233 Matrix HDU - 5015（矩陣快速冪）

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

題意：

給出矩陣的第0行（233,2333,23333,…）和第0列a1,a2,…an（n<=10，m<=10^9），給出式子： A[i][j] = A[i-1][j] + A[i][j-1]，要求A[n][m]。

分析：

這個題主要是轉移矩陣不好寫

第一列元素爲：

⎛⎝⎜⎜⎜⎜⎜⎜0a1a2a3a4⎞⎠⎟⎟⎟⎟⎟⎟$$\left(\begin{array}{c}0\\ a_1\\ a_2\\ a_3\\ a_4\end{array}\right)$$

轉化爲：

⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜23a1a2a3a43⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟$$\left(\begin{array}{c}23\\ a_1\\ a_2\\ a_3\\ a_4\\ 3\end{array}\right)$$

則第二列爲：

⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜23×10+323×10+3+a123×10+3+a1+a223×10+3+a1+a2+a323×10+3+a1+a2+a3+a43⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟$$\left(\begin{array}{c}23\times 10+3\\ 23\times 10+3+a_1\\ 23\times 10+3+a_1+a_2\\ 23\times 10+3+a_1+a_2+a_3\\ 23\times 10+3+a_1+a_2+a_3+a_4\\ 3\end{array}\right)$$

根據前後兩列的遞推關係有等式可得關係轉移矩陣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)$$\begin{array}{}\text{(112)}& \left(\begin{array}{c}{a}_{0,m}\\ a_{1,m}\\ a_{2,m}\\ a_{3,m}\\ .\\ .\\ .\\ a_{n,m}\\ 3\end{array}\right)=\left(\begin{array}{c}1000\cdots 01\\ 1010\cdots 01\\ 1011\cdots 01\\ ...\cdots ..\\ ...\cdots ..\\ ...\cdots ..\\ ...\cdots ..\\ 1011\cdots 11\\ 000\cdots 01\end{array}\right)\left(\begin{array}{c}{a}_{0,m-1}\\ a_{1,m-1}\\ a_{2,m-1}\\ a_{3,m-1}\\ .\\ .\\ .\\ a_{n,m-1}\\ 3\end{array}\right)\end{array}$$

code:

#include <bits/stdc++.h>
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;
}