HDU 5015 233Matrix （构造矩阵）

233 Matrix

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1399    Accepted Submission(s): 826

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 ≤ 10⁹). The second line contains n integers, a_1,0,a_2,0,...,a_n,0(0 ≤ a_i,0 < 2³¹).

 

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
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#define LL long long
using namespace std;
const long long MAXN = 15;
const long long mod = 10000007;
struct Matrix
{
    long long mat[MAXN][MAXN], n;
    Matrix(){memset(mat, 0, sizeof(mat));}
    Matrix operator * (Matrix & rhs)
    {
        Matrix res; res.n = n;
        for(long long i=1;i<=n;i++)
        {
            for(long long j=1;j<=n;j++)
            {
                for(long long k=1;k<=n;k++)
                {
                    (res.mat[i][j] += (mat[i][k] * rhs.mat[k][j]) % mod) %= mod;
                }
            }
        }
        return res;
    }
};
Matrix pow_mod(Matrix a, long long b)
{
    Matrix res; res.n = a.n;
    for(long long i=1;i<=a.n;i++) res.mat[i][i] = 1;
    while(b)
    {
        if(b & 1) res = res * a;
        a = a * a;
        b >>= 1;
    }
    return res;
}
long long a[MAXN], n, m;
int main()
{
    while(scanf("%I64d%I64d", &n, &m)!=EOF)
    {
        for(long long i=1;i<=n;i++) scanf("%I64d", &a[i]);
        Matrix ans; ans.n = n + 2;
        for(long long i=1;i<=n + 1;i++)
        {
            ans.mat[i][1] = 10;
            for(long long j=2;j<=i;j++)
            {
                ans.mat[i][j] = 1;
            }
        }
        for(long long i=1;i<=n+1;i++) ans.mat[n+2][i] = 0;
        for(long long i=1;i<=n+2;i++) ans.mat[i][n+2] = 1;
        ans = pow_mod(ans, m);
     /*   for(long long i=1;i<=n+2;i++)
        {
            for(long long j=1;j<=n+2;j++)
                cout << ans.mat[i][j] << ' ';
            cout <<endl;
        }*/
        a[0] = 23, a[n+1] = 3;
        long long rs = 0;
        for(long long i=1;i<=n+2;i++) (rs += a[i-1] * ans.mat[n+1][i]) %= mod;
        printf("%I64d
", rs);
    }
    return 0;
}