233 Matrix

Time limit

5000 ms

Memory limit

65536 kB

OS

Windows

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

解析：这道题构造矩阵的方法比较特殊，很是恶心啦，也是看完大神的思路博客才懂，https://blog.youkuaiyun.com/gwq5210/article/details/39449343，大神的代码

这篇博客中的彩色的那个图就是关键，懂得矩阵构造的原理才能构造矩阵，最终的出的矩阵如下：

10      0       0         0       ,,,      1                23                                   233

10      1       0         0       ,,,      1                f(1,m-1)                         f(1,m)

10      1       1         0       ,,,      1         *     f(2,m-1)             =          f(2,m)

10      1       1         1       ,,,      1               f(3,m-1)                         f(3,m)

,,,       ,,,      ,,,         ,,,     ,,,      1                ,,,                                   ,,,

10      1       1         1       ,,,      1                f(n,m-1)                        f(n,m)

0        0       0          0       0      1                3                                     3

自己可以手推一下就懂了，

代码如下：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
#define ll long long
#define mod 10000007
struct node
{
    ll t[15][15];
} a;
int n,m;
ll pp[15];
void init()
{
    memset(a.t,0,sizeof(a.t));
    for(int i=0; i<=n; i++)
    {
        a.t[i][0]=10;
        a.t[i][n+1]=1;
    }
    a.t[n+1][n+1]=1;
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=i; j++)
            a.t[i][j]=1;
    }
}
node juzhen(node x,node y)
{
    node c;
    memset(c.t,0,sizeof(c.t));
    for(int i=0; i<=n+1; i++)
    {
        for(int j=0; j<=n+1; j++)
            if(x.t[i][j])
            {
                for(int k=0; k<=n+1; k++)
                    c.t[i][k]=(c.t[i][k]+x.t[i][j]*y.t[j][k]%mod)%mod;
            }
    }
    return c;
}
node mix(node b,int m)
{
    node ans;
    memset(ans.t,0,sizeof(ans.t));
    for(int i=0; i<=n+1; i++)
        ans.t[i][i]=1;
    while(m)
    {
        if(m%2)
            ans=juzhen(ans,b);
        m/=2;
        b=juzhen(b,b);
    }
    return ans;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%lld",&pp[i]);
        }
        pp[0]=23;
        pp[n+1]=3;
        init();
        node ans=mix(a,m);
        ll sum=0;
        for(int i=0; i<=n+1; i++)
            sum=(sum+(pp[i]*ans.t[n][i]%mod))%mod;
        printf("%lld\n",sum);
    }
    return 0;
}