HDU - 5015 233 Matrix

233 Matrix

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 a _i,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?

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).

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

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

Output

234
2799
72937

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define maxn 10010
#define INF 0x3f3f3f3f
#define eps 1e-8
#define mod 10000007
#define ll __int64
using namespace std;

struct Node{

    ll mat[15][15];
    Node()
    {
        memset(mat,0,sizeof mat);
    }
};
ll n;
Node mul(Node A,Node B)
{
    Node C;
    for(int i=1;i<=n+2;i++)
    {
        for(int j=1;j<=n+2;j++)
        {
            for(int k=1;k<=n+2;k++)
            {
                C.mat[i][j]=(C.mat[i][j]+A.mat[i][k]*B.mat[k][j]%mod+mod)%mod;
            }
        }
    }
    return C;
}
Node Pow(Node A,int k)
{
    Node B;
    for(int i=1;i<=n+2;i++)
        B.mat[i][i]=1;
    while(k>=1)
    {
        if(k&1)
             B=mul(B,A);
        A=mul(A,A);
        k=k/2;
    }
    return B;
}


int main()
{
    int m;
    while(~scanf("%d%d",&n,&m))
    {
        Node A,B;
        A.mat[1][1]=23;
        for(int i=1;i<=n;i++)
            scanf("%d",&A.mat[i+1][1]);
        A.mat[n+2][1]=3;
        for(int i=1;i<=n+1;i++) B.mat[i][1]=10;
        for(int i=1;i<=n+2;i++) B.mat[i][n+2]=1;
        for(int i=2;i<=n+1;i++)
            for(int j=2;j<=i;j++)
                B.mat[i][j]=1;
        B=Pow(B,m);
        A=mul(B,A);
        cout<