233 Matrix

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 749 Accepted Submission(s): 453

Problem Description 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?
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

Sample Output

234
2799
72937

Hint

  
   
  
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#include"iostream"
#include"stdio.h"
#include"string.h"
#include"algorithm"
#include"queue"
#include"vector"
using namespace std;
#define N 15
#define LL __int64
const int mod=10000007;
int n;
int b[N];
struct Mat
{
    LL mat[N][N];
}a,ans;
Mat operator*(Mat a,Mat b)
{
    int i,j,k;
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    for(i=0; i<=n+1; i++)
    {
        for(j=0; j<=n+1; j++)
        {
            c.mat[i][j]=0;
            for(k=0; k<=n+1; k++)
            {
                if(a.mat[i][k]&&b.mat[k][j])
                {
                    c.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
                    c.mat[i][j]%=mod;
                }
            }
        }
    }
    return c;
}
void mult(int k)
{
    int i;
    memset(ans.mat,0,sizeof(ans.mat));
    for(i=0;i<=n+1;i++)
        ans.mat[i][i]=1;
    while(k)
    {
        if(k&1)
            ans=ans*a;
        k>>=1;
        a=a*a;
    }
}
void inti()
{
    int i,j;
    b[0]=23;
    b[n+1]=3;
    for(i=1; i<=n; i++)
        scanf("%d",&b[i]);
    memset(a.mat,0,sizeof(a.mat));
    for(i=0; i<=n; i++)
    {
        a.mat[i][0]=10;
        a.mat[i][n+1]=1;
    }
    a.mat[n+1][n+1]=1;
    for(i=1; i


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