233 Matrix

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 ≤ 10 ⁹). 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 

矩阵快速幂

比较裸 ，首先找到原态和现态的关系，233，2333，23333，23333， 这些数都遵循 a0=a0'*10+3  ，

a1=a0'*10+3+a1'  ;   a2=a0'*10+3+a1'+a2'  ;  a3=a0'*10+3+a1'+a2'+a3'  ;

即 an=a0'*10+3+

 

代码如下：

#include<iostream>
#include<stdio.h>
#include<cstring>
const int MOD=10000007;
using namespace std;
typedef long long LL;
const int M=15;
struct Matrix{
    LL matrix[M][M];
};
int n;//矩阵的阶数
void init(Matrix &res){
    for(int i=0;i<=n;i++)
    {
        for(int j=0;j<=n;j++)
            res.matrix[i][j]=0;
        res.matrix[i][i]=1;
    }
}
Matrix multiplicative(Matrix a,Matrix b){
    Matrix res;
    memset(res.matrix,0,sizeof(res.matrix));
    for(int i = 0 ; i < n ; i++)
        for(int j = 0 ; j < n ; j++)
            for(int k = 0 ; k < n ; k++)
                res.matrix[i][j] =(res.matrix[i][j]%MOD+a.matrix[i][k]%MOD*b.matrix[k][j]%MOD)%MOD;
return res;
}
Matrix pow(Matrix mx,int m){
    Matrix res,base=mx;
    init(res); //初始为单位矩阵，即除主对角线都是1外，其他都是0
    while(m)
    {
        if(m&1)
            res=multiplicative(res,base);
        base=multiplicative(base,base);
        m>>=1;
    }
    return res;
}
void lk_init(Matrix &b)
{
    for(int i=0;i<n-1;i++)
    {
        b.matrix[i][0]=10;
        b.matrix[i][n-1]=1;
    }
    b.matrix[n-1][n-1]=1;
    for(int i=1;i<n-1;i++)
        for(int j=1;j<=i;j++)
            b.matrix[i][j]=1;
}
int main()
{

    int m,a[M];
    while(~scanf("%d%lld",&n,&m))
    {
        for(int i=1;i<=n;i++)
            scanf("%lld",&a[i]);
        a[0]=23;
        a[n+1]=3;
        n=n+2;
        Matrix base={0};
        lk_init(base);
        base=pow(base,m);
        LL ans=0;
        for(int i=0;i<n;i++)
            ans=(ans%MOD+base.matrix[n-2][i]%MOD*a[i]%MOD+MOD)%MOD;\
        printf("%lld\n",ans);
    }
    return 0;
}