233 Matrix 矩阵快速幂

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?

InputThere 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 < 2 ³¹).OutputFor 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<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 18
#define N 33
#define MOD 10000007
#define INF 1000000009
const double eps = 1e-;
const double PI = acos(-1.0);
/*
组合数学 找规律
递归显然不行，列数太多
只需考虑每个点被加上的次数
a(i,0) = a(i,1) 到 a(n,m) 路径条数（向左和向下两个方向） C(n+m-i-1,n)
发现列数太多没办法打表 再换一种方法
矩阵快速幂
从第一列向后考虑 找出他们的转移矩阵（这里很巧妙的加了一条边 凑2333后面的3）十分巧妙！~
*/
LL a[MAXN], n, m;
struct mat
{
    LL data[MAXN][MAXN];
    mat()
    {
        memset(data, , sizeof(data));
    }
    mat operator*(const mat& rhs)
    {
        mat ret;
        for (int i = ; i <= n + ; i++)
        {
            for (int j = ; j <= n + ; j++)
            {
                for (int k = ; k <= n + ; k++)
                    ret.data[i][j] = (ret.data[i][j] + data[i][k] * rhs.data[k][j]) % MOD;
            }
        }
        return ret;
    }
};
mat fpow(mat a, LL b)
{
    if (b <= ) return a;
    mat tmp = a, ret;
    for (int i = ; i <= n + ; i++)
        ret.data[i][i] = ;
    while (b!= )
    {
        if (b & )
            ret = tmp*ret;
        tmp = tmp*tmp;
        b = b / ;
    }
    return ret;
}
int main()
{
    while (cin >> n >> m)
    {
        a[] = ;
        for (int i = ; i <= n + ; i++)
            cin >> a[i];
        a[n + ] = ;
        mat ans;
        for (int i = ; i <= n + ; i++)
        {
            ans.data[i][] = ;
            ans.data[i][n + ] = ;
            for (int j = ; j <= i; j++)
                ans.data[i][j] = ;
        }
        ans.data[n + ][n + ] = ;
        ans = fpow(ans, m);
        LL result = ;
        for (int i = ; i <= n + ; i++)
            result = (result + a[i] * ans.data[n + ][i]) % MOD;
        cout << result << endl;
    }
    return ;
}