sdut-B - 233 Matrix_n our daily life we often use 233 to express our f-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_43811809/article/details/87860491

本文探讨了一种特殊的矩阵运算——233矩阵的构建与求值问题。通过矩阵相乘与快速幂运算，文章提供了一个高效算法，用于计算特定形式的矩阵中指定元素的值，特别适用于大规模数据计算场景。

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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

Sample Output

	234
    2799
    72937

23333333……

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
struct node
{
    long long data[15][15];
    node()
    {
        memset(data, 0, sizeof(data));
    }
};
int n;
const int mod = 10000007;
node mul(node B, node A)//矩阵相乘
{
    int i, j, k;
    node C;
    for (i = 1; i <= n + 2; i++)
        for (j = 1; j <= n + 2; j++)
            for (k = 1; k <= n + 2; k++)//此处k是可以放在最外层，
                C.data[i][j] = (C.data[i][j] + B.data[i][k] * A.data[k][j]) % mod;
    return C;
}
node f(node A, int m)
{
    node ans;
    for (int i = 1; i <= n + 2; i++)//这是一个单位矩阵。
        ans.data[i][i] = 1;
    while (m > 0)
    {
        if (m & 1)//满足条件，相乘一波
        {
            ans = mul(ans, A);
        }
        A = mul(A, A);//无论如何，自乘一波
        m >>= 1;
    }
    return ans;
}
int main()
{
    int m;
    while (scanf("%d%d", &n, &m) != EOF)
    {
        node A, B;
        {//对矩阵赋初值
            A.data[1][1] = 23;
            for (int i = 1; i <= n; i++)
                scanf("%lld", &A.data[i + 1][1]);
            A.data[n + 2][1] = 3;
            for (int i = 1; i <= n + 1; i++)
                B.data[i][1] = 10;
            for (int i = 1; i <= n + 2; i++)
                B.data[i][n + 2] = 1;
            for (int i = 1; i < n + 2; i++)
                for (int j = 2; j <= i; j++)
                    B.data[i][j] = 1;
        }
        B = f(B, m);
        A = mul(B, A);
        cout << A.data[n + 1][1] << endl;
    }

    return 0;
}