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

题目大意：
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 .
利用公式求出f(k) % m并输出。注意 k<2*10^9 , m < 10^5.

解题思路：利用矩阵快速幂进行求解，重点在于怎样构造矩阵。

利用矩阵快速幂所求第一行即是答案。
node ans=a^(n-9)*b;

Code:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int m;
ll k;

struct node {
	ll materix[50][50];
};
node a,b;

node mul(node a,node b){//矩阵乘法
	node ans;
	memset(ans.materix,0,sizeof(ans.materix));
	for(int i=1;i<=10;i++){
		for(int j=1;j<=10;j++){
			for(int k=1;k<=10;k++){
				ans.materix[i][j]=(ans.materix[i][j]+a.materix[i][k]*b.materix[k][j]%m)%m;
			}
		}
	}
	return ans;
}

node ksm(node a,ll b){//矩阵快速幂
	node ans;
	memset(ans.materix,0,sizeof(ans.materix));
	for(int i=1;i<=10;i++) ans.materix[i][i]=1;
	while(b){
		if(b&1) ans=mul(ans,a);
		a=mul(a,a);
		b>>=1;
	}
	return ans;
	
}

int main(){
	while(cin>>k>>m){
		memset(a.materix,0,sizeof(a.materix));
		for(int i=1;i<=10;i++){
			cin>>a.materix[i][1];
		}
		for(int i=1;i<=10;i++){
			a.materix[i][i+1]=1;
		}
		if(k<10){
			cout<<k%m<<endl;
			continue;
		}
		else {
			node b;
			for(int i=1;i<=10;i++){
				b.materix[1][i]=10-i;
			}
			node ans=ksm(a,k-9);
			ans=mul(b,ans);
			cout<<ans.materix[1][1]%m<<endl;
		}
	}
	return 0;
}