A Simple Math Problem(矩阵快速幂)----------------------蓝桥备战系列

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

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>
const int maxn=1e5+5;
typedef long long ll;
using namespace std;

ll mod;
ll op[15];
struct mat
{
	ll a[15][15];
};
mat Mul(mat a,mat b)
{
	mat ans;
	memset(ans.a,0,sizeof(ans.a));
	for(int t=0;t<10;t++)
	{
		for(int j=0;j<10;j++)
		{
			for(int k=0;k<10;k++)
			{
				ans.a[t][j]=(ans.a[t][j]+a.a[t][k]*b.a[k][j])%mod;
			}
		}
	}
	return ans;
}
mat ans;

ll quick(ll n)
{
	mat res;
	memset(res.a,0,sizeof(res.a));
	for(int t=0;t<10;t++)
	{
		res.a[0][t]=op[t];
	}
	res.a[1][0]=1;
	res.a[2][1]=1;
	res.a[3][2]=1;
	res.a[4][3]=1;
	res.a[5][4]=1;
	res.a[6][5]=1;
	res.a[7][6]=1;
	res.a[8][7]=1;
	res.a[9][8]=1;
	while(n)
	{
		if(n&1)
		{
			ans=Mul(res,ans);
		}
		res=Mul(res,res);
		n>>=1;
	}
	return ans.a[0][0];
}
int main()
{
	ll n;
	while(cin>>n)
	{
		cin>>mod;
		for(int t=0;t<10;t++)
		{
			scanf("%lld",&op[t]);
		}
		if(n<10)
		{
			printf("%lld
",n%mod);
		}
		else{
		
		for(int t=9;t>=0;t--)
		{
			ans.a[9-t][0]=t;
		}
		printf("%lld
",quick(n-9)%mod);
	    }
	}

   return 0;
}