有些重点需要注意的点，矩阵乘积之后要开 long 类型接收数据， 单纯按照题意计算会超时，分析时间复杂度可以知道，

复杂度为前面矩阵的行列的乘积与后面矩阵的列数的积；所以从左往右为 dn2 从右往左为 nd2

import java.util.*;

class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int d = in.nextInt();
        int[][] Q = new int[n][d];
        int[][] Kt = new int[d][n];
        int[][] V = new int[n][d];
        int[] W = new int[n];
        for (int i = 0; i < n; i ++) {
            for (int j = 0; j < d; j ++) {
                Q[i][j] = in.nextInt();
            }
        }
        for (int j = 0; j < n; j ++) {
            for (int i = 0; i < d; i ++) {
                Kt[i][j] = in.nextInt();
            }
        }
        for (int i = 0; i < n; i ++) {
            for (int j = 0; j < d; j ++) {
                V[i][j] = in.nextInt();
            }
        }
        for (int i = 0; i < n; i ++) {
            W[i] = in.nextInt();
        }
        // 输入结束

        long[][] nums1 = qualify1(Kt, V);


        for (int i = 0; i < Q.length; i ++) {           // 先将权重向量乘入 还是 存粹的 从右往左算，都不影响答案；
            for (int j = 0; j < Q[0].length; j ++) {
                Q[i][j] = W[i] * Q[i][j];
            }
        }

        long[][] nums2 = qualify2(Q, nums1);

        for (int i = 0; i < nums2.length; i ++) {
            for (int j = 0; j < nums2[0].length; j ++) {
                System.out.print(nums2[i][j] + " ");
            }
            System.out.println();
        }
    }

    public static long[][] qualify1(int[][] Q, int[][] Kt) {        // 入参为两个整型数组的； 后者是懒得写了，直接复制粘贴
        int row = Q.length;
        int col = Kt[0].length;
        long[][] ans = new long[row][col];
        for (int i = 0; i < row; i ++) {
            for (int j = 0; j < col; j ++) {
                long res = 0L;
                for (int k = 0; k < Q[0].length; k ++) {
                    res += Q[i][k] * Kt[k][j];
                }
                ans[i][j] = res;
            }
        }
        return ans;
    }

    public static long[][] qualify2(int[][] Q, long[][] Kt) {       // 入参前者为整型数组的，后者为long
        int row = Q.length;
        int col = Kt[0].length;
        long[][] ans = new long[row][col];
        for (int i = 0; i < row; i ++) {
            for (int j = 0; j < col; j ++) {
                long res = 0L;
                for (int k = 0; k < Q[0].length; k ++) {
                    res += Q[i][k] * Kt[k][j];
                }
                ans[i][j] = res;
            }
        }
        return ans;
    }
}

// 这题目重点需要注意，矩阵内部的乘积和 很容易 爆 int ，所以需要开 long 类型