hdu 5015 - 233 Matrix（矩阵乘法）-优快云博客

本文介绍了一种特殊的矩阵——233矩阵，并提供了一个算法来解决如何根据已知的第一列数值计算出特定位置的矩阵值的问题。该算法利用矩阵乘法和快速幂运算，实现了高效计算。

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

Sample Output

234
2799
72937

Hint

矩阵的第一列为：
0
a[1]
a[2]
a[3]
a[4]

可以变换一下：

23
a[1]
a[2]
a[3]
a[4]

我们可以一列一列转移，这样就可以轻松求得a[n][m]

那么由第一列转移到第二列则为
23*10+3
a[1]+23*10+3
a[2]+a[1]+23*10+3
a[3]+a[2]+a[1]+23*10+3
a[4]+a[3]+a[2]+a[1]+23*10+3
3

那么矩阵A就是

10 0 0 0 0 1
10 1 0 0 0 1
10 1 1 0 0 1
10 1 1 1 0 1
10 1 1 1 1 1
0 0 0 0 0 1

答案就是第n行（下标从0开始）的每一个数分别乘以a数组第一列的每一个数的和。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;

const int mod = 10000007;

struct Matrix
{
    long long m[12][12];
    int n;
    Matrix(int x)
    {
        n = x;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                m[i][j] = 0;
    }
    Matrix(int _n,int a[12][12])
    {
        n = _n;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
            {
                m[i][j] = a[i][j];
            }
    }
};
Matrix operator *(Matrix a,Matrix b)
{
    int n = a.n;
    Matrix ans = Matrix(n);
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
            for(int k=0;k<n;k++)
            {
                ans.m[i][j] += (a.m[i][k]%mod)*(b.m[k][j]%mod)%mod;
                ans.m[i][j] %= mod;
            }
    return ans;
}
Matrix operator ^(Matrix a,int k)
{
    int n = a.n;
    Matrix c(n);
    int i,j;
    for(i=0;i<n;i++)
        for(j=0;j<n;j++)
            c.m[i][j] = (i==j);
    for(;k;k>>=1)
    {
        if(k&1)
            c=c*a;
        a = a*a;
    }
    return c;
}

int a[15];
int main(void)
{
    int n,m,i,j;
    while(scanf("%d%d",&n,&m)==2)
    {
        a[0] = 23;
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        a[n+1] = 3;

        Matrix A = Matrix(n+2);
        for(i=0;i<=n;i++)
        {
            A.m[i][0] = 10;
            for(j=1;j<=i;j++)
            {
                A.m[i][j] = 1;
            }
            for(j=i+1;j<=n;j++)
                A.m[i][j] = 0;
            A.m[i][n+1] = 1;
        }
        for(j=0;j<=n;j++)
            A.m[n+1][j] = 0;
        A.m[n+1][n+1] = 1;

        A = A^(m);
        LL ans = 0;

        for(i=0;i<=n+1;i++)
        {
            ans = (ans + A.m[n][i]*a[i]%mod)%mod;
        }
        printf("%lld\n",ans);
    }

    return 0;
}