HDU5015-233 Matrix

233 Matrix

                                                                          Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                                       Total Submission(s): 1910    Accepted Submission(s): 1141

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

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

       题意：给出矩阵的第0行（233,2333,23333,...）和第0列a1,a2,...an（n<=10，m<=10^9），给出式子： A[i][j] = A[i-1][j] + A[i][j-1]，要求A[n][m]。

          解题思路：建立12*12的矩阵：

#include 
#include 
#include 
#include 

using namespace std;

#define ll long long
#define mod 10000007

int n;
struct Matrix
{
    ll x[12][12];
    Matrix()
    {
        memset(x,0,sizeof x);
    }
}a,t;

Matrix mul(Matrix c,Matrix d)
{
    Matrix sum;
    for(int i=0;i<=n+1;i++)
    {
        for(int j=0;j<=n+1;j++)
        {
            for(int k=0;k<=n+1;k++)
            {
                sum.x[i][j]+=c.x[i][k]*d.x[k][j];
                sum.x[i][j]%=mod;
            }
        }
    }
    return sum;
}

Matrix mypow(Matrix l,int m)
{
    Matrix sum;
    for(int i=0;i<=n+1;i++)
        sum.x[i][i]=1;
    while(m)
    {
        if(m&1) sum=mul(sum,l);
        m>>=1;
        l=mul(l,l);
    }
    return sum;
}

int main()
{
    int m;
    while(~scanf("%d %d",&n,&m))
    {
        for(int i=1;i<=n;i++)
            scanf("%lld",&a.x[i][0]);
        if(m==0) {printf("%lld\n",a.x[n][0]);continue;}
        Matrix k;
        a.x[0][0]=23;
        for(int i=0;i<=n;i++)
        {
            k.x[i][0]=10;k.x[i][n+1]=1;
            for(int j=1;j