hdu 5015 233 Matrix （矩阵快速幂）

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

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

 

Sample Output

234 2799 72937

Hint

思路：

第一列元素为：

0

a1

a2

a3

a4

转化为：

23

a1

a2

a3

a4

3

则第二列为：

23*10+3

23*10+3+a1

23*10+3+a1+a2

23*10+3+a1+a2+a3

23*10+3+a1+a2+a3+a4

3

根据前后两列的递推关系，有等式可得矩阵A的元素为：    

#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