题解 | #矩阵乘法#

import sys

# 获取a,b数组的行/列数量
a_row = int(input())
a_column = b_row = int(input())
b_column = int(input())

# 根据2个矩阵行数，获取输入组成a\b数组
a_arr = [list(map(int, input().split())) for i in range(int(a_row))]
b_arr = [list(map(int, input().split())) for i in range(int(b_row))]
# print(a_arr, b_arr)

# 初始化乘积详情数组
product_arr = list()

# a数组的行数做主循环, b数组的列数做次循环
for i in range(a_row):
    for j in range(b_column):
        # k来自b_row循环，即满足公式A(ik) * B(kj)
        summary = sum([a_arr[i][k]*b_arr[k][j] for k in range(b_row)])
        product_arr.append(str(summary))

# a_row做循环，b_column作为数组分割间隔
for k in range(1, a_row+1):
    print(" ".join(product_arr[(k-1)*b_column:(k)*b_column]))