2011 成都 I Isabella's Message

最新推荐文章于 2019-07-07 13:50:38 发布

原创最新推荐文章于 2019-07-07 13:50:38 发布 · 384 阅读

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本文介绍了一个涉及矩阵旋转和字符串操作的解密问题，通过Python实现了解密算法，演示了如何从给定的矩阵中提取已知单词以还原原始消息。

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=9
#Description
首先给你两个N*N的矩阵,然后告诉你认识的单词,具体题目就会懂.如果看过冒险小虎队那就更懂了.
#Algorithm
就是模拟,然后一点字符串操作,当然了,这些对python来说都不是事
一行矩阵旋转90度代码
g = list(zip(*g[::-1]))
#Code

def solve():
    n = int(input())
    g1 = []
    for i in range(n):
        g1.append(list(input()))
    g2 = []
    for i in range(n):
        g2.append(list(input()))
    m = int(input())
    st = set()
    for i in range(m):
        st.add(input())
    ss = []
    for i in range(4):
        s = []
        for j in range(n):
            for k in range(n):
                if g2[j][k] == '*':
                    s.append(g1[j][k])
        g2 = list(zip(*g2[::-1]))
        ss.append(s)
    sss = []
    for i in range(4):
        s = ""
        for j in range(4):
            s += "".join(ss[(i + j) % 4])
        y = list(filter(None, s.split('.')))
        sss.append(y)
    sss.sort()
    for ee in sss:
        flag = True
        for e in ee:
            if e not in st:
                flag = False
                break
        if flag:
            return " ".join(ee)
    return "FAIL TO DECRYPT"
t = int(input())
for i in range(1, t + 1):
    print("Case #{}: {}".format(i, solve()))