0];
        if(n<10){
            cout<<n<<endl;
            continue;
        }
        for(i=1;i<10;i++)
            e.f[i-1][i]=1;
        e=pow_mod(e,n-9);
        ll ans=0;
        for(i=0;i<10;i++)
            ans=(ans+(9-i)*e.f[i][0])%m;
        cout<<ans<<endl;
    }
    return 0;
}