question can 關系 fin output matrix test case have padding

233 Matrix

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 749 Accepted Submission(s): 453


Problem Description 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 ≤ 10⁹

). The second line contains n integers, a_1,0,a_2,0,...,a_n,0(0 ≤ a_i,0 < 2³¹).
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

思路：

第一列元素為：

0

a1

a2

a3

a4

轉化為：

23

a1

a2

a3

a4

3

則第二列為：

23*10+3

23*10+3+a1

23*10+3+a1+a2

23*10+3+a1+a2+a3

23*10+3+a1+a2+a3+a4

3

依據前後兩列的遞推關系，有等式可得矩陣A的元素為：

#include"iostream"
#include"stdio.h"
#include"string.h"
#include"algorithm"
#include"queue"
#include"vector"
using namespace std;
#define N 15
#define LL __int64
const int mod=10000007;
int n;
int b[N];
struct Mat
{
    LL mat[N][N];
}a,ans;
Mat operator*(Mat a,Mat b)
{
    int i,j,k;
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    for(i=0; i<=n+1; i++)
    {
        for(j=0; j<=n+1; j++)
        {
            c.mat[i][j]=0;
            for(k=0; k<=n+1; k++)
            {
                if(a.mat[i][k]&&b.mat[k][j])
                {
                    c.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
                    c.mat[i][j]%=mod;
                }
            }
        }
    }
    return c;
}
void mult(int k)
{
    int i;
    memset(ans.mat,0,sizeof(ans.mat));
    for(i=0;i<=n+1;i++)
        ans.mat[i][i]=1;
    while(k)
    {
        if(k&1)
            ans=ans*a;
        k>>=1;
        a=a*a;
    }
}
void inti()
{
    int i,j;
    b[0]=23;
    b[n+1]=3;
    for(i=1; i<=n; i++)
        scanf("%d",&b[i]);
    memset(a.mat,0,sizeof(a.mat));
    for(i=0; i<=n; i++)
    {
        a.mat[i][0]=10;
        a.mat[i][n+1]=1;
    }
    a.mat[n+1][n+1]=1;
    for(i=1; i<n+1; i++)
    {
        for(j=1; j<=i; j++)
        {
            a.mat[i][j]=1;
        }
    }
}
int main()
{
    int i,m;
    while(scanf("%d%d",&n,&m)!=-1)
    {
        inti();
        mult(m);
        LL s=0;
        for(i=0;i<=n+1;i++)
            s=(s+(ans.mat[n][i]*b[i])%mod)%mod;
        printf("%I64d\n",s);
    }
    return 0;
}

hdu 5015 233 Matrix （矩陣高速冪）