CTF 数表 乘积

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你拿到了一个15*15的数表，请你求得数表中在一条线上（行、列、两个对角线）连续5个数的乘积的最大值，并将这5个数的10个数字按数字从小到大从左到右排列形成密码串，并分别提交。

file.jpg

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python复制代码import pytesseract
from PIL import Image, ImageEnhance

# 打开图像文件
img = Image.open("file.jpg")

# 增强图像对比度，使文本更清晰
enhancer = ImageEnhance.Contrast(img)
img_enhanced = enhancer.enhance(1.5)  # 调整对比度的级别

# 调整图像大小并应用阈值过滤器进行二值化处理
img_resized = img_enhanced.resize((1024, 768))
img_binarized = img_resized.convert("L").point(lambda x: 0 if x < 128 else 255, "1")

# 配置 Tesseract 以将图像视为单一块文本
custom_config = r"--oem 3 --psm 6"
text = pytesseract.image_to_string(img_binarized, config=custom_config)

# 替换识别结果中的度数符号为一个空格
text = text.replace("°", " ")
# print(text)


# 处理识别的文本，将其转换为二维数组
def convert_text_to_matrix(text):
    # 根据换行符分割文本为多行
    lines = text.strip().split("\n")

    # 初始化二维数组
    matrix = []
    for line in lines:
        # 过滤掉非数字字符，并以空格分割每一行中的数字
        numbers = line.strip().split()
        # 将字符串中的数字转换为整数列表
        number_list = [int(num) for num in numbers if num.isdigit()]
        # 如果列表非空，则添加到二维数组中
        if number_list:
            matrix.append(number_list)

    return matrix


# 调用函数，将文本转换为二维数组
matrix = convert_text_to_matrix(text)

# 打印转换后的二维数组
print("矩阵：")
for row in matrix:
    print(row)


# 辅助函数：计算列表中连续五个元素的乘积
def product_of_five(numbers):
    from functools import reduce
    from operator import mul

    return reduce(mul, numbers)


# 定义函数计算最大乘积
def max_product(matrix):
    n = len(matrix)
    max_prod = 0  # 初始化最大乘积为 0
    max_numbers = []  # 初始化最大乘积对应的数字序列

    # 遍历矩阵的行和列
    for i in range(n):
        for j in range(n - 4):
            # 计算每行的连续五个数字的乘积
            row_prod = product_of_five(matrix[i][j : j + 5])
            # 更新最大乘积和数字序列
            if row_prod > max_prod:
                max_prod = row_prod
                max_numbers = matrix[i][j : j + 5]

            # 计算每列的连续五个数字的乘积
            col_prod = product_of_five([matrix[j + k][i] for k in range(5)])
            if col_prod > max_prod:
                max_prod = col_prod
                max_numbers = [matrix[j + k][i] for k in range(5)]

    # 遍历矩阵的对角线
    for i in range(n - 4):
        for j in range(n - 4):
            # 计算主对角线上连续五个数字的乘积
            diag1_prod = product_of_five([matrix[i + k][j + k] for k in range(5)])
            if diag1_prod > max_prod:
                max_prod = diag1_prod
                max_numbers = [matrix[i + k][j + k] for k in range(5)]

            # 计算副对角线上连续五个数字的乘积
            diag2_prod = product_of_five([matrix[i + k][j + 4 - k] for k in range(5)])
            if diag2_prod > max_prod:
                max_prod = diag2_prod
                max_numbers = [matrix[i + k][j + 4 - k] for k in range(5)]

    # 对最大乘积对应的数字序列中的每一位数字进行排序
    password = "".join(sorted("".join(map(str, max_numbers))))
    return max_prod, password  # 返回最大乘积和排序后的密码字符串


# 调用函数计算最大乘积和密码串
max_prod, password = max_product(matrix)
print("MAX  SUM:", max_prod)  # 打印最大乘积
print("PassWord:", password)  # 打印排序后的密码串
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csharp复制代码矩阵：
[71, 51, 67, 63, 89, 41, 92, 36, 54, 22, 40, 40, 28, 66, 33]
[60, 99, 13, 45, 22, 44, 75, 33, 53, 78, 36, 84, 20, 35, 17]
[28, 64, 23, 67, 10, 26, 38, 40, 67, 59, 54, 70, 66, 18, 38]
[68, 32, 62, 12, 20, 95, 63, 94, 39, 63, 68, 40, 91, 66, 49]
[25, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89]
[89, 75, 10, 76, 44, 20, 45, 35, 14, 30, 61, 33, 97, 34, 31]
[28, 22, 75, 31, 67, 15, 94, 63, 80, 44, 62, 16, 14, 79, 53]
[42, 96, 35, 31, 47, 55, 58, 88, 24, 70, 17, 54, 24, 36, 29]
[48, 35, 71, 89, 67, 95, 54, 61, 37, 44, 60, 21, 58, 51, 54]
[68, 15, 64, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77, 34, 89]
[83, 90, 35, 90, 16, 87, 97, 57, 32, 16, 26, 26, 79, 33, 27]
[87, 57, 62, 20, 72, 23, 46, 33, 67, 46, 55, 12, 32, 63, 93]
[73, 38, 25, 39, 11, 24, 94, 72, 18, 58, 46, 29, 32, 40, 62]
[41, 72, 30, 23, 88, 34, 62, 92, 69, 82, 67, 59, 85, 74, 84]
[29, 78, 31, 90, 31, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57]
MAX  SUM: 2171903490
PassWord: 1455677899

本文转载自: 掘金

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