233 Matrix(矩阵快速幂+思维)

In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it would be 233, 2333, 23333... (it means a 0,1 = 233,a 0,2 = 2333,a 0,3 = 23333...) Besides, in 233 matrix, we got ai,j = a i-1,j +a i,j-1( i,j ≠ 0). Now you have known a 1,0,a 2,0,...,a n,0, could you tell me a n,m in the 233 matrix?

Input

There are multiple test cases. Please process till EOF. 

For each case, the first line contains two postive integers n,m(n ≤ 10,m ≤ 10 9). The second line contains n integers, a 1,0,a 2,0,...,a n,0(0 ≤ a i,0 < 2 31).

Output

For each case, output a n,m mod 10000007.

Sample Input

1 1
1
2 2
0 0
3 7
23 47 16

Sample Output

234
2799
72937

这个题的难点在于如何去构造矩阵，我们一般的构造矩阵是一维递推式，这个我们也可以通过改变一下就我们让第一行为23，最后一行为3，然后根据递推关系判断

如图：

代码:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<cmath>
const long long mod=10000007;

const int maxn=1e5+5;
typedef long long ll;
using namespace std;
int n,m;
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<=n+1;t++)
	{
		for(int j=0;j<=n+1;j++)
		{
			for(int k=0;k<=n+1;k++)
			{
				ans.a[t][j]=(ans.a[t][j]+a.a[t][k]*b.a[k][j])%mod;
			}
		}
	}
	return ans;
}
mat anss;
ll quickPow(int k)
{
	mat res;
	memset(res.a,0,sizeof(res.a));
	for(int t=0;t<=n;t++)
	{
		res.a[t][0]=10;
	}
	for(int t=0;t<=n;t++)
	{
		for(int j=1;j<=t;j++)
		{
		   res.a[t][j]=1;
		}
		res.a[t][n+1]=1;
	}
	for(int t=0;t<=n;t++)
	{
		res.a[n+1][t]=0;
	}
	res.a[n+1][n+1]=1;
	while(k)
	{
	  	if(k&1)
	  	{
	  	 anss=Mul(res,anss);
	  	}
	  	res=Mul(res,res);
	  	k>>=1;
	}

	return anss.a[n][0]%mod;
}

int  main()
{

	while(cin>>n>>m)
	{
	   memset(anss.a,0,sizeof(anss.a));
	   anss.a[0][0]=23;
	   for(int t=1;t<=n;t++)
	   {
	   	scanf("%lld",&anss.a[t][0]);
	   }
	   anss.a[n+1][0]=3;
	   cout<<quickPow(m)<<endl;

	}
	return 0;
}