233 Matrix HDU - 5015（矩阵快速幂）

233 Matrix HDU - 5015

 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

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]。

分析：

这个题主要是转移矩阵不好写

第一列元素为：

⎛⎝⎜⎜⎜⎜⎜⎜0a1a2a3a4⎞⎠⎟⎟⎟⎟⎟⎟ ( 0 a 1 a 2 a 3 a 4 )

转化为：

⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜23a1a2a3a43⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟ ( 23 a 1 a 2 a 3 a 4 3 )

则第二列为：

⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜23×10+323×10+3+a123×10+3+a1+a223×10+3+a1+a2+a323×10+3+a1+a2+a3+a43⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟ ( 23 × 10 + 3 23 × 10 + 3 + a 1 23 × 10 + 3 + a 1 + a 2 23 × 10 + 3 + a 1 + a 2 + a 3 23 × 10 + 3 + a 1 + a 2 + a 3 + a 4 3 )

根据前后两列的递推关系有等式可得关系转移矩阵A：

⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜a0,ma1,ma2,ma3,m...an,m3⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟=⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜10       0       0      ⋯     0     110       1       0      ⋯     0     110       1       1      ⋯     0     1.       .       .      ⋯     .     ..       .       .      ⋯     .     ..       .       .      ⋯     .     ..       .       .      ⋯     .     .10       1       1      ⋯     1     1 0        0       0      ⋯     0     1⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜a0,m−1a1,m−1a2,m−1a3,m−1...an,m−13⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟(112) (112) ( a 0 , m a 1 , m a 2 , m a 3 , m . . . a n , m 3 ) = ( 10               0               0             ⋯           0           1 10               1               0             ⋯           0           1 10               1               1             ⋯           0           1 .               .               .             ⋯           .           . .               .               .             ⋯           .           . .               .               .             ⋯           .           . .               .               .             ⋯           .           . 10               1               1             ⋯           1           1   0                 0               0             ⋯           0           1 ) ( a 0 , m − 1 a 1 , m − 1 a 2 , m − 1 a 3 , m − 1 . . . a n , m − 1 3 )

code:

#include 
using namespace std;
typedef long long ll;
const int N = 15;
const int mod = 1e7+7;
int b[N];
int n;
struct Matrix{
    ll mat[N][N];
    Matrix operator * (const Matrix &b)const{
        Matrix ans;
        for(int i = 0; i <= n+1; i++){
            for(int j = 0; j <= n+1; j++){
                ans.mat[i][j] = 0;
                for(int k = 0; k <= n+1; k++){
                    if(mat[i][k] && b.mat[k][j]){
                        ans.mat[i][j] += mat[i][k] * b.mat[k][j];
                        ans.mat[i][j] %= mod;
                    }
                }
            }
        }
        return ans;
    }
}a;

Matrix q_pow(Matrix a,int b){
    Matrix ans;
    memset(ans.mat,0,sizeof(ans));
    for(int i = 0; i <= n+1; i++){
        ans.mat[i][i] = 1;
    }
    while(b){
        if(b & 1)
            ans = ans * a;
        b >>= 1;
        a = a * a;
    }
    return ans;
}

void init(){
    b[0] = 23;
    b[n+1] = 3;
    for(int i = 1; i <= n; i++){
        scanf("%d",&b[i]);
    }
    memset(a.mat,0,sizeof(a.mat));
    for(int i = 0; i <= n; i++){
        a.mat[i][0] = 10;
        a.mat[i][n+1] = 1;
    }
    a.mat[n+1][n+1] = 1;
    for(int i = 1; i < n+1; i++){
        for(int j = 1; j <= i; j++){
            a.mat[i][j] = 1;
        }
    }
}
int main(){
    int m;
    while(~scanf("%d%d",&n,&m)){
        init();
        Matrix ans = q_pow(a,m);
        ll s = 0;
        for(int i = 0; i <= n+1; i++){
            s = (s + ans.mat[n][i] * b[i] % mod) % mod;
        }
        printf("%lld\n",s);
    }
    return 0;
}